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INTRODUCTION TO COMPLEX
NUMBER SYSTEM
Unit Structure
1.0. Objectives
1.1. Introduction
1.2. The Field of Complex Numbers
1.3. Extended Complex Plane, The Point at Infinity , Stereographic
Projection
1.4. Summary
1.5. Unit End Exercises
1.0 OBJECTIVES
After going through this unit you shall come to know about
• The field of complex numbers denoted by /g932.
• Representations of complex numbers in polar forms .
• The Euclidean two dimensional plane 2 along with the
point at infinity forms the extended complex p lane.
• The extended complex plane is in one to one corr espondence
with the unit sphere in 3 and such a correspondence is known
to be the stereographic projection.
1.1 INTRODUCTION
Numbers of the form zab i= + , where a a n d b a r e r e a l
numbers and 1i=−are called as Complex Numbers. The identities
involving complex numbers lead to solutions to many problems
in the theory of real valued functions .The wi der acceptance of
complex numbers is because of the geometric represe ntation of
complex numbers, which was fully developed and stud ied by
Gauss. The first complete and formal definition of complex
numbers was given by William Hamilton. We shall begin with
this definition and then consider the geometry of complex
numbers. munotes.in
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1.2 THE FIELD OF COMPLEX NUMBERS
A complex number z is an ordered pair (),xy of real numbers.
i.e. (),, zx y x y= ∈ ∈. Complex number system, denoted by
is the set of all ordered pairs of real numbers (i. e. ×) with the
two operations of addition and multiplication (i or ×) which satisfy : ()()()() ( ) ( )() ( )11 2 2 1 2 1 2
11 2 2
11 2 2 1 2 1 2 1 2 2 1(i) , ,
,,,
(ii) , . , ,xy x y x x y y
xy x y
xy x y x x y y x y xy + = + + + ∨ ∈ = − +
The word ordered pair means ()11,xy and ()11,yx are distinct unless 11xy=.
Let (),, zx y x y= ; ∈ ∈. ‘x’ is called Real part o f a c o m p l e x
number z a n d i t i s d e n o t e d by Rexz=, (Real part of z) and ‘ y’ is
called Imaginary part of z and it is denoted by Imyz=.
T w o c o m p l e x n u m b e r s ()11 1,zx y= and ()22 2,zx y= are said
to be equal i f f 12xx=and 12yy= i . e . r e a l p a r t a n d i m a g i n a r y p a r t
both are equal.
About Symbol ‘i’ :
T h e c o m p l e x n u m b e r ()0, 1 is denoted by ‘ i’ and is called the
imaginary number.
() ()
()
()2.0 , 1 . 0 , 101 , 001, 0ii i= = = − + = − b y p r o p e r t y ( i i ) a b o v
2i⇒ = − 1
Similarly,
() ( ) ( ) ()32.1 , 0 . 0 , 1 0 0 , 1 0 0 , 1ii i= = − = − − + = −
3ii⇒ = −
41i⇒ =
Using this symbol i, we can write a complex number (),xy as xi y+
(Since ()()()()()(),0 0 ,1 ,0 ,0 0 , ,xi y x y x y x y+= + = + = The complex
number (),zx y= can be written as zx i y=+
Note: (The set of all complex numbers) C forms a field.
() ( ) ( ) ( )43.0 , 1 . 0 , 1 0 1 , 0 0 1 , 0 3ii i= = − = + + = ∴munotes.in
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Propeties of complex numbers
Let ()11 1,zx y= , ()22 2,zx y= and ()33 3 , zx y= ∈.
1) Closure Law : 12zz+∈ and 12.zz ∈
2) Commutative Law of addition : 12 21zz z z+=+ ()()()()() ( )12 1 1 2 2 12 12 21 2122 1 1 2 1,, , ,
,,zz x y x y xxyy x x y yxy x y z z+= + = + + = + +
= + = +
3) Associative Law of addition : ()()12 3 1 23zz z z z z++=++
4) Existence of additive Identity : The Complex Number ()00 , 0= i . e . 00zi=+ i s c a l l e d t h e i d e n t i t y w i t h r e s p e c t t o
addition.
5) Existence of additive Inverse :
F o r e a c h c o m p l e x n u m b e r 1,z∈ ∃ a unique complex number
z ∈ s.t. 110zz z z+=+ = i.e. 1zz=− . The complex number z
is called the additive inverse of 1z and it is denoted by 1zz=− .
6) Commutative law of Multiplication : 12 21..zz z z =
()()12 11 22., . ,zz xy xy = ()12 12 12 21,xx yyxy x y= ⋅ −⋅ ⋅+⋅ …..(1)
and
()()()21 22 11 21 2 121 1 2., . ,., . .zz xy x y xx y y xy x y = = − +
()12 2112 21.. , ..xx y y xy x y= − + 12.zz= from (1)
7) Associative Law of Multiplication : ()()12 3 1 23.. ..zz z z z z =
()()()()12 3 1 1 2 2 3 3.. , , ..zz z x y x y x y = ()[]11 2 3 2 32 3 3 2,.. , ..xy x x y yx y x y= − +
()()()()123 2 3 123 2 3 123 3 2 123 2 3.. . , . . . .xx x yy yx x y y xx x x y yx x y y= − − − + + −
()123 12 3 131 12 3 123 132231 123.. .. .. ,
.. ( * )xx x xy y xx y yy y xx x xx yxxy yyy= − − . . + . . + . . + + . .
()()()()() () ( )
() () () ( ) ()
()12 3 1 1 2 2 3 311 2 3 2 3 1 2 3 1 2 312 12 33 1 1 2 2 3 3
12 3,, ,
,, ,
,, , ,zz z x y x y x yxy x x y y x x x y y yxx yy xy x y x y x y
zz z++ = + +
= + + + = + + + +
= + + + + = + +
= + +
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()()()()()()12 3 11 22 33 1 2 1 2 1 2 2 1 33123 312 123 12 3 132 213 123 123.. ,. , , , ,
,zz z xy xy xy x x y y x y x y xyxx x x yy xx y yy y xx y x yy xx x yy y = = − + = − − + + + −
()123.zz z= f r o m ( * )
8) Existence of Multiplicative Identity : 11 1.1 1.zz z = =
T h e c o m p l e x n u m b e r ()11 , 0= ( i . e .10zi=+) is called the
identity with respect to multiplication.
9) Existence of Multiplicative Inverse : For each complex
number 10,z≠ there exists a unique complex number z in s.t.
11.. 1zz z z = = i.e. 11zz= is called the multiplicative inverse of
complex number 1z and it is denoted by 11zz= or 1z − .
L e t (),zx y= and ()11 1,zx y= ∴ 12.1zz = ∴ ()()()11,, 1 , 0xy x y = ⇒()()11 1 1,1 , 0xx yy xy x y− + = ⇒ 11.. 1xx yy − = ……..(i) and 11.. 0xy x y + = …......(ii)
E q u a t i o n ( i i ) 1x× - Equation (i) 1y× , we get
()2
11 1
211 1 12211 10xx y x yxx y yy yyx y y + = − − = − + − + = − ∴ 1
22
11yy
xy− = + ( i i i )
Substitute equation (iii) in equation (ii) i.e. 11.. 0xy yx + = ∴ 11..xy yx = − 1122
11yxxy− = − + 11
221111xyxyxy ⇒ = × +
1
22
11xx
xy = +
∴ 1122 2211 12,xyzxy xy − = + +
z is the multiplicative inverse of complex number ()11 1,zx y= .
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10) Distributive Law :
()12 3 12 13..zz z zz zz+= +
Subtraction: T h e d i f f e r e n c e o f t w o c o m p l e x N u m b e r s ()11 1,zx y= and ()22 2,zx y= is defined as :
()()12 1 1 2 2,,zz x y x y−= − ()12 12,xxyy= − −
Division: It is defined by the equality 11122.zzzz− = 20z≠
()22 1 2 1 2 1 2 2 11122 22 22 22
22 22 22 22..,, ,xy x x y y x y x yxy
xy xy xy xy − + − + = = + + + +
Geometrical Representation of a Complex Number :
Consider a complex numberzx i y=+. ∵ C o m p l e x n u m b e r i s d e f i n e d a s a n o r d e r e d p a i r e d o f real
numbers.
i . e . (),zx y=
xP ( ,y)xY
Xy
Fig. 1.1
T h i s f o r m o f a c o m p l e x n u m b e r z s u g g e s t t h a t z c a n b e
represented by point (say) P whose Cartesian co-ordinates are x and
y r e fe r r e d ( r e l a t i n g ) t o r e c t a n g u l a r a x i s X and Y, usually called the
Real and Imaginary axis respectively.
To each complex number there corresponds points in the
plane and conversely, one and only one each point i n the plane there
exist one and only one complex number.
A plane whose points are represented by the complex
numbers is called Complex Plane o r G a u s s i a n P l a n e o r A r g a n d
Plane . Gauss was first who formulated that complex numb ers are
represented by points in a plane in 1799 then in 18 06 it was done by
Argand. munotes.in
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Vector Representation of a Complex Numbers :
If P i s t h e p o i n t i n t h e C o m p l e x P l a n e c o r r e s p o n d i n g t o
complex number z c a n b e c o n s i d e r e d a s v e c t o r OP w h o s e i n i t i a l
point is the origin ‘O’ and terminal point is (, )Pz x y = = as shown in
the figure 1.2.
Y
XP=z ≡(x,y ) θ z
Fig 1.2
Conjugate :
If zxi y = + ∈ then the complex number xi y− is called the
conjugate of a complex number z or complex conjugate and it is
denoted byz.
e.g. 43 43zi zi = + ⇒ = −
3345 45iiwe we− = + ⇒ = +
Geometrically:
The complex conjugate of a complex number (, )zx y = is the image
or reflection of z in the real axis.
Y
X Oxiyz + =
xi y z - =1θ
2θ
Fig 1.3
Let zx i y = +
∵Rexz = andImzz = .∴Re Im22zz zzxz a n d yz + − = = = =
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Definition: The modulus o r a b s o l u t e v a l u e o f a c o m p l e x n u m b e r zx i y = + is defined by22zx y=+ .
Y
Xz OPx y () ,= z
Fig 1.4
The distance between Two Complex Numbers :
Let ()()11 1 2 2 2,, ,zx y z x y= = in complex plane is given by ()12 1 2,dz z z z = − () ( )2212 12xx yy = − + −
Y
Xz1xy ()11,z2xy ()22,
Fig 1.5
Polar form of a Complex Numbers :
If P i s a p o i n t i n t h e C o m p l e x P l a n e c o r r e s p o n d i n g t o
complex number (),zx i y x y=+ = a n d l e t (),r θ b e t h e p o l a r c o -
ordinates of point (),xy from figure 1.3, cosxr= θ and sin ,yr= θ
where 22rx y= + is called the modulus o r a b s o l u t e v a l u e o f z
(denoted by z a n d 1tanyx − θ= is called the argument o r
amplitude of z (denoted by arg )zθ = . Here θ is the angle between
the two lines OP and the real axis (x- axis) .
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Y
Xxy(),
xyy
xP=rz=
θ
O
Fig 1.6
∴zxi y = + ∴cos sinzr i r = θ + θ ∴(cos sin )zr i = θ + θ
This form is called the polar form of a the c omplex number z.
P
XY
O2θ− π
2θ+π
4θ+πθ
Fig 1.7
Any complex number 0z ≠ has an infinite number of distinct
arguments.
Any two distinct arguments of z differ each other b y an integral
multiple of2π.
If one of the value of argument of z is θ then arg 2zn = θ + π where 0, 1, 2, ...n= ± ±
The value of θ which lies in the interval − π<θ≤π or ()02<θ≤ π
is called the principal value of argument of z and it is denoted
byArg z = θ. munotes.in
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The relation between Arg z and argz is given by arg 2Arg z z n = + π
where 0, 1, 2, ...n= ± ±
Exponential form of complex number : A complex number can be
written in the form ofizr eθ= , w h e r e 22rx y=+ and
1tanyx − θ= .This is known as the exponential form.
(Note: cos sinieiθ= θ+ θ,known as Euler’s Identity)
Note : 1) 1ieθ = , 500
42) 1i
eπ =
Solved Examples :
1. Let 12 31, 1 2 , 1 3zi z i z i=+ = − =+ . Find i) z 1.z2 ii) z 1/z2 iii) z2 iv) 1z v) arg(z 1) vi) Express z 1 in polar and exponential form.
Solution:
i) 2
12(1 )(1 2 ) 1 2 2 1 2 2 3 3zz i i i i i i i i=+ −=−− − =−− += −
ii) 12211 1 2 1 2 2 1 3/12 12 12 514iiii i izzii ii++ ++ + −− +== × ==−− +−
iii) z2=1+2i
iv) 111, 1xy==
∴22 2 2
111 2zx y=+ = + =
v) 111, 1xy==
∴ 111
11tan ( ) tan ( )14y
x−−πθ== =
vi) 11 1 1(cos sin )zr= θ+ θ
12(cos sin )44zππ∴=+
2. Find the principal value of arg ' 'i
Arg z θ= − π<θ≤π
{}1tan ( . )yArg i z i z o x iyx− = θ = = ⇒ = + ∵
()111tan tan0− − == ∞ 2 π =
3. Find the principal value of ()arg 1i +
∵()1zi x i y = + = +
∴1, 1xy= =
11 1rg tan tan1yAzx−−π = θ = = = 4
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4. Express the Complex Number 13zi=+ in polar form.
Solution : ∵()cos sinzr i= θ+ θ ( 1 )
a n d ∵13zx i y i=+ = +
∴1x= and 3y=
∵22rx y=+ 13 4= + = ∴2r =
∵()11 1 3tan tan tan 31y
x− − − θ== = 1tan tan3− π= ∴3π θ= ∴2c o s s i n33ziππ=+
Results : ()()tan 45 tan 180 45 1=+ =
()()tan 45 tan 180 45 1− + − = −
()()tan 60 tan 180 60 3=+ =
()()tan 60 tan 180 60 3− = − =−
5. Express the Complex Number 1zi=− + in polar form
Solution : ∵ ()cos sinzr i= θ+ θ ∵ 1zi=− + (given)
Comparing with zx i y=+ ∴1, 1xy=− = ∴() ( )222211rx y=+ = − + 11= + =2 ∵111tan tan1y
x− − θ== − ()1tan 1− = − ()1tan tan4−π = π−44π3π=π− = 332c o s s i n44ziππ= +
Basic Properties of Complex Numbers :
1) zz= iff z is purely a real number
Proof: Let ,zx i y zx i y=+ = −
Let , zz= ⇒xiy x+=iy− 20 0iy y= ⇔ = zx⇔ =z⇔ is real number.
2) 0z = iff 0z= munotes.in
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Proof: 22 2 200 0zx y x a n d y = = ⇔ = = 00xa n d y⇔ = =
i.e. 0z =
3) zz =
Proof: zx i y = + and zx i y=−
∴22zx y = + ( 1 )
()222 2zx y x y = + − = + ( 2 )
∴zz = f r o m ( 1 ) & ( 2 )
4) Re Rezz z ≤ ≤
5) Im Imzz z ≤ ≤
6) Let , zw∈ ∈ then
zwzw+= +
L e t ,zx i y w u i v=+ =+ ∴() ()zw xi y ui v+= + + + () ()xu i yv=++ + ()()xu i yv=+ −+ ()()xi y ui v=−+− zw=+
∴zwzw+= +
7) .zw z w=
() ()zw x iy u iv = + + () ()xu yv i xv yu= − ++()()xu yv i xv yu= −− +()()xu ixv yv iyu= −−− ()()xu i v i yu i v= −− −()()ui vxi y= − −.zw=
8) 2.zz z =
2222 22zx y x y = + = + ( 1 )
()().zz x i y x i y = + +2xi x y= − ixy+22iy−22xy=+ ( 2 )
∴2.zz z = f r o m ( 1 ) a n d ( 2 )
9) zw z w =
()()2zw zw zw = ()()zz w w= = 22zw zw z w⇒ =
10) zzww =
11zzzww w = × = × zw =
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11) zz=
()zx i yx i y x i y=+ = −=+
_____
()zx i yxi y xi yz= = − = − − = + = ∵
Addition of two Complex Numbers :
Let ;zxi y w ui v = + = +
Now, zwO AO B+= +OA AC=+OC= Y
XOB C
AW
ZZ+W
Fig 1.8
Triangle Inequality :
1) ,zw ∈ then zw z w + ≤ +
Proof:
()()2.zw zw zw + = + + ()()..zw z w=+ zz zw zw ww=+ ++
22 2.zw z w z wz w + = + + + ( 1 )
Now, ()()()()..zw zw x i y u i v x i y u i v + = + −+− + .zw z w= + + ()2Rezw=
()2. R ezw z z≤ ≤ ∵
2zw=
.2zwz w zw+ + = ( 2 )
Substitute (2) in equation (1), we get
22 22zw z w zw + ≤ + + ()2zw≤ + zw z w⇒ + ≤ +
Geometrically , in any triangle, the sum of the two sides of a tr iangle
is greater than or equal to the third side(t he points are
collinear, in case it is equal). munotes.in
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Y
XOB
Ayxy+x
Fig 1.9
2) Let ,zw ∈ then zw z w − ≤ −
Proof: Let zzw w=+ −
Taking mod on () both the sides ∴ zz w w = + −
zw w≤ − +
∴zwz w − ≤ − ( i )
Interchanging z and w, we get
wzw z z w − ≤ − = − ( i i )
()zw z w− − ≤− −
zw z w − ≥− − (iii)
From equation (i) and (ii), we get
()zw z w− − ≤ − zw≤ − ∴ zwz w − ≤−
3) Let ,zw ∈ then
22 22R ezw z z w w + = + +
22 22R ezw z z w w − = − +
4) Parallelogram Law: The sum of the squares of the lengths of the
diagonals of a parallelogram is equal to the sum of s q u a r e s o f
lengths of its sides. i.e. prove that ()22 2 22 zw zw z w + + − = +
Proof: Let ,zw ∈
() ()22zw xu i yv + = + + +222R e( z )zw w−= + +
22 22R ezzw z ww− − = − + ∵
∴()22 2 22 zw zw z w++ − = + munotes.in
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Y
X O B C A z z z
Fig 1.10
5) Let ,zw ∈ then zw z wzw − ≤ ± ≤ +
Proof: i) T.P.T. zw z w ± ≤ +
C a s e ( i ) zw z w + ≤ + by triangle inequality
C a s e ( i i ) ()zw z w − = + − zw≤ + − zw≤ +
F r o m a b o v e b o t h c a s e s ,
zw z w ± ≤ + ( * )
i i ) T . P . T . zwz w − ≤ ±
C o n s i d e r zz w w = + − zw w≤ + + − zw w≤ + +
∴zwz w − ≤ + ( a )
C o n s i d e r ww z z = + −
wz z≤ + + − wz z≤ + +
∴wzw z − ≤ +
∴()zw w z− − ≤ + ( b )
∴zw z w − = ± ( * * )
F r o m ( * ) a n d ( * * ) , w e g e t
zw z wzw − ≤ ± ≤ +
Theorem: The field is not a linearly totally ordered field
OR
The field is not partially ordered field (Total ordering or partial
ordaring means that if ab≠ then either ab< or ab>).
Proof: Suppose that such a total (partial) ordering exists .
Then for i.e. , we have either 0i> or 0i< if 0i> ∴ 1.0ii− = >
or if it ()00ii< − >
()()10ii− = − − > munotes.in
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We get 10− >, which is not true in . ∴ Our supposition is not true. ∴ is not linearly totally ordered field.
Properties of polar form and exponential form
1) Let ()111 1 1 1cos sinizr e r iθ= = θ+ θ, ()222 2 1 2cos sinizr e r i θ== θ+ θ
then ()()12 1 2 1 1 2 2.. c o s s i n c o s s i nzz r r i i = θ+ θ θ+ θ ()1212.irre θ+θ=
∵()2 2,1 ,in ii nee e n θ+π θπ= = ∈
()1212 1 2.. .izz r re θ+θ = ()1212.1irr e θ+θ= =∵
and () ( )12 1 2 arg . arg arg mod 2zz z z = + π in the sense that they are
same but for an integral multiple of 2π.
Note : 12 1 2arg . arg arg 2 where 0,1 or 1zz z z k k = + + π = −
2. Let 111izr eθ = and 222izr eθ= and 20z≠
∴()()11 1122 2 2cos sin
cos sinri z
zr iθ+ θ=θ+ θ() 1122irer θ− θ=
∴1
12
2arg arg arg (mod 2 )zzzz = − π
Let 11z=− and 2zi= − , ()−π ≤ θ ≤ π ∵111zx i y x= − = + ⇒ = − and 0y= ∴ ()()1
10 arg arg 1 tan1z− = − =− () ( )11tan 0 tan tan− − == π=π ∵ 20zi x i y x=− = + ⇒ = and 1y = − ∴ ()()()11
21arg arg tan tan0yzix− − − = − = = ()1tan− = ∞2π=− ()()()12arg . arg 1. argzz i i = − − =()()111 tan tan0− − == ∞2π= ∴ ()12 1 2arg . arg arg 2zz z z k = + + π w h e r e 0k=
Let 11z= − and 2zi= ∵ 1111 , 0zz x i y x y=− ⇒ = + ⇒ = − = ∴ ()()()11
10 arg arg 1 tan tan1yzx− − = − = =− ()1tan 0− = =π ∵ 20, 1zi x i y x y== + ⇒ = = ∴ ()()()()11 1
21 arg arg tan tan tan0yzix− − − == = = ∞2π= munotes.in
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∴()( )( )()()11
121arg . arg 1. arg tan tan0yzz i ix− − − = − = − = =()tan= − ∞ 2π= − ∴ ()12 1 2arg . arg arg 2 where 1zz z z k k = + + π = −
In this case, we get correct answer by adding 2− π to bring within
the interval(),− π π.
When principal argument are added together in multi plication
problem, the resulting argument need not be the pri nciple value.
De-Moivre’s Theorem :
Theorem: If n is any integer or fraction then
()cos sin cos ( ) sin ( )nin i nθ + θ = θ + θ
Proof:
LHS= () ( )()cos sin ( ) cos sinn in i nie e n i nθθθ + θ== = θ + θ =RHS
∴()cos sin cos sinnin i n θ+ θ= θ+ θ
e.g. i) ()5cos sin cos 5 sin5iiθ+ θ= θ+ θ
ii) 252(cos sin ) cos sin5ii2 θ + θ= θ + θ 5
Note: () ( )sin cos sin cosnii nθ+ θ≠ θ + θ
But, ()sin cos cos sin22nnin i n ππ θ+ θ= −θ+ −θ
cos sin22ni nππ = −θ+ −θ (by above thm)
e.g. 1) ()2323 sin cos cos sin22ii ππ θ+ θ= −θ+ −θ
22cos sin32 32iππ = −θ+ −θ
2) ()cos sin cos sinnin i nθ− θ= θ− θ
e . g . ()4544cos sin cos sin55iiθ− θ= θ− θ
∵zx i y=+ 222xyr+= is equation of circle with centre at the origin & radius
equal to r. munotes.in
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Or
Fig 1.111
∵ zx i y=+∴ 22 2zx y r r = + = = ∴ 222xyr+=
The equation of the circle with the centre a t cai b=+ ∈ and
radius equal to r.
r
c
o
Fig 1.12 ∴ zc r − =
e.g. 1) ()21zi −+ =
This is equation of the circle with centre ()2,1 and radius 1. 13z − =, circle with centre ()1, 0 and radius = 3 2zi + =, circle with centre ()0, 1 − and radius = 2
Roots of Complex Number :
Definition: A number w is called the nth root of complex number z if
nwz= or 1nwz= .
Theorem: In , given 0z≠, the equation expansion nwz= has n-
distinct solution given by 2., 0 , 1 , . . . , 1kin nkwr e k nθ+π = = − where rz= and Arg .zθ =
plane−
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Proof: Given, z∈ and 0z≠ ∴The polar form of complex number z is ()cos sinzr i= θ+ θ where rz= and argzθ=. ∴()cos sinnwz r i== θ+ θ ( g i v e n )
()()cos 2 sin 2rk i k= θ+π+ θ+π
OR
() ()111cos 2 sin 2 nnnwz r k i k== θ+π+ θ+π
22cos sinn kkrinn θ+πθ +π = + ( by D e - M oi v e r’ s
theorem)
2kin nk ww r eθ+π == where ()0,1, 2, ..., 1kn= −
Note : It is sufficient to take ()0,1, 2, ... 1kn= − s i n c e a l l o t h e r
values of k lead to repeated roots.
Example : Find all the fourth roots of 1zi= + and locate these
roots in plane.
Solution : Let 41wz i== + ∴1, 1xy= = ∴ 2211rx y=+ = + ∴2r = ()()()11 1 1tan tan tan 11y
x− − − θ== = ∴4π θ= ∴ 42c o s s i n44wi ππ =+ (polar form)
()()2c o s 2 s i n 244ki kππ=+ π+ + π
1148822c o s s i n44kkwi π+ππ +π =+
18822c o s s i n16 16kkwi π+ππ +π =+ where ∴ F o u r t h r o o t s o f e q u a t i o n s a r e
For 0k=, 180 cos sin16 16wz i ππ =+
1k=, 18199cos sin16 16wz i ππ =+ munotes.in
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2k=, 18217 17cos sin16 16wz i ππ =+
3k=, 18325 25cos sin16 16wz i ππ =+
Which are the required four fourth root of 1zi=+
OYX1ω
2ω
3ω0ω
16π916π
1716π
2516πplane−
Fig 1.13
Example : Find all the fifth roots of 32z=− and locate these
roots in -plane.
Solution : Let 532wz== − ∴ 32x=− and 0y= ∴ ()2 2232 32rx y=+ = − = ()()()11 1 0tan tan tan 032y
x− − − θ== =⇒ θ= π ∴ ()532 cos sinwi= π+ π ∴ () () ()1155 32 cos 2 sin 2kwk i k= π+π+ π+π
222c o s s i n55kki π+ππ +π =+
For 0k=, 02c o s s i n55wi ππ =+
1k=, 1332c o s s i n55wi ππ =+
2k=, 2552c o s s i n55wi ππ =+ []2c o s s i ni= π+ π munotes.in
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3k=, 3772c o s s i n55wi ππ =+
4k=, 4992c o s s i n55wi ππ =+ Y
X1ω
5ω3ω0ω
5π35π
π
95π75π
4ω
Fig 1.14
Example : Solve 85310zzz+++ = ∵ 85310zzz+++ = ∴ ()()53 311 1 0zz z++ += ∴ ()()5311 0zz+ + =
Consider, 310z+= ∴ 13 311zw z=− ⇒ == − ∴ 1, 0xy=− = ∴ 2211rx y r=+ = ⇒ =
and ()()110tan tan 01− − θ== − −
θ=π ∴ ()31c o s s i nwz i== π+ π i n p o l a r
() ()3cos 2 sin 2 wz k i k== π+π+ π+π
() ()13cos 2 sin 2kwz k i k== π+π+ π+π
b y D e - M o i v r e ’ s t h e o r e m
22cos sin33kkkwz iπ+ππ +π == + where 0,1, 2k= munotes.in
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For 0k=, 0cos sin33wiππ =+
1k=, () () 133cos sin cos sin33wi iππ =+ = π+ π
2k=, 255cos sin33wiππ =+
Now, consider 510z+= ∴51z=− ∴151wz== − ∴1a n dr = θ=π ∴cos sinwi= π+ π ∴()()()cos 2 sin 2wk i k= π+π+ π+π ∴()152cos sin 25kkwi kπ+π =+ π+π w h e r e 0,1, 2, 3, 4k= ∴22cos sin55kkkwi π+ππ +π =+
For 0k=, 0cos sin55wiππ =+
1k=, 133cos sin55wiππ =+
2k=, ()()2cos sin wi= π+ π
3k=, 377cos sin55wiππ =+
4k=, 499cos sin55wiππ =+
Example: F i n d a l l r o o t s o f ()1488 3i+ a n d r e p r e s e n t t h e m
graphically. (2009)
Solution: Let 88 3zi=+
() ()2181 3 1 6 , t a n 33rπθ−=+ = = =
141416 cos 2 sin 2 0,1,2,3.33zk i k kππππ ∴ =+ + + =
()()61 612c o s s i n12 12kkiππ ++=+
munotes.in
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0 02 c o s s i n12 12for k w iππ== + 1312π 712π
17712 c o ss i n12 12for k w iππ== + 12π
213 1322 c o ss i n12 12for k w iππ== + 1912π
319 1932 c o s s i n12 12for k w iππ== +
Q. Find all the roots of ()133i + and locate them graphically.
1.3. EXTENDED COMPLEX PLANE, THE POINT AT
INFINITY AND STEREOGRAPHIC PROJECTION
Construction of the Stereographic Projection Map. (2012) Let
be the Complex plane. Consider a unit sphere S (rad ius 1) tangent to at a point 0z=. The diameter NS is perpendicular to and we
call points N a n d S the north and south poles of the sphere S
corresponding to any point z on the Complex Plane , we can
construct a straight line NZ intersecting sphere S at a point ()PN≠.
PN
ZSOS = The Unit Spherex
Fig 1.15
Thus to each point of the Complex Plane , there
corresponds one and only one point of the sphere S and conversely,
to each point of the sphere S (except N), there corresponds one and
only one point on the plane. For completeness, we say that the point
N itself corresponds to the point at infinity of the plane . This one-
to-one correspondence between the points of the pla ne ∞ and the
points of the sphere S is called the Stereographic Projection. The
sphere is called the Riemann Sphere (because Comple x Number can
also be represented by point on the Sphere.) munotes.in
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()()() ()2 2211 1txy t t− + = − +Suppose Complex Plane p a s s e s t h r o u g h c e n t r e o f t h e u n i t
sphere S.
Let 222
12 31xxx++= be the equation of unit sphere S.
()0, 0,1N≡
Also, identify with (){}12 1 2,, 0 : ,xx x x ∈ ∈.
Put (),zx y= and ()123,,px x x= . We will find equations expressing 123,,xx x in terms of x and y.
The equation of straight line Nz in 3 passing through points N and
Z is given by
(){}1:tz t Nt−+ ∈()()(){}1, 0 , 0 , :tx y t t = − + 1 ∈() ()(){}1, 1,tx tyt t = − − ∈……(1) ∵ S t r a i g h t l i n e Nz intersects sphere S.
() ()2222 211 1tx ty t− +− += ∴ ()()222 211txy t− += − ∴ ∴ () () ()2 211 1tz t t− = − + ∴ ()()11tz t− = +
This equation holds if PN≠ … (∵if PN= then 1t= and z=∞) ∴ 221zt z t − = + ∴ ()2211zz t −=+ ∴ 2211ztz − =+ ( f o r PN≠)
22 2
2211 111
11zz zt
zz − + − + −=− =+ + 22
1z = + ∵ P o i n t s N, P, Z are collinear. ∴ F r o m e q u a t i o n ( 1 ) ,
munotes.in
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()12221
11xz zxt xzz + =− = = + +
()()22221
11iz zyxt y
zz −− =− = = + + ( 2 )
2
3 211zxt
z − = = + ∴ P o i n t zx i y=+ ∈ corresponds to point P.
()2
22 21,,11 1iz z z zzP
zz z − − − += + + +
Again from equation (2), 11311xxxtx == − − , 22311xxytx == − − , 1231xi xzx i yx + =+ =−
Point ()123,,Px x x S≡ ∈ corresponds to point z.
12
31xi xzx + = ∈−
Note : From figure (Fig 1.20)
The straight line Nz in 3 intersects sphere and in exactly one point PN≠.
If 1z >, then point P is in the Northern hemisphere and if 1z <,
then point P i s i n t h e s o u t h e r n h e m i s p h e r e . A l s o , i f 1z =, then Pz= and as z→∞, P approaches N.
Distance function :
Let z and z′ be any two points on the Complex Plane . Suppose
point (),zxy corresponds to point ()123,,Px x x S≡ ∈. Suppose point (),zxy′′′ corresponds to point ()123,,Px x xS′′ ′ ′≡ ∈.
We define distance function as
()( ) ( ) ( ) ()222
11 2 2 3 3 ,,dz z dP P x x x x x x ′′ ′ ′ ′ = = − + − + −
() ( ) ( ) ()222211 2 2 3 3,dz z x x x x x x′′ ′ ′ = − + − + −
22 22 2211 1 1 22 1 233 3 322 2xx x x xx x xxx x x′′ ′′ ′′=+ − ++ − ++ −
222
12 311Since and xxx x x x22212 3 + + = + + =′′′ 11 2xxxx x x222 2 2212 3 1 2 3⇒ +++ + += + =′′ ⇒() ()211 2 2 33,2 2dz z x x x x x x′′ ′ ′ = − ++ munotes.in
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Put ()2
12 3 22 21,,11 1iz z z zzxx x
zz z −− − += = = + + +
()2
12 3 22 21,,11 1iz z z zzxxx
zz z ′′+ ′ − ′′ +′′′= = =
′′ ′ + + +
()
()12
2
22 2
22
22 2,2 2
11 1
11
11 1iz zzz z zdz z
zz z
iz z zz
zz z ′′ − − ′′ + + ′ = − + ′′ + + + − − ′ − − + ′ + + +
1.4 SUMMARY
1) A Complex Number Z is an ordered pair (),xy of real numbers.
2) The distance between Two Complex Numbers : L e t ()()11 12 2 2,, ,Zx y Z x y= = b e t w o c o m p l e x n u m b e r s . T h e
distance between them in complex plane is given by ()12 1 2,dZ Z Z Z = − () ( )2212 12xx yy = − + −
3) If P is a point in the Complex Plane corresponding to C omplex
Number (),Zx i y x y=+ = and let (),r θ be the polar co-ordinates of
point (),xy from figure cosxr= θ and sinyr= θ where
22rx y = + is called the modulus or absolute value of Z (denoted
by Z and
1tanyx − θ= is called the argument or amplitude of Z (denoted by argZθ= )
Here θ is the angle between the two lines OP & the real a xis (axis –
X)
4) The modulus or absolute value of a Complex Number Zx i y=+
is defined by 22Zx y = + .
5) De-Moivre’s Theorem :
If n is any integer or fraction then ()cos sin cos sinnin i nθ+ θ= θ+ θ
6)Theorem : In , given 0z≠, the equation expansion nwz= has
n-distinct solution given by 2., 0 , 1 , . . . , 1kin nkwr e k nθ+π = = −
where rz= and argzθ=. munotes.in
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1.5 UNIT END EXERCISES
1) Find two square roots of2i.
(Hint: Let xi y+ be a square root of 2i ⇒ ()2x+iyzi=. ()()2222xy x y ii⇒ − +=, comparing real and imaginary parts
on both the sides, we get two equations in x,y.
220;xy −= 22xy=.()1xi y i⇒ += + or ()1xi y i+= −−
2) Describe the set {}:1 1zz+< in the Complex plane /g932.
Solution: Let ;R e ( ) ,I m ( )zx i y x z y z=+ = = 2211 ( 1 )zx i y xy+⇒ += + += + .
Hence 11z+< describes all real number pairs (, )xy in 2 such
that ()2 211 .xy++ <
22(1 ) 1xy⇒ ++< T h i s i s a n e q u a t i o n o f t h e o p e n d i s c w i t h
centre
at and radius equal to 1, which can be des cribed as follows :
r = 1 B ((-1,0),1)
(-1,0)
3) Find polar form of the Complex Number 1.i+
4) Show that the n th roots of 1 satisfy the “ cy clotomic “
equation
1210nnzz z−−++ ⋅⋅⋅++=.
(Hint : Use the identity ()()1211 . . . 1 .nn nzz z z z−−−= − ++ + +
(2009)
***** munotes.in
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2
SEQUENCES OF COMPLEX NUMBERS
Unit Structure
2.0 Objectives
2.1. Introduction
2.2. Convergent Sequences
2.3. Topological Aspects of the Complex Plane ( L i m i t s ,
Continuity, Uniform Continuity)
2.4. Summary
2.5. Unit End Exercises
2.0 OBJECTIVES
This unit shall make you understand :
• Cauchy and convergent sequences of complex nu mber
.
• The connection between the convergence of real a n d
imaginary parts of a sequence nnnzxy=+, namely nx and ny with the convergence of nz in. We shall also see that
under what conditions a given sequence of co mplex
number nnnzxy=+ is a Cauchy sequence. Can we relate to
fdour findings for real values sequence nx andny.
2.1. INTRODUCTION
We have already associated the meaning to a sequ ence of
real numbers as a function , :aZ+→, denoted by (() )nNan∈. On
a similar line , we shall define a sequence of complex numbers ,
where each term of a sequence is a complex nu mber . For
example 1nnzz= i s a s e qu e nc e o f C om p l e x N um be r s w i th t e rm s
123111,,, . . .zz zetc. In this Unit, we shall consider the topolog ical
aspects of the Complex plane. The concept of abso lute value can
be used to define the notion of a limit of a seq uence of complex munotes.in
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numbers . We shall begin with the definition of a complex valued
sequence .
Definition: A function whose domain is a set of natural number () and range is a subset of , is said to be Real sequence .
Any function whose domain is a set of nature number s () a n d
range is subset of complex numbers , is said to be complex
sequence .
Generally , we denote it by {}nz. nz is the nth term of the sequence.
e.g. 1) The set of numbers 23 2 0 0,,, . . . ,ii i i . This is finite sequence
and its nth term is nnzi=, 1, 2, ..., 200n=
2) The set of numbers () ()2322 2,, , . . .23ii i ++ + 1
It is the infinite sequence and its nth term is ()2nnizn+= .
Sequences : Definition : A function whose domain is a set of
natural number () and range is a subset of , is said to be Real
sequence .
A n y f u n c t i o n w h o s e d o m a i n i s a s e t o f n a t u r e n u m b e rs ()
and Range is subset of , is said to be Complex sequence .
G e n e r a l l y , we denote it by {}nz. nz i s t h e nth t e r m o f t h e
sequence.
e.g. 1) The set of numbers 23 2 0 0,,, . . . ,ii i i . This is finite sequence
and its nth term is nnzi=, 1, 2, ..., 200n=
2 ) T h e s e t o f n u m b e r s () ()2322 2,, , . . .23ii i ++ + 1
It is the infinite sequence and its nth term is ()2nnizn+= .
2.2 CONVERGENT SEQUENCES
A s e q u e n c e {}nz is said to converge to a point 0z [ o r a
sequence {}nz has to limit 0z] if for every 0ε>, there is an N s.t.
0 nzz n N −<ε ∨ ≥ and we write 0limnnzz→∞ =. munotes.in
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Geometrically , 0nzz→ i f e v e r y ε-nbd of 0z c o n t a i n s a l m o s t a l l
terms of the sequence {}nz.
z
01
2
34
5zz
zz
zε
Fig 2.1
Divergent Sequences : A sequence is said to be divergent if it is not
convergent.
Theorem: Prove that any convergent sequence has a unique lim it.
Solution : Let 1limnnzz →∞= and 0limnnzz →∞=
If 10zz≠ then for 1002zz−∈ = > 1N∃ s.t. 1,/ 2n nz z1≥ Ν ⇒ −<ε
and 2N∃ s.t. 0,/ 2n nz z2≥ Ν ⇒ −<ε
choose N = max {}1, 2 1NN N N⇒ ≥and 2NN≥ 12Nzz⇒ −<εand 0/2Nzz−<ε
12 1 0 NN zz zz z z⇒ −≤− + −<ε
10 0zz⇒ < −<∈ a contradiction. 10zz∴=.
Theorem : Suppose nnnzx i y=+ and 00 0zx i y=+ then 0limnnzz→∞ =
iff 0limnnxx→∞ = and 0limnnyy→∞ =.
Proof : Suppose 0limnnzz→∞ = 0,⇒ε> ∃, an integer N s.t.
0 nzz − <ε nN∨ ≥.
Now, 00 0nn nzz xi yx i y − = + −−00nnxx i yy≤ − + −
0
0 and n
nxxnNyy − <ε ∨ ≥
− <ε∴ munotes.in
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0limnnxx→∞⇒ = and 0limnnyy→∞ =
Conversely, suppose 0limnnxx→∞ = and 0limnnyy→∞ = 10,N∨ ε> ∃
and 2N s.t. 012 nxx n Nε − < ∨ ≥
and 022 nyy n Nε − < ∨ ≥
Choose {}12max ,NN N=
if nN≥ then ∴00 0nn nzz xi yx i y − = + −− 00nnxx yy≤ − + − ()Re and Imzz zz ≤ ≤ ∵
22εε<+= ε ∴0 nzz − <ε 0limnnzz→∞ =
Theorem : If 0limnnzz→∞ =, then 0limnnzz→∞ = and the sequence {}nz is bdd.
Proof : Suppose 0limnnzz→∞ = 0∨ ε>, ∃ a n i n t e g e r N s.t.
0 nzz − <ε nN∨ ≥. ∵00nnzzz z − ≤ − ∴0 nzz n N − < ε ∨ ≥ ( 1 ) ∴limnnnzz→∞ = ∴ from equation (1)
00 n zz z −ε< < + ε ∴ Sequence {}nz is bounded.
Example : If 0limnnzz→∞ = and 0limnnww→∞ = prove that
i) []00limnnnzw zw→∞ + =±
ii) []00lim .nnnzw z w→∞ =
iii) 00limn
n nzzww→∞ = provided 00w≠
Definition : Cauchy Sequence : A sequence {}nz is said to be a
Cauchy sequence for every 0ε> t h e r e i s a n i n t e g e r N s.t.
nmzz−<εnN∨ ≥, and mN≥. munotes.in
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Note : From equation (1) ,mnzz n N − <ε ∨ ≥
Put mnp=+ for 1, 2, 3, ...p= ∴ np nzz n N+ − <ε ∨ ≥ and 1p≥.
Theorem : Every convergent sequence is a Cauchy Sequence.
Proof : Suppose {}nz is a convergent Sequence. ∴A sequence {}nz has a limit of 0z. 0limnnzz→∞ = ∴For every 0ε>, there is an integer N s.t.
0 2 nzz n Nε − < ∨ ≥
If mN≥ and nN≥ then
0 2 mzz m Nε − < ∨ ≥
0 2 nzz n Nε − < ∨ ≥ ∴00nm n mzz zzzz − = −+−00nmzzzz≤ −+− 22εε< + = ε ∴nmzz − <ε if nN≥ and mN≥ ⇒ {}mz is a Cauchy sequence.
Theorem : is complete. [i.e. T.P.T. every Cauchy sequence in
is convergent.]
Proof : Let {}{}nn nzx i y=+ be a Cauchy sequence in . {}nx⇒ and {}ny are Cauchy sequence in . ∵ is complete. ∴0nxx→ and 0nyy→ for 00,xy ∈ ∴()lim limnn nnnzx i y→∞ →∞ = + lim limnnnnxi y→∞ →∞= + 00xi y=+0z= ⇒ sequence {}nzis convergent.
Hence is complete.
Note : A sequence is convergent iff it is a Cauchy sequenc e (Cauchy
Criteria for convergence of a sequence.)
Theorem : Let ()nna∈ b e a s e q u e n c e o f p o s i t i v e r e a l n u m b e r s . I f
1
00lim 1, lim 0.n
nnnnalt h e naa+
→→=< = munotes.in
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OY
XεZ0εProof : We have 112ll+<<, and by data there exist m∈ such that
112n
nalla++<< for all n≥m
Put r=12l+. Then 0 < r < 1. Then
()212 ,mm m m r m maa aa r a r r a r++ +<< < = and so on. We get
k
mk maa r k+< ∀∈. Put mmacr= . Then 0n
nac r n<< ∀∈. Since 01r<<, 0ncr→ as n→∞. So 0naa s n→→ ∞.
2.3 TOPOLOGICAL ASPECTS OF THE COMPLEX
PLANE
Topology in the -plane :
A f u n c t i o n ×→ , (),zz z z′′ →− h a s t h e f o l l o w i n g
properties.
i) 0zz′− ≥, if zz′≠ and 0zz ′− = if zz′=
ii) zz z z′′− = −
iii) ,, zz zw wz z z w ′′ ′− ≤− +− ∈
T h u s , i s a m e t r i c s p a c e w i t h E u c l i d e a n m e t r i c ( d i s t a n c e )
defined by ()12 1 2 1 2,, ,dz z z z z z = − ∈
1) Let 0z∈ and 0ε>, then the set (){}00,: Bz z z z ε=∈ − < ε is
called an open disk or open ball with centre at 0z and radius ε (This
is also called the ε-nbd of 0z or nbd 0z).
Geometrically, ()0,Bz ε is an open disk, consisting of all points at a
distance less than ε from the point 0z.
Fig 2.2 munotes.in
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2) A set of the form, (){}{}00 0,: 0 Bz z z z z ε = ∈ < − < ε is called
the deleted neighbourhood of 0z or punctured disk.
3) The set of the form (){}00,:Bz r z z z r∂ = ∈ − = i s t h e c i r c l e
with centre at 0z and radius r and is called the boundary of circle .
4) Let G∈, A set of G is said to be open i n i f f o r e v e r y
0,0zG r ε ∃ >, s.t. ()0,Bzrc G
OY
X r
Z0
plane−
Fig 2.3
e.g. i) Interior of circle is an open set
ii) The entire plane is an open set
iii) Half planes: Re 0, Re 0, Im 0, Im 0zzzz > < > < are open set.
Y
X0 Re z>Y
X
0 Im z<
plane−plane−
Fig 2.4
Thoerem :. Any open disk is an open set
Proof : Let 0,0zr∈ > a n d (){}00,:Bz r z z z r = ∈ − < b e a n
open disk. munotes.in
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Let ()00,aB zr az r∈ ⇒ − < ( i )
If za d−< then
00zz zaaz − = −+− 0za az≤ − + −
r<δ+−δ ( F r o m ( i ) )
r< ∴ 0zz r − <
i.e. ()()0,,zB a zB zr∈ δ⇒ ∈ ⇒ ()()0,,Ba Bz r δ⊂ ⇒ A n y o p e n d i s k i s a n o p e n s e t
5) The complement of a set S⊂ is denoted by cS, and defined by
{} :cSz z S=∈ ∉
6) A set F⊂ is said to be closed if its complement i.e. cF is open.
OR
A set F is said to be closed if it contains all its limit p oints.
7) A set of the form (){}00,:Bz r z z z r = ∈ − ≤ i s c a l l e d t h e
closed disk or closed ball .
e . g . i ) is closed set
i i ) ∅ is closed set
i i i ) {}:I m 4Ez z=∈ =
i v ) {}:2Sz z=∈ ≤
v ) {}:2Sz z z=∈ − ≤
8) Interior point : Let S⊂, then the point zS∈ is said to be an
interior point of set S if 0r∃ > s.t. (),Bzr S ⊂.
9) The point CS∈ i s s a i d t o b e e x t e r i o r p o i n t o f t h e s e t S if ∃ a (),Bcr which does not contain any point of set S.
10) A point pS∈ is said to be a boundary point of set S if it is a
neither a interior point nor an exterior point.
OY
XExterior
BoundaryInterior
plane−
Fig 2.5
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11) The set of all interior point of the set S is said to be interior of S.
12) A set G⊂ is said to be open if each point of G is an interior
point of G.
13) Closure set : The closure of the set S⊂, denoted by ()Cl S ()Cl S S S= δ∪(where Sδ is boundary element is always closed.)
14) A subset S of is said to be Dense if ()Cl S=
e . g . i ) is dense in .
i i ) {}, xi y x y+ ∈ ∈ is dense in .
15) An open set G is said to be connected if for any two points 1z
and 2z can be joined by a curve that lies entirely in G.
OR
A metric space (),Xd i s s a i d t o b e c o n n e c t e d i f t h e o n l y
subset of X which are both open and closed are X a n d ()the empty set∅ = .
Y
Xa
bB(0,1)plane−
Fig 2.6
e.g. 1) Open disk is a connected set.
2) The unit disk (){}0,1 1Bz z = ∈ < is a connected set.
3) The annulus {}:1 2Bz z=∈ ≤ < is connected Fig. 2.6
4) The set {}:2 1Sz z=∈ − < or 21z + < is not connected
Fig. 2.6(b).
Y
Xplane−z21zY
Xplane−01 2 3 -1 -2 -3
fig (a) fig (b)
21z−< 21z+<
Fig 2.7 munotes.in
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16) A domain is an open connected set .
17) A domain together with some none or all of its boundary
point is referred to as a region .
18) Bounded Set : A s e t i s s a i d t o b e b o u n d e d i f 0R∃ > s . t . (){},:SB O R z z R⊂ = ∈ ≤ .
19) A set which is closed as well as bounded is cal led compact
set.
20) A set that cannot be enclosed by any closed dis k is called
unbounded set.
21) Let 1z∈ a n d 2z∈. These we denote the line segment
from 1z to 2z by [](){}12 1 2,1 : 0 1zz t z t z t = − + ≤≤
Y
X1z2z
Fig 2.8
Function, limits and continuity :
Definition : Let A and B be two non-empty subset of complex
numbers. A function from A to B is a rule, f, which associates to
each 00 0zx i yA=+ ∈ a unique 00 0wui vB=+ ∈
T h e n u m b e r 0w i s t h e v a l u e o f f at 0z a n d w e w r i t e ()00fz w=. If z varies in A then ()fz w= varies in B. We say that
f is a complex valued function of a complex variable.
H e r e w is the dependent and z is the independent variable.
Let :fA B→ b e a f u n c t i o n a n d SA⊂
then() ( ) / , }fS fz z S = { ∈ where ()fS is called the image of S under
‘f’ and the set {}() /Rf z z A = ∈ is called range of ‘f’.
Single and Multiple Valued Function :
Let {}0z∈−, then we write the polar form of a complex number z
is izr eθ= where rz= and [],θ∈ − π π i.e. (),izz r r eθ= θ= . munotes.in
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θY
X0z
Fig 2.9
If we increase θ to 2θ+π ∴ ()()2,2izr r e θ+π θ+π=2.iire eθπ= ireθ=(),zr= θ r e t u r n i n g t o
its original value.
Definition : A function f is said to be a single valued if f satisfies () ((,) ) (, 2)fz fz r fr = θ = θ +π.
Otherwise, f is said to be a multiple valued function .
e.g. () ,nfz z n= ∈ is a single valued function.
Solution : ∵() ( ) ()(),nifz fz r r eθ= θ=
()()() 2,2ninfz r r eθ+π θ+π= ()2in nnreθ+π=2.ni n i nre eθπ=
{}21,ni n i nre e nθ π= = ∈∵
()nireθ=()(),fz r= θ
Note : If n∉ then ()nfz z= is a multiplied valued function. ∵21ineπ≠, when n∉
Let :fA B→ be a function.
i) If the elements of A are complex numbers and tho se of B are Real
Numbers then we say that f is a real valued function o f c o m p l e x
variable.
ii) If the elements of A are Real Numbers and those o f B are
complex numbers then we say that f is a complex valued function of
real variable.
Let :f→ be a function then the graph of f is a subset of ×
and it is two dimensional object and we can represe nt it very well on
the two dimensional page. However the graph of the :f→ is a munotes.in
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subset of × (××× C a r t e s i a n p r o d u c t ) i . e . a f o u r
dimensional object and we cannot represent it on tw o dimensional
plane. In this case we consider two plane, one pla ne is z-plane and
other one is w-plane.
Fig 2.10
Limit Point :
L e t D be a subset of i.e. D⊂ then we say that a point 0z
is a limit point of D if every neighbourhood of 0z contains a point of
D other than 0z i.e. {}00(( , ) )Bz r z− for any 0r>.
Definition : Let f be a complex valued function defined on D and let ()0zC l D∈ . We say that a number ℓ is a limit of ()fz as 0zz→
and we write ()0lim
zzfz
→ =ℓ. ⇔ i f f 0, 0∨ ε> ∃ δ> s . t . ()fz − <εℓ w h e n e v e r zD∈ a n d
0 0 zz< − <δ. ()(),fz B⇔∈ εℓ wherever (){}00,zD B z z∈ δ−∩
()0fz ≠ℓ0zzδ
Fig 2.11 Y
X.Z pl an e −z plane −V
Uplan e −plane ω−ωmunotes.in
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Note : 1) f may not be defined at 0zz =
2 ) 0z need not be in D.
3 ) e v e n i f 0zD∈, ()0fz≠ℓ
4) In real variable theory, if 0x∈ then 0xx→ has only two
possible ways, either from left or from right. In complex case, 0zz→, in any manner in the Complex Plane.
Theorem : Let f be a complex valued function defined on D and let ()0zC l D∈ . If ()0lim
zzfz→ exists, then this limit is unique.
Proof : Let ()10lim
zzfz
→=ℓ and ()20lim
zzfz
→=ℓ
T.P.T. 12=ℓℓ
By definition for a given 0ε>, 120, 0∃ δ> δ>
s.t. () 12fzε − <ℓ , whenever (){}01 0,zD B z z∈ δ−∩ a n d
() 22fzε − <ℓ , whenever (){}02 0,zD B z z∈ δ−∩ .
Let {}12min ,δ= δ δ.
If (){}{}00,zD B z z∈ δ− ∩ then 12 1 2() ()fz fz−= − + −ℓℓ ℓ ℓ()()12fz fz≤ − + − ℓℓ
22εε<+= ε ∵ ε is arbitrary. ∴ 12=ℓℓ
i.e. limit is unique.
Theorem : Let f be a complex valued function defined on D.
suppose, ()()(),,fz u x y i v x y= + , 00 0zx i y=+ , 000wui v=+ a n d ()0zC l D∈ .
Then 00lim
zzw→= iff ()00lim ,
zzuxy u→ = and ()00lim ,
zzvxy v→ =.
Proof : Direct part –
Let 00lim ( )
zzfz w→ = and 00 0wui v= +
By definition, 0, 0∨ ε> ∃ δ> s.t. ()0 fz w − <ε w h e n e v e r (){}00,zD B z z∈ δ−∩ .
Now, ()()()()00 0,,fz w u x y i v x y u i v − = + −+ ()()00,,uxy u ivxy v= −+ − munotes.in
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()()00,
,uxy uvxy v − ≥
− Re Re
Im Imzz zzz z ≤ ≤ ≤ ≤ ∵
If (){}00,zD B z z∈ δ−∩ t h e n ()0 ,uxy u − <ε a n d ()0 ,vxy v − <ε ⇒ ()00lim ,
zzuxy u→ = and ()00lim ,
zzvxy v→ =
Conversely, assume that 0limzz →0(, )ux y u = & 0limzz →0(, )vxy v = ∵By definition given 120, 0 .∈ > ∃ δ δ >
s.t. 0(, ) / 2ux y u − < ∈ and 0(, ) / 2vx y v − < ∈ whenever 01zz −<δ
and 02zz−<δ.
let {}12min ,δ = δδ. ∵whenever 0,zz −<δ
Consider 00 0() (, ) (, )fz w u x y i v x y u i v − = + −− 00(, ) (, )uxy u vxy v≤ − + −22∈∈<+= ∈ 00()
zzlin f z w→⇒ =
Examples : If ()2izfz= i n t h e o p e n d i s k ()0,1B , prove that
1lim22ziz i→=.
Solution : Given ()2izfz = ∴We must prove that for every 0, 0ε< ∃ δ>, s.t. 22iz i − < ε
whenever ()0,1zB∈ and ()01 , zf z< − <δ ≠ℓ
if 01z < − < δ, then 122 2 2i iz iz δ − = − <
choosing 2δ=ε, we see that
22iz i − <ε whenever ()0, 1zB∈ and
01 z< − <δ. ∴1lim22ziz i→ =
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zp l a n e−01
z1
wp l a n e−()2izfz =()2ifz =
2δ=∈
Fig 2.12
Problem : If ()2zifz − i n t h e o p e n d i s k 1z < p r o v e t h a t
()
1lim2zifz
→ = .
Solution : Given ()2izfz =
We must prove that for every 0ε>, for given any 0ε> we can find 0δ> s.t.
22iz i − <ε whenever 01 z< − <δ.
If 01 z< − <ε 2iz i − ⇒ < ε 12iz⇒ − < ε 12z⇒ − < ε
1i =∵
Choosing, 2δ=ε, we see that
22iz i − <ε, whenever 01 z< − <δ ∴
1lim22ziz i→ =
zp l a n e−wp l a n e−0z
12δ=∈()fz2i
ε
Fig 2.13 munotes.in
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Problem : Prove that
0lim
zzz→ does not exist.
Solution : We know that the function ()fz→ℓ (a unique limit) as 0zz→ in any manner in the -plane.
Let ()zfzz =
Let 0z→, along the real axis. ∴0,yz x= = ()zx i y=+∵ ∴()
00 0lim lim lim 1zz zzxfzzx →→ → = = =
Let 0z→, along the imaginary axis. ∴0,xz i y= = ()zx i y=+∵ ∴()
00 0lim lim lim 1zz zzi yfzzi y →→ → − = = = − ⇒ limit is not unique along real and imaginary axis. ∴
0lim
zzz→ does not exist.
Problem : If 2()fz z=, prove that 20lim ( )
zzfz z→=.
Solution : L e t 0∈>given, to find 0δ> s.t. 22zz−<ε whenever
0 0zz<−<δ
consider 2200() ()zz z z z z−=+ −
00zzzz+ −0zz<δ+ 002zz z=δ− +002zz z≤δ − +δ0 .2 z<δδ+δ=∈ ∵Choose 0δ> s.t. min
0,112 z∈+
220 zz⇒ − <∈. 20lim ( )
zzfz z o→⇒ = .
Theorem : Let f and g be defined in the neighbourhood of 0z except
possibly at 0zz=.
If ()0lim
zzfz
→ =ℓ and ()0lim
zzgz m→ =
Then 1) ()()0lim
zzfz g z m→ ± = ±ℓ
2 ) ()()0lim .
zzfz g z m→ =ℓ munotes.in
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3 ) ()() 0lim
zzfzgz m→ = ℓ
Continuity
Definition : A function :fD→ is said to be continuous at a point
0zD∈, iff ()0lim
zzfz→ exists and 00lim) ( ) ( )zzfz fz→ = .
OR
Definition : A function :fD → is said to be continuous at a point
0zD∈ iff 0, 0∨ ε> ∃ δ> s.t. ()()0 fz fz − <ε whenever zD∈
and 0zz − <ε
OR
Definition : A function f is said to be continuous at a point 0zD∈
iff the following 3 conditions hold true :
i) f is defined at 0z i.e. ()0fz exists.
ii) ()0lim
zzfz→ exists
iii) ()()00lim
zzfz fz→ =
OR
Definition : A function :fD→ is continuous or f is continuous
on D if f is continuous at every point of D.
Example : If ()2fz z= t h e n p r o v e t h a t f is continuous at a point
zi=∈.
Solution : Given, ()2
0,fz z z i= = ∴ ()21fi i== − ∴ 22lim 1ziz
→ = = − ℓ ∴ ()2lim 1
zizf i→ = − = ⇒ f is continuous at a point zi=.
Example : Let ()2
0zz ifzzi ≠=
= prove that f is not continuous at
a point zi=.
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Solution : ()0fi= ( g i v e n ) ∴ ()2lim lim 1zi zifz
→→ = = − ℓ
∴ ()()lim 1
zifz fi→ = − ≠
∴ f is not continuous at 0zz=
Problem : Discuss the continuity of2
42()31zfzzz=++at 4izeπ=
Solution : 24 421ii
ze z e i zππ= ⇒ == ⇒ =−
13()13 1ifzi∴ ==−++
∴ the limit exist 4izeπ= .
∴()fz is continuous at4izeπ=
Uniformly Continuous : A function :fD→ i s s a i d t o b e
uniformly continuous on D iff the following cond itions holds for
every 0,ε> ∃ δ> s.t. for any two points 1z and 2z in D. ()()12 1 2zz f z f z − <δ⇒ − <ε
Example : Let ()2fz z= i n t h e o p e n d i s k ()0,1B , prove that f is
uniformly continuous on ()0,1B .
Solution : Given, ()2fz z=. ∴ We must prove that for a given 0ε>, we can find 0δ>, s.t. for
any two points 1z and 2 in ()0,1B and ()()12 1 2zz f z f z − <δ⇒ − 2212zz= − 12 12zz zz= − +
()12zz<δ + b y t r i a n g l e i n e q u a l i t y
2< δ ()121a n d 1zz < <∵
Choosing 2εδ=, we see that 22
12zz − <ε w h e n e v e r 12,zz − <δ ()12,0 , 1zz B∨ ∈ . ∴ f is uniformly continuous on ()0,1B .
Definition : Unbounded set :
A set E is said to be unbounded if 0R∃ > s.t.zE∈ vzE∈.
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Definition : Limit at Infinity
L e t f be defined on an unbounded set E . I f f o r e a c h 0ε>, 0R∃ > s.t. ()fz − <εℓ whenever zE∈ and zR > then we say
that ()fz→ℓ as z→∞ and we write ()lim
zfz
→∞ =ℓ.
e.g. 1lim 0zzz→∞=for given 0∈> a b o v e 0R>s.t. 1R>∈ ⇒11 10 zRzz−=< ε > >∈
Infinite Limit :
L e t f be defined out D except possible at 0zD∈. If for every 0, 0R> ∃ δ> s.t. ()fz R > whenever 0 0 zz< − <δ then we say
that ()fz→∞ as 0zz→ and we write ()0lim
zzfz
→ = ∞.
e.g. 211lim1zz→ = ∞−
2.4 SUMMARY
1) Let f be a complex valued function defined on D and let ()0zC l D∈ . If ()0lim
zzfz→ exists, then this limit is unique.
2) Let f a n d g b e d e f i n e d i n t h e n e i g h b o u r h o o d o f 0ze e x c e p t
possibly at
If ()0lim
zzfz
→ =ℓ and ()0lim
zzgz m→ =
Then 1) ()()0lim
zzfz g z m→ ± = ±ℓ
2 ) ()()0lim .
zzfz g z m→ =ℓ
3 ) ()() 0lim
zzfzgz m→ = ℓ
3) A function f is said to be continuous at a point 0zD∈ i f f t h e
following 3 conditions holds
i) f is defined at 0z i.e. ()0fz exists.
ii) ()0lim
zzfz→ exists
iii) ()()00lim
zzfz fz→ =
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4) A function :fD→ is said to be Uniformly continuous iff the
following conditions holds for every 0,ε> ∃ δ> s.t. for any two
points 1z and 2z in D then()()12 1 2zz f z f z − <δ⇒ − <ε
5) Every convergent sequence is a Cauchy Sequence.
6) is complete.
2.5 UNIT END EXCERCISES
1) Find the limit of a sequence
for 1z<.
Solution : Consider 00 0nn
nzzz−= −= → as n→∞, for 1z<.
2) Check whether the sequence nnzni=+ is convergent or not.
Solution : nnzni=+∵, then 2534n
niazi= −<+, because
2110()1ni
ni nin−=− = →+++ as n→∞ .
3) Which of the following subsets of are connected , if not
connected then what are it’s components ?
(a) {}:1Xz z= ≤Ans : X is connected .
(b) {}{}:1 : 2 1Xz z z z= ≤ −≤∪
Ans: X is not connected , because {}{}:1 : 2 1Xz z z z = ≤− ≤∪
is a disjoint union of nonempty closed subset s ( Components) of
X.
4) Let ,nzz be points in and let d be the metric on ∞.
Show that 0nzz−→ if and only if (),0ndz z→ as n→∞ .
(Hint: F o r ,zz′ ∈, 222(, )((1 )(1 ))zzdzz
zz′− ′ = ′++ and 1222(, )(1 )dz
z∞=
+
5) Let ()Pz b e a n o n c o n s t a n t p o l y n o m i a l i n
. Show that
()Pz→∞ as z→∞. munotes.in
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6) Suppose :fX→Ω is uniformly continuous , show that if {}nx is a Cauchy sequence in , then {}()nfx is a Cauchy
sequence in Ω.
7) Show that if f a n d g a r e b o u n d e d u n i f o r m l y c o n t i n u o u s
functions from X into then fg is also bounded and uniformly
continuous function from X into .
(Hint: () ( ) () ( ) ( ) ( ) ( ) ( )fg x fg y f x g x f y g y− = −() () () () () ()fx fy g x fy g x gy≤− + − .)
8) Verify the continuity of the following function f of the extended
complex plane {}∪∞ at the point 34a−= ( 2 0 1 2 )
()fz=∞ if 34z−≠
= 143zz++ if 34z−≠
*****
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3
SERIES OF COMPLEX NUMBERS
Unit Structure
3.0. Objectives
3.1. Introduction
3.2. Convergence of Series
3.3. Tests for determining the Convergence of Power Series
3.4. Summary
3.5. Unit End Exercises
3.0 OBJECTIVES
This unit shall make you construct a seri es of
complex numbers by understanding the definition of a series of
real’s . Basically, we are going to define a power series of the
form 0nnnaz∞
=∑ . We shall check for the conditions , under
which the given power series is convergent or not. Hence, we
shall employ certain tests in order to det ermine the
convergence of the given power series.
3.1 INTRODUCTION
An infinite series of real’s is the expressio n of the form
1kka∞
=∑ , where kais a real number for all 1k≥. Similarly we
construct an infinite series of complex nu mbers as 1nnz∞
=∑ ,
where nzs′ are complex numbers for all 1n≥. For example
12kkiki∞
=∑+is an infinite series of complex numbers. To c h e c k
whether the sum exists or not, in other wo rds whether a given
series of complex numbers is convergent or not, we employ
certain tests for convergence and we shall convert the given
problem of checking convergence of the series of com plex
numbers to checking convergence of the se ries of real
numbers. For example,
12k
kiki∞
=∑+ is a convergent series of real
numbers, because
2 411ki
ki k=
++ and we know that 1411k
k∞
=∑+ munotes.in
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converges. Let us start with defining a ser ies of complex
numbers.
3.2 COVERGENCE OF SERIES
Definition: Let {}nz b e a s e q u e n c e o f c o m p l e x n u m b e r s , F o r m a
new sequence defined by 11 212 12,, . . . , . . .nnSz S zz S z z z= =+ =++ ,
where nS is called the sequence of nth partial sums of sequence {}nz.
The sequence {}nS i s s y m b o l i z e d b y 121...nnzz z∞
=++ =∑ c a l l e d a n
Infinite series .
If limnnSS→∞ = exists then the series is said to be convergent and S is
its sum i.e.
11lim limnk nnnnkzz S S∞∞
→∞ →∞==== =∑∑
A series is said to be divergent if it is not convergen t sequence. (The
necessary condition for the convergence of the ser ies.)
Theorem : If the series
1nnz∞
= ∑ is convergent then lim 0nnz
→∞ =.
Proof : Given series is
1nnz∞
= ∑.
Let 12 1 ...nn nSz z z z−=++ + + … … … . . ( 1 )
be the nth partial sum of series.
Given that the series is convergent
Let S be the sum
1nnz∞
=∑ ∴limnnSS→∞ = ∴ from equation (1) 1nn nzSS−= − ()12 1 1...nn zz z S−− + ++ =∵
Taking limit on both sides, ∴1 lim lim limnn nnnnzSS S S−→∞ →∞ →∞ = − = − ∴lim 0nnz
→∞ =
Consider the infinite series 01
0...n
nzz z∞
= = ++∑
If 12...nn n nRzz z++=+ + + t h e n nR i s c a l l e d r e m a i n d e r o f i n f i n i t e
series. If S is sum of infinite series then nnSS R=+ or nnRS S=−.
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Theorem : A series
0nnz∞
= ∑ o f c o m p l e x t e r m s i s c o n v e r g e n t i f f f o r
every 0ε>, ∃ an integer N s.t. 1...nn n pzz z++ + + + < ε nN∨ ≥
and 0p≥. (Cauchy criteria for convergence of series)
Proof : Suppose
0nnz∞
=∑ is convergent
Let 012 1...nnSz z z z−=+ ++ + be the nth partial sum of series and let S
be a sum of series. ∴limnnSS→∞ = 0,∨ ε> ∃ an integer N s.t. nSS − <ε (1) nN∨ ≥
Let 12...nn n nRzz z++=+ + + be the remainder of an infinite series. ∴nnSS R=+ or nnRS S=− ∴ From equation (1)
nn nSS S S R n N − = − = < ε ∨ ≥
i.e. 12 ...nn nzz z n N++ + + + < ε ∨ ≥ ∴12 ...nn n n pzz z z n N++ + + + + + < ε ∨ ≥ and 0p≥
Converse :
Given 0ε>, there is an integer N s.t.
12 ...nn n n pZz z z++ + + + + + < ε … … … … … … … . ( 2 ) nN∨ ≥ and 0p∨ ≥
We know that, 01
0... ...nn
nzz z z∞
==+ + ++∑
If S is its sum then we write nnSS R= − ∴nnRSS= − nnRS S⇒ < −
But 1...nn n n pRz z z + + = + + + < ε g i v e n f r o m ( 2 )
nN∨ ≥ and 0p∨ > ⇒ nSS n N − <ε ∨ ≥ ⇒ limnnSS→∞ = ∴
1nnz∞
= ∑ is convergent. munotes.in
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Definition : Let nz∈. For every 0n≥ the series
1nnz∞
= ∑ converges
to 0z iff for every 0,ε> ∃ an integer N s.t.01n
kkzz n N=∑ −<ε ∀≥.
Definition : A series
1nnz∞
=∑ converges absolutely if
1nnz∞
=∑ converges.
Proposition : If the series
1nnz∞
= ∑ converges absolutely then
1nnz∞
= ∑
converges.
Proof : Let 0ε>, consider an infinite series
1nnz∞
= ∑.
Let 12 ...nnSz z z=++ + be the partial sum of series given that
1nnz∞
= ∑
convergent absolutely. ∴ F o r a g i v e n 0,ε> ∃ an integer N s.t.
1k
knz∞
=+ < ε∑ nN∨ ≥ ( 1 )
If nmN>≥ then ∴12 ...nm m m nSS z z z + + − = + + +
1nkkmz =+= ∑
1nkknz=+≤ ∑
1kknz∞
=+≤ ∑<ε from (1) {}nS⇒ is a Cauchy sequence. {}nS⇒ is a convergent Sequence. (∵by Cauchy criteria) ∴0z∃ ∈ s.t. 0limnnSz→∞ =
Thus,
1nnz∞
= ∑ is convergent.
Examples :
1) Prove that 2
011. . .1n
nzz zz∞
= = + ++ = − ∑ if 1z <
Solution : Given 2
01. . .n
nzz z∞
= = + + +∑
Let 211. . .nnSz z z−=++ + +
23...nnzS z z z z = +++ + (multiplied by z, 0z≠) munotes.in
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∴() 11 1nnnn nSz S z S z z− = − ⇒ −=− ⇒ 11nnzSz − =− (1 )z≠
T.P.T. 11nSz= − … … . . ( 1 )
i.e. T.P.T. lim 0n
nz
→∞ =
Given any 0ε>, we must find integer N s.t. nz < ε nN∨ ≥ [If 0z=, then the result is true].
Let 0z≠ ∴ n nzz = < ε ⇒ log lognz < ε ⇒ loglognz ε > {logz∵ is negative when 1z<}
Choosing log
logNzε= , we see that nzn N < ε ∨ > ∴ lim 0n
nz
→∞ = ∴ F r o m e q u a t i o n ( 1 )
1lim lim1nnnnzSz→∞ →∞ − = −11z= − ()0a snzn = →∞∵
Hence,
011n
nzz= = − ∑ w h e n 1z <
Note : 1) Geometric series (G.S.)
0nnz∞
= ∑ i s c g t w h e n 1z < a n d
divergent when 1z ≥.
Uniformly converges for series :
For each n∈, let ()nfz b e a c o m p l e x f u n c t i o n o f c o m p l e x
variable.The series ()nfz∑ c o n v e r g e s t o ()fz p o i n t w i s e fo r e a c h zD∈ iff () ()nfz f z∑ = and for each zD∈ [This means that for each zD∈ and for each 0,ε> ∃ an integer N (depends on z) and ε], s.t. ()()nSz f z nN − <ε ∨ ≥.
Definition : The series ∑()nfz is said to be uniformly convergent
on D to f(z) if for every 0ε>, ∃ an integer N (depends only on ε)
s.t. ()()nSz f z − <εzD∨ ∈, and nN∨ ≥. A power series about munotes.in
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0z i s a n i n f i n i t e s e r i e s o f t h e
form () () ()200 0 1 0 20n
n
nazz a zza zz a∞
=− =+ − + − ∑ + … … … , w h e r e
constants na and 0z are called complex numbers and z is a complex
variable .
Note : If 00z= then 01
0...n
n
naz a a∞
= = + +∑
This is power series about origin (i.e. 0z=)
e.g. Geometric Series (G.S.) 2
01. . .n
nzz z∞
==+ + +∑
i) If 1z < t h e n 1lim1nnSz→∞ = − a n d t h e G . S . c o n v e r g e s w i t h
011n
nzz∞
= = − ∑ .
ii) If 1z > then limnnS
→∞ = ∞ and the G.S. diverges.
3.3 TESTS FOR DETERMINING THE CONVERGENCE
OF A POWER SERIES
Weierstrass M-test :
Statement : Let :nfD ⊂→ be a complex function defined on D
s.t. ()nnfz M zD < ∨ ∈ a n d nN∈. If
1nM∞ ∑ i s c o n v e r g e n t
series of positive real numbers then series
1nf∞ ∑ i s u n i f o r m l y
convergent.
Proof : Given ()nnfz M ≤ vzD∈and nN∈ ( 1 )
Let 0ε>
Given that
1nM∞ ∑ is convergent ∴∃ an integer N s.t.
1n
knMn N∞
=+<ε ∨ ≥∑ (2)
Given series ()
1fn z∞
∑
Let ()()()()12 ...nnSz f z fz fz=++ + if nmN>≥, then munotes.in
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()()()()()12 ...nm m m nSz S z f z f z fz++ − = + + +
() ()
11nn
k
km kmfz f n z=+ =+= ≤ ∑∑
1nkkmM=+≤ ∑ f r o m ( 1 )
1n
k
kmM
=+≤ < ε∑ f r o m ( 2 ) ∴ ()()nmSz S z − <ε, ,nm N∨ ≥ ⇒ (){}nSz is a Cauchy sequence. ⇒ S e q u e n c e (){}nSz is a convergent sequence ∴ w∃ ∈ s.t. ()limnnSz w→∞ =
Define ()wf z= this gives a function :fD→ for each zD∈ and
for each nN≥ ()()()()nnSz f z f z Sz − = − . ()()()()nnSz f z f z Sz − = − ()()12...nnSz f z++= + +
() ()
11k
kn knfk z f z∞∞
=+ =+= ≤ ∑∑
1k
knM∞
=+≤ < ε∑ f r o m ( 1 ) a n d ( 2 )
Hence ()()nSz f z zD − <ε ∨ ∈ where n≥ ⇒ S e r i e s 1nf∞
∑ is uniformly convergent on D.
Examples : 1) Prove that the series
11nn
nzn∞
= +∑ i s u n i f o r m l y
convergent on a set {}:1Dz z=∈ ≤ .
Solution : Given series
11nn
nzn∞
= +∑
Let ()1nnnzfz
n=+ ∴()111n
nnz
fz
n nn
= ≤+ + {}1z ≤∵
321nMn≤ =
()1nn +∵ munotes.in
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∴ 321nM
n =
∑∑ is a p-series and it is convergent
312p = >∵ ∴ B y W e i e r s t r a s s M - t e s t ,
T h e g i v e n s e r i e s
11n
nz
nn∞
= + ∑ is uniformly convergent.
2) Given series ()
11n
nzz∞
= −∑
i ) P r o v e t h a t t h e s e r i e s c o n v e r g e s f o r 1z < a n d f i n d i t s
sum.
i i ) P r o v e t h a t t h e s e r i e s c o n v e r g e s u n i f o r m l y t o t he sum z for 12z ≤.
i i i ) D o e s t h e s e r i e s c o n v e r g e s u n i f o r m l y f o r 1z ≤? Explain.
Solution :
i) The given series is ()
11n
nzz∞
=−∑ ∴ () () () ()23
111 1 1 . . .n
nzz z z zz zz∞
= −= −+ −+ −+∑
22334 1...nnzz z z z z z z+=−+ −+ −++ −
()11nnzz z z+=− = − ∴ Let ()1nnSz z z+=−
We must prove that given any 0ε>, we can find an integer N s.t. ()nSz z nN − <ε ∨ ≥ ∴()1n
nSz z z z z+ − = −− <ε1nz+⇒ − < ε1nz+⇒ < ε()1l o g l o gnz⇒ + < ε
log1, 0lognzzε⇒ +> ≠ log1lognNzε⇒ > −=
Choosing, log1logNzε= − ( 1 ) ∴ nSZ − <ε nN∨ ≥ ∴ 1lim limn
nnnSz z z+
→∞ →∞ = − =
Hence, the series converges for 1z < and z is its sum. munotes.in
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56
ii) Since from (i) the series converges to some z f or 1z < a n d
hence it converges for 12z ≤
log1logNzε= − f r o m ( 1 )
If 12z =, then log11log2N ε= − is the largest value of log1log zε− . ∴ log,11log2nSz n Nε − <ε ∨ ≥= − where N depends only on ε and not on z. ∴ T h e g i v e n s e r i e s c o n v e r g e s u n i f o r m l y t o s u m z f o r 12z ≤.
iii) If 1z ≤ ∴ log1log 1Nε= − =∞ f r o m ( 1 )
Hence, the series does not converges uniformly for 1z ≤.
Ratio Test for series (2012)
Statement: Let kz∑be an infinite series for non-zero complex term
s.t.
12
11
1.. . . . . . . .
0nNNN n
nN N
NN n
p
Nn n
pzz zzz zzz z
zz z−++
+ −
≥=< λ
λ∴ → ∑∑ = L then
i ) I f 1L<, the series converges absolutely.
i i ) I f 1L>, the series diverges.
i i i ) I f 1L=, the series may converge or diverge.
Proof: Suppose L<1. Then for λ with L<λ<1, there exist an integer
N s.t. 1nnzz+<λ nN∀≥ so that,
12
11.. . . . . . . .nNNN n
nN N
NN nzz zzz zzz z−++
+ −=< λ pNp Nzz+∴ < λ for p1≥.
Since
1pN
pz
≥λ∑ is a convergent (geometric series),
1Nnz≥∑ is
covergent by comparision test. This proves (i) munotes.in
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57
If L>1, then there an integer N s.t. L>k>1 a n d 1nnzz+ > k for all nN≥ ∴for all n > N 12
11.. . . . . . . .nN NN n
nN N
NN nzz zzz z kzz z− ++
+ −=> →∞
Hence, 0nz∴→as n→∞ and so nz∑ diverges. Hence (ii).
Example:
1) Prove that the power series
0!nnzn∞
= ∑ converges for all values of z.
Solution : Given power series is
1!nnzn∞
=∑
Here, ()111!1 !nnaann+ = ⇒ = +
111 (1 ) !lim lim lim 0 11 1!n
nn nna n
ann+
→∞ →∞ →∞+== = <+
.
Therefore the the series is convergent.
Comparison Test :
If the series nv ∑ c o n v e r g e s a n d nnuv ≤ t h e n nu ∑
converges absolutely. Also, nu ∑ converges.
Abels’ theorem :
Statement : If the power series nnaz∑ converges to particular
value 00z≠ o f z t h e n i t c o n v e r g e s a b s o l u t e l y z∨ s . t . 0zz < .
Proof : Given that the power series nnaz∑ u s c o n v e r g e s f o r a
particular value 00z≠of z. ∴0nnaz ∑ converges. ∴0lim 0n
nnaz
→∞ = ⇒ sequence {}0nnaz is bounded. ∴∃ positive number M s.t. 0n
naz M≤
0nnMa
z ≤ munotes.in
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or
0n
n
nzaz M nz ≤ ∨ ∵0.nzMz ∑ is a geometric series and convergent for
01z
z <
i.e. 0zz < ∴ By comparison test nnaz ∑ c o n v e r g e s a b s o l u t e l y f o r 0zz < .
Cauchy-Hadaward Theorem :
Statement: For a given power series
0nn
naz∞
= ∑ define a number R,
0R≤≤ ∞, by 11lim .supnnaR= then
i) If zR <, then the series converges absolutely
ii) If 0rR<<, then the series converges uniformly on {}:zz r∈ ≤
iii) If zR > , then the series diverges [Here R is radius of
converges of power series.] (2008)
Proof : Given, 11lim supnnaR = ( 1 )
[Note : A n u m b e r L is said to be a limit superior of the sequence {}nu i f i n f i n i t e l y m a n y t e r m s o f t h e s e q u e n c e nu a r e g r e a t e r t h a n
L−ε, while finite number of terms greater than L+ε where 0ε>.]
i) Let zR <, then 0r∃ > s.t. zr R < < ∴ 11rR> ∴ B y d e f i n i t i o n o f t h e l i m s u p a n d f r o m e q u a t i o n ( 1 ) ⇒ 1nnar <
n
nnzaz n Nr < ∨ ≥
∵ nz
r ∑ i s a G . S . a n d i t i s c o n v e r g e n t f o r 1zr < i . e . zr <. munotes.in
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59
∴ B y c o m p a r i s o n t e s t , ∴ nnaz∑ converges for zR <. ∴
0nn
naz∞
= ∑ converges absolutely for zR <.
ii) Let 0rR<< choosing ,0rr r R′′ ∃ < < <.
B y u s i n g p a r t ( 1 ) , w e h a v e
11 1nnnnaa vr< ′ nN≥ ⇒
()1nna
r<′ ∴ nnn
nzrazrr< ≤ ′′ ∵ nrr∑′ is a G.S. of positive real numbers and it converg ent
for rr′< . ∴ By Weierstrass M-test, the power series
0nn
naz∞
= ∑ converges
uniformly on {}:zz r∈ ≤ .
iii) Let zR >, then 02rr R∃ > ∋ > > ⇒ 11rR< . ∴ B y d e f i n i t i o n lim sup and from equation (1)
1 1nnan Nr > ∨ ≥ ⇒ 1nnar > ⇒ 1n nn
nnnz razrr > > = ∴ 1n
naz n N > ∨ ≥ ⇒ 0n
naz → a s n→∞{∵nz ∑ c o n v e r g e s t h e n lim 0, 0nnnzz
→∞ = → as n→∞} ⇒ P o w e r s e r i e s
0nn
naz∞
= ∑ is divergent for zR >.
munotes.in
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Definition : The radius of convergence R of the power series
0nn
naz∞
= ∑ i s d e f i n e d a s R = s u p {r′ t h e s e r i e s c o n v e r g e s vzsatisfying zr≤}.
i) If 0R=, then power series nnaz ∑ cgs only for 0z=.
ii) If R=∞, then the power series converges , for all values of z.
iii) If 0R<< ∞, the power series converges for all (),nzf z R< and
diverges v(),nzf z R>.
T h e p o w e r s e r i e s m a y c o n v e r g e o r d i v e r g e o n t h e c i rcle zR =. The circle zR = is then called the circle of convergence .
Note : If
0nn
naz∞
= ∑ is power series with radius of converges R then 1limn
n naRa →∞+ = ,provided this limit exist.
Theorem : If 0na≠ f o r a l l b u t f i n i t e l y m a n y v a l u e s o f n t h e n t h e
radius of convergence R of
0nn
naz∞
= ∑ i s r e l a t e d b y f o l l o w i n g , 111lim inf lim supnnnnaa
aR a++ ≤≤ .
In particular, if 1limn
nnaa+→∞ e x i s t s , t h e n 11 1lim lim supnnnn naaRa+
→∞= = . (2007,2008,)
Proof : Given R is radius of convergence of the series
0nn
naz∞
= ∑ .
Suppose, 1lim supn
naLa+ = and 1lim infn
na
a+ =ℓ ∴ B y t h e d e f i n i t i o n o f l i m i t s u p , ∃ an N s.t.
1n
naLn Na+ < + ε ∨ ≥ ∴ 12
11,, . . . ,NN n
NN naa aLL Laa a+ +
+ − < + ε < + ε < + ε
munotes.in
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Multiplication of these inequalities gives
()12
11...nNNN n
NN naa aLaa a−+ +
+ − < + ε ∴ ()nNn
NaLa− <+ ε ∴ ()nNnNaa L− < + ε ∴ ()11 nNnnnNaa L− < + ε ∴ () ()11 N nnnNaa L L−
< + ε +ε ∵ ε is arbitary as n→∞, we get
1lim sup nnnaL→∞ ≤ ( 1 )
()lim 1 if 0npn a p = → >∵
Similarly, by the definition of lim in f 1lim infnna ≤ ℓ ( 2 )
From equation (1) and (2), we get ∴ 11lim inf lim sup nnnnaa L≤ ≤ ≤ℓ ⇒ 1LR≤≤ ℓ 11lim sup nnaR = ∵
If 1limn
n na
a+ →∞ exists then L=ℓ ∴ 111lim inf lim supnnnnaaR= = ∴ 11 1lim lim supnnnnn naaRa+ →∞ →∞ = =
Note : 1limn
n naRa →∞+ =
Theorem: Let nnaz∑be a power and 1nnna z−∑be the power series
obtained by differentiating nnaz∑ t e r m b y t er m . T he n t h e d e r i v e d
series has same radius of convergence as the origin al series .
(2009)
Proof : Suppose for R and R’ be the radii of the convergenve of the
series nnaz∑ and 1nnna z−∑ respectively.
Then we have, munotes.in
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11 111lim limnn nn naa n d n aRR==′
In order the desired result we have to show that 1lim 1n
nn
→∞=
Suppose 1nn=1+h . Then we have
n = (1+h)n=1+nh+2 (1 )..........2!nnnhh−+ >21(1 )2nn h− i . e . h2
<21n−
Thus ln122iiππ=+= so that 1lim 1n
nn
→∞=.
Hence R=R’
Proposition : Let ()
0nn
nfz az∞
== ∑ have radius of convergence 0R>
then for each 1K≥, the series
() ( ) ( )12 . . . 1nKn
nKnn n n K a z∞−=− −− + ∑ has radius of convergence R
. (2009)
Proof : Let R b e a r a d i u s o f c o n v e r g e n c e o f t h e p o w e r s e r i e s
()
0nn
nfz az∞
== ∑ .
Let R′ be a radius at cgs. of a power series ()!!nKnnK∞
= − ∑ .
T.P.T. RR′= ∴()11!lim sup!nn naRn k= ′ −
()11!lim sup . lim sup!nnnnank= −
()11! 1lim sup .!nnRn k R= ′ − … . ( 1 )
11lim sup nnaR =∵ munotes.in
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Now, 11lim sup limnnnnn naaa+ →∞ →∞ = ∴ ()()
()()11! !!lim sup lim!1 !n
nnn k n
nk n k n →∞+ − = × − + −
()1!lim
nnn
→∞ += ()()!.1nk
nk− + − () ! nk −!n 11 1lim lim1 1 1 nnn n
k nknn→∞ →∞+ += = + − + −
Substituting in equation (1), we get
111.RR= ′
RR′ =
Example: 1) Prove that the power series
0!nnzn∞
= ∑ converges for all
values of z.
Solution : Given power series is !nzn ∑ .
Here, ()111!1 !nnaann+ = ⇒ = + ∴()11!
lim lim1!1n
nn na nRan→∞ →∞ += = +()()1! 1 !lim lim! nnnn nn →∞ →∞++= =
!n lim 1
nn
→∞= + = ∞ ∴ R=∞ ∴
0!nzn∞ ∑ converges for all values of z.
2) Prove that the power series
0!nnz∞ ∑ converges only for 0z=.
Solution : Given the power series
0!nnz∞ ∑ ∴ ()1!1 !nnan a n+= ⇒ = + munotes.in
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∴ () 1!lim lim1!n
nn na nRan →∞ →∞ + = = +
!lim
nn
→∞= ()1!nn +11lim1 n n →∞ = = + ∞ 0=
∴ 0R= ∴
0!nnz∞ ∑ converges only for 0z=.
3) Find the radius of convergence of series ()()2
35nnzn+ +∑.
Solution : The given series is ()()2
35nnzn+ +∑. ∴ ()()1122335 3831 5nnn nnaann n+++++= ⇒ == + + + + ∴ ()()()() 123lim lim35 38n
nn nnn aRnn a →∞ →∞ + + += = + +
()
()()
()()()() ()8 213 23 8lim lim53 35 3 31 nnnn nnnnnn nnnn→∞ →∞ + + + + = = + + + + ()()() ()2
28 213
lim5331 nnnnn nnn
→∞ + + = + +
()()() ()() ()
() ( )8 213 10303533010 331n + + + + ∞== =+ + + + ∞∞ ∴1R=
4) Find the radius of convergence of the power seri es21()35nnzn+∑+.
Solution : The given series is 21()35nnzn+∑+ ∴ ()()121 121 2335 3 1 5 38nnn nnaann n+ + + + += ⇒ ==++ + + ∴ ()()()() 121 38lim lim .35 23n
nn nnn aRan n →∞ →∞ + ++= = + + ()()() ()2
28 123
lim5332 nnnnnnn→∞+ + = + + munotes.in
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()()() ()() ()
() ( )8 1232030533+0 2 032+ + + + ∞∞== + + + ∞∞1= ∴1R=
5) Find the radius of convergence of the series ()nnci d z +∑ where
,cd ∈.
Solution : The given series ()nnci dz+∑ ∴ () ()11nnnnac i d a c i d++ =+ ⇒ =+ ∴ ()
()11limn
n
nn nci d aRaci d+ →∞ + += = +
()
() ()1lim limn
nnnci dci dci d ci d→∞ →∞+= = + + +
By Rationalizing
22221lim
nci d
cdcd→∞−==
+ +
6) Find the radius of convergence of the power seri es 141nnz + ∑ .
Solution : The given power series is 141nnz + ∑ ∴ 111141 4 1nnnnaa++= ⇒ = + + ∴ 144
4lim141
4nnn n
nR
→∞+=+lim 4n→∞= 4=
7) Find the radius of convergence of the series 211nnzn +∑ .
Solution : The given series is 2
11nnzn+ ∑ . Let 21(1 )nnan=+ 11lim supnaR= 2
1lim sup 1n
n= + 1n1lim sup 1nn= + 1lim 1nnn→∞= +e= 1Re = munotes.in
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8) Find the domain of region of convergence of the power series
()
11.3.5.... 2 1 1
!nnnznz∞
= − − ∑ a n d s h o w t h e d o m a i n o r r e g i o n
graphically.
Solution : The given power series is ()
11.3.5.... 2 1 1
!nnnznz∞
= − − ∑ .
Put 1zz − =ξ, we get ⇒ ()11.3.5.... 2 1
!nnn
n∞
= − ξ∑
Now, ()1.3.5.... 2 1!nnan − = ⇒()()()11.3.5.... 2 1 2 11!nnnan+ − + =+ ∴1limn
n naRa →∞ += ()
() ()
()1.3.5.... 2 1
!lim1.3.5.... 2 1 2 1
1!nn
n
nn
n→∞ − = − + +()()() ()1.3.5.... 2 1 1 !lim1.3.5.... 2 1 2 1 !nnnnn n→∞ − += − +
()()()1! 1 !lim lim21 ! nnnn n
nn →∞ →∞++= = + ()21 !nn + 1lim lim21 nnnn
n →∞ →∞+= = +()11nn+
()12n +
11
lim12 nnn→∞+= + 1112 + ∞= + ∞
12R = ∴ D o m a i n o f c o n v e r g e n c e o f p o w e r s e r i e s 12 ξ <
i.e. 11 1122zzzz− < ⇒ − <
Taking square on both sides. 22114zz − < munotes.in
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∴ () ()1114zzz z− −< {}2.zz z =∵ ∴ ()41zzz z z z−− +<
44 4 4zzz z z z⇒ −− +<44 4 3 0zzz z⇒ −− +<()44033zz z z⇒ − ++ <
Put zx i y=+ zx i y⇒ =+
22 84033xy⇒ + −+<
2281 6 4 1 6039 9xx y ⇒ − ++ + − <3
2
2 44039xy ⇒ − + −< 24439xy ⇒ − += ∴ ()4,03C= and 23r=
Centre = 4,03 , Radius = 23
23r=
4,03c
Fig 3.1
Given series converges inside the circle.
9) Find the domain of convergence of the power seri es
()
01
34nnizi∞
= − +∑ .
Solution : The given power series is
0134nnizi∞
= − +∑()()2
0134n
niz iii∞
= + = = − +∑ ∵
()
034nnnizii∞
==+ + ∑ ∴ 34nniai=+ munotes.in
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∵11lim supnnaR= 1lim sup34nni
i= + lim sup34ii= + 34 34i iii = = + + ∴ 22zx i y z x y =+ ⇒ = +
00 1 1 1ii y i=+ ⇒ = + = = and
2234 3 4 91 6 2 5 5i + = + = + = = ∴ 1134 5i
Ri == + ∴
Fig 3.2 ∴ 5R = ∴ Domain of convergence of power series is 5zi + <
∴ Centre =(0,1) , Radius = 5 ∴ The given series converge inside the circle.
10) Find the radius of converges of the series () ( )32 2nnz + − ∑ .
Solution : The given series is () ( )32 2nnz+ − ∑ .
Put ()2z−=ξ
()32nn⇒ + ξ∑ ∴ ()132 3 1 2 35nnan a n n+=+ ⇒ =+ + = +
Now, ()() 12332lim lim lim5 35 3n
nn n nna nnRan nn→∞ →∞ →∞ + ++= = = + + 1=
∴1R = ∴ Domain of convergence of power series is 25z−<
i.e a circle with C=(2,0) and r =5
11) Find the region of convergence of power series
()
()13
1214nnnz
n−∞
= + + ∑ . (0,1)
munotes.in
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Solution : The given series is ()
()13
1214nnnz
n−∞
= + + ∑ ∴
() ()1 33111
14 24nnnnaa
nn++= ⇒ = + +
Now,
()1
311lim sup lim sup
14nnnaRn= = +1n
()31lim sup
1n n= + 4 14=
4R = ∴ T h e r e g i o n o f c o n v e r g e n c e o f t h e s e r i e s i s 2zR + <
i.e. 24z + < 221 6z⇒ + < ()222 1 6 .zz z z z⇒ + + < = ∵ ().2241 6zz z z⇒ +++< ()24 1 6zz z z⇒ ++ + <
Put zx i y zx i y=+ ⇒ =−
222. 21 2 0xy x⇒ ++ −< ()22.a n d 2zz x y z z x = + + =∵
2241 2 0xy x⇒ ++ −< 2244 1 2 4 0xx y⇒ ++ + −− <
()2 221 6 0xy⇒ ++ −< ()2 221 6xy⇒ ++ < ∴ (2 , 0 ) 4cr=− =
0-1-24
(-2,0)YX
Fig 3.3 ∴ The given series converges inside the circle.
(12) Find the radius of convergence of (i)()221( / 2)(! )nnz
n−∑ (2009)
(ii)2
2(1 )jjz
jj∞
=∑− (2009) (iii)nnnz∑ (2008) (iv)nzn∑ (2008) munotes.in
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(13) Find the power series for the function 1()fzz= about the point
z=2 and find its radius of convergence. (2007 )
14) Check for the convergence of the seri es
0nnnz∞
=∑ .
Solution : Here after comparison with
0,nn
naz∞
=∑
1nan n = ∀ ≥ ∵.
/g1178 11lim sup lim 1 1.nnnn n an L→∞ →∞== ∴=
/g1101 11.RL== /g1178 The series 0nznn ∞= ∑ converges for 1z< and
diverges for 1z>.
For 1,nzn z n= = →∞ , as n→α.
0nnnz∞
=∑ diverges for 1z=.
15) Find the radius of convergence of the following series .
(i)
1nnzn∞
=∑ (ii) 21nnzn∞
=∑ (iii)
1!nnzn∞
=∑ (iv)
12nnz∞
=∑
Solution : (i) /g1179 1
21lim sup 1.nnn naL an→∞= ⇒ = =
/g117811.RL== The radius of convergence of the power ser ies
12nnzn∞
=∑ is equal to 1.
(iv) /g1179 1na= if /g1179 2nk= for some integer k = 0 otherwise.
/g1178 Consider /g1179 {}1lim sup sup 1,0 1.nnn La→∞ = = =
/g1178 11.RL==The radius of convergence of the power series
02nnz∞
=∑ is equal to 1.
munotes.in
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3.5 SUMMARY
1) If the series
1nnz∞
= ∑ is convergent then lim 0nnz
→∞ =.
2) A series
0nnz∞
= ∑ o f c o m p l e x t e r m s i s c o n v e r g e n t i f f f o r e v e r y 0ε>, ∃ a n i n t e g e r N s.t. 1...nn n pzz z++ + + + < ε nN∨ ≥ a n d 0p≥. (Cauchy criteria of convergence of series)
3) The series nf∑ is said to be uniformly converges on D to f if
for every 0ε>, ∃ a n i n t e g e r N (depends only on ε) s.t. ()()nSz f z − <ε zD∨ ∈, and nN∨ ≥.
4) Let :fn D⊂→ b e a c o m p l e x f u n c t i o n d e f i n e d o n D s.t. ()nnfz M zD < ∨ ∈ and nN∈. If
1nM∞ ∑ is convergent series
of positive Real numbers then series
1nf∞ ∑ is uniformly convergent.
5) Let nu∑ b e a n i n f i n i t e s e r i e s f o r n o n - z e r o c o m p l e x t e r m s . t.
1limn
nnuLu+
→∞= then
i ) I f 1L<, the series converges absolutely.
i i ) I f 1L>, the series diverge.
i i i ) I f 1L=, the series may converges or diverge.
6) A power series a b o u t 0z i s a n i n f i n i t e s e r i e s o f t h e f o r m
2
00 0 1 0 2
0() () () . . . . .n
n
nazz a zza zz a∞
=− =+ − +− +∑ w h e r e c o n s t a n t s na
and 0z are called complex numbers and z is a complex variable .
7) The radius of convergence R of the power series
0nn
naz∞
=∑ i s
defined as {}sup : the series cgs satisfyingRr z z r= ∨ ≤.
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3.6 UNIT END EXCERCIES
1) Check for the convergence of the series 0nnz∞
=∑ .
Solution: If 1z< t h e n 1111. . .11n
n zzzzz+−++ + = →−− a s n→∞
011n
nzz∞
= ∴ = ∑−.
If 1z>, then limn
n z→∞=∞. ∴ The series 0nnz∞
=∑ diverges.
2) Show that the radius of convergence of the power series (1 )1nn
nz+∞
≥∑ is equal to 1.
Solution :
(1 ) 2 6 1 2
111..23nn
nzz z z+ ∞
≥=−+ − +∑ ∴0na= if (1 )nk k≠ +for some integer
(1 )nnan−= otherwise. ∴{}1lim sup sup 0,1 1. 1nnn La RL→∞1== = = =∵ ∴The series has the radius of convergence eq ual to 1.
3) Find whether 12kkiki∞
≥∑+ converges or not.
Solution : ∴ 2 211ki
ki k=
++ a n d w e k n o w t h a t 1211k
k∞
≥∑+
converges. ∴The series 121kkik∞
≥∑+ converges.
4) Check whether 11kki∞
=∑+ i s c o n v e r g e n t o r n o t . ( H i n t : C h e c k
whether 11Rekik∞
=+∑is convergent or not .)
5) Show that ()kfz k z∞
κ=1 = ∑ is continuous in 1z<.( Hint :
Here the convergence is uniform . Show that 1kkkz∞
=∑ i s
convergent in 1z<. Let ()kkfz k z= w h i c h i s u n i f o r m l y
continuous for all 1k≥)
6) Show that the series 0!nnzn∞
=∑ is convergent everywhere in
the complex plane.
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7) Show that the functions () c o s () , () s i n ()fz z g z z = =are analytic
in the whole complex
plane. (Hint: Show that each of the series
0(1 )() c o s ()(2 )1nz nnzfz zn∞
=−== ∑ a n d 210(1 )() s i n ()(2 1)!nnnzgz zn+∞
=−== ∑+
have infinite radius of convergence. )
8) Let ()nabe a sequence of positive real’s and
1limn
n
naLa+
→∞=. Show that
1
limnnnaL→∞= .
9) Find the radius of convergence of the s eries 210(2 1)n
nz
n+∞
=∑+ .
Solution: 21 3 5
0... ...(2 1)! 3! 5!n
nzz zzn+
∞
==+ + ++∑+ ∴Comparing with 0nnnaz∞
=∑ , we get 0na= if 2, 4, 6, ...n=
1if 3, 5, 7 ...!nn= =
{}1lim sup sup 0,0 0nnn La→∞∴ = = =.
10R∴ = = ∞. The power series 210(2 1)n
nz
n+∞
=∑+has infinite radius of
convergence.
10) Find the domain of convergence of the series
0(1 )(1 )!nnn zn∞
=−+ ∑ .
Solution: Put 1zθ=+.
00(1 ) (1 ) (1 )(1 ) .!! !nn nnn
nn n zann n∞∞
==−− −∴ += θ = ∑∑ .
1 1lim lim 01n
nn
naLan+
→∞ →∞∴ = = =+
10R∴== ∞. The given power series converges for all θ.
But 1zθ=+. As θ varies over , ∴
also varies over . ∴The given power series converges for all co mplex numbers.
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11) Find the domain of convergence of the p ower series
0()nnzi∞
= − ∑
(Ans : The series converges for z∈ such that 1zi−<.
12) Show that 11nn→ as n→∞. ( Hint: Put 1lognnan=
for 1n≥.)
13) If nnaz∑ h a s r a d i u s o f c o n v e r g e n c e R, what is the radius of
convergence of
2nnaz∑ and of 2?nnaz∑
14) Prove that the series ()()131214nnnz
n−∞
=+
∑
+ converges for 24z+≤ .
15) Find the radius of convergence of the power ser ies
2
0,nn
naz a ∞=∈∑C.
16) Find the domain of convergence of the series 23 11 . 3 1 . 3 . 5..............22 . 5 2 . 5 . 8zz z + + +
17) Find the radii of convergence of the following power series
211nzni + + ∑
18) Show that the ROC for the power series nzn∑
is 1. Discuss the
convergence of this series of the points on the bou ndary or the
disc {/ 1 }zz∈ <
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4
DIFFRENTIABILITY
Unit Structure
4.0. Objectives
4.1. Introduction
4.2 . Differentiability in complex
4.3. Summary
4.4. Unit End Exercises
4.0. OBJECTIVES
After going through this chapter you shall com e to know about :
• Defining a polynomial with complex coefficient s and in an
indeterminant z, which can take any complex number value.
• An infinite series of the form 0nn naz∞=∑ is called a power
series.
We shall investigate for the differentiability of a power series as
a function of a complex variable
, at the same time we shall
also check for the condition , under which t wo power series
are one and the same , that is both the pow er series represent
the same complex valued function.
4.1. INTRODUCTION
Through this Unit , we shall examine the notion of “ a
function of
“, where
is a Complex Number of the form .zx i y=+
A Complex Number z can be viewed as an ordered pair of real
numbers x and y as (, )zx y = . The point of view taken in this Unit
is to understand some functions , which are dire ct functions of zx i y=+ and not simply functions of the separate part s x and y
Consider for example the function 222xy i x y−+ i s a d i r e c t
function of xi y+, since 22 22() .xy i x yx i y−+ = + munotes.in
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()2()fz z= but 222xy i x y+ − is not expressible as a polynomial
in variable xi y+. Therefore we are compelled to consider a
special class of functions , given by direct/ ana lytic expressions in xi y+. We shall name such direct functions as the a n a l y t i c
functions . Let us start this Unit by definin g an c polynomial ()pz in a Complex variable z.
Definition : A polynomial P(x,y) in a Complex variable z =(x,
y) is an expression of the form (, )Pxy = 01() . . .()nnxi y xi yα + α + + + α + w h e r e 01,. . ,nα α α a r e
complex constants e.g. 22() ( , ) 2iP x y x y i x y = −+ 22() 2ii x y ixy + − is
not a polynomial in zxi y = + .
4.2 DIFFERENTIABILITY IN COMPLEX
Differentiation: Let G be an open set in and :fG→ can be a
function, we say that f is differentiable at a point 0z in G if the limit ()()00 0lim
zzfz fzzz → − −……(1)
exists, this limit is denoted by ()0fz′ a n d i s c a l led d e r i v a t iv e o f f
at 0z.
Put0zz h=+, (complex number) then equation (1) becomes
()()()000
0lim
zzfz h fzfzh → + − ′ ⇒ = .
In terms of ‘ε−δ’ notation limit in equation (1) exists iff 0, 0,∨ ε> ∃ δ> ∋()()()0
0
0fz fzfzzz − ′ − < ε− whenever0 0zz< − <δ.
If f is differentiable at each point of G then f is differentiable on G.
Notice that if f is differentiable on G, ()0fz′ d e f i n e s a f u n c t i o n
:fG′ →.
If f′ continuous then we say that f is continuously differentiable . If f′ i s d i f f e r e n t i a b l e , t h e n f is twice differentiable continuing, a
different function ∋ each successive derivative is differentiable is
called infinitely differentiable .
Proposition: If:fG→ is differentiable at 0zG∈ then f is
continuous at0z (2012)
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Proof : Given :fG→ is differentiable at 0zG∈. ∴ ()()()000 0lim
zzfz fzfzzz → − ′ = − exists. ∴ () ()()()00 00 00lim lim .
zz zzfz fzfz fz z zzz →→ − − = − − ()0.0fz′= 0= ∴ ()()00lim
zzfz fz→ = ∴ f is continuous at 0z.
Theorem : If f and g are differentiable at 0zG∈ t h e n (),. , , 0fg f g f gg± ≠ are also differentiable at 0zG∈.
The Increment Theorem: Let :fG→ b e a c o m p l e x v a l u e d
function 0zG∈ and 0r>, ()0,Bz r G∃ ⊂. Then f is differentiable
at 0z i f f ∃ a c o m p l e x n u m b e r α a n d a f u n c t i o n
:( ; ) ( , , )Bos osr s u c ht h a tvη→ (;)hB o s∈, ()()()00fz h fz h h h+= + α+ η and ()0lim 0hh
→ η=.
Proof : Let f is differentiable at point 0z.
(Let) put ()()()00fz h fzhh + − η= − α
So that, ()()()00fz h fz h h h+= + α+ η
Let ()0fz′=α ( * ) ∵ f i s d i f f e r e n t i a b l e a t 0z. ∴ ()()()00
00lim lim
hhfz h fzhh →→ + − η= −α
()fz′= −α=α−α f r o m ( * )
= 0
Conversely,
Let ()0lim 0hh
→ η= and (**)
()()()00fz h fz h h h+= + α+ η ∴ ()()()00fz h fzhh − − = α+η
Taking limn→∞ on both sides. munotes.in
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()()[]00
00lim lim 0hhfz h fz
→→ −− = α+η ∴ ()()00
0lim 0hfz h fz
n → + − = α+
f r o m ( * * )
()n
ndwfz a zdz= → ∑ ∴ f is differentiable at point 0z and ()0'fz=α.
Composite Function : Let G∈ a n d Ω∈ b e o p e n s e t s . L e t
:fG→ and :gΩ→ be functions ()fG∋ ⊂Ω. Then for each ZG∈, the association gf d e f i n e d b y []()()gf z gf z= i s a
function called of composite function .
Note : In generalfg gf≠
Chain Rule :
Theorem: Let , G∈ Ω∈ b e o p e n s e t s a n d l e t f a n d g b e
differentiable on G and Ω ( r e s p e c t i v e l y ) . S u p p o s e ()fG⊂Ω t h e n gf is differentiable on G and ()()()().gf z gf z fz zG′′= ∨ ∈
Proof : Fix a point zG∈, choose ()0,rB z r G> ∋ ⊂.
Let 0h≠∈ and ()hr z h < ≠
Given that f is differentiable on G. ∴ f i s d i f f e r e n t i a b l e a t a p o i n t ZG∈. ∴ B y i n c r e m e n t t h e o r e m , ()()()()fz h fz h f z h h′ +− =+ η w h e r e ()hη i s c o n t i n u o u s f u n c t i o n
and ()0lim 0hh
→ η=
Put ()()Kf z h f z=+ − , where ()()Kh fz h h′=+ η
Also g is differentiable at ()fz∈Ω ∴by increment them,
()()()gfzh gfz K+= + ()()()gfz K g fz K K′=+ + ψ
where ()Kψ is continuous function and ()0lim 0kK
→ ψ =. ∴()()()()()().'gfzh gfz h f z h h g fz K′ += + + η + ψ
()()()()()() ( ) () ( )..
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()()()()gfz h g fz f z h h′′ =+ + δ
Where, ()()()()()()()hh g f z f z Kh K′′ δ=η + ψ +ηψ
T.P.T.
0lim ) 0.hh
→ δ( = ∵()0hη→ as 0h→
As ()()0, 0hK f z h f z→ = + −→ ∴()0Kψ→ as 0h→
Hence, ()lim 0nh
→∞ δ= ∴by increment theorem, ()gf is differentiable at zG∈ ∵Z was arbitrary . ∴gf i s d i f f e r e n t i a b l e o n G and ()()()()gf z gf z fz zG′′= ∨ ∈
Let zx i y G=+ ∈ a n d :fG→ b e d e fi n e d by , ()()()fz u z i v z=+ ,
where u and v are real valued function
OR ()()(),,,fx y u x y i v x y = + .
Definition : If ()()0,,lim
hux hy uxyh → + − e x i s t s t h e n i t i s c a l l e d
partial derivative of u w.r.t x as the point (),xy and is denoted by
(),uxyx ∂ ∂ or (),xux y .
Theorem : Let ()
0nn
nfz az∞
== ∑ have the radius of convergence 0R> then
1) The function f is infinitely differentiable on ()0;Br and
()() ( 1 ) ( 2 ) . . . ( 1 )k nkn
nkfz n n n n ka z∞−= = −− − + ∑ for zR < and 1k∀≥.
2) If 0n≥ then ()(0)!n
nfan=
Proof 1) For zR <, we will write () () ()
0n
nn n
nfz az S z R z∞
== = +∑
where ()
0nnn
nSz a z∞
== ∑ and ()
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KnRz az∞
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Put ()11
1nn
ngz n a z∞−−
==∑()limnnSz→∞′ =
Fix a point 0z in ()0;BR
(Choose 0 0,rz r R> ∋ < < and zr R < < ()0zz≠
We will prove that ()()00fz g z′=
Let 0δ> be arbitary ()()0;0 ,Bz B r∋ δ⊂
Let ()0,zB z∈ δ then []00 00 00 0() () ( ) ( ) () ( )() ()nn n nSz Rz Sz Rz fz fzgz gzzz zz + − + −− = −−−
()()() ()()()0 000
0 0nn nn
nnSz Sz Rz RzSz Szzz zz − − ′′ = − ++−−
Taking modulus on both the sides. ∴()()()()()() () ()0000 0 000nn
nnfz fz S z S zgz S z S z gzzz zz − − ′ − ≤ − + − −− ()()0
0nnRz Rz
zz − + − (1)
Let 0ε>, be given
Now, ()()()0000 11 nnKKK
KnRz Rzaz zzz zz∞
=+ − = −− − ∑
()12 2 100 0 0
0 11...KK K K
K
Knaz z z z z z z zzz∞−− − − =+= − + + + + −∑ ()()012 2 100 0
0 1...nn KK K K
K
KnRz Rzaz z z z z zzz∞−− − −=+ − ≤ + + + + −∑11 11...KK K
K
Knar r r∞−− −=+≤ + + + ∑ 11.KK
KnaK r∞−=+= ∑ ∵The derived series 11nn
nna z∞−= ∑ is convergent at zr=. ∴The power series 11.KK
KaK r∞−
= ∑ converges for rR<. munotes.in
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∴For the above 0,ε> ∃ a n i n t e g e r 1N ∋
111..3K
K
KnaK r n N∞−
=+ε < ∨ ≥∑ . (by Cauchy criteria)
Thus, ()()0
03nnRz Rz
zz − ε <− 1nN∨ ≥ ( 2 ) ∵()()00limnnSz g z→∞′ =
For the above 0ε>, ∃ an integer 2N s.t. ()()003nSz g zε′ − < ( 3 )
Choose {}12 max , .NN N f o r n = ≥ Ν ∴ F o r t h i s n, we can find 0δ> s.t.
()()()0
0
03nn
nSz SzSzzz − ε ′ − <− ( 4 )
whenever 0 0 zz< − <δ.
From equation (1), (2), (3), (4) we get
()()()0
0
0333fz fzgzzz − ε ε ε − < + +−
=ε whenever 0 0 zz< − <δ ⇒ f is differentiable at ()00,zBR∈ ∵ z is arbitary. ∴ f is differentiable on ()0;BR
A repeated application of this argument shows that the heigher
derivatives (),, . . . , . . .Kff f′′ ′ exists, so that
()() ( ) ( ) 1. . . 1KnKn
nKfz n n n K a z∞−== − −+ ∑ e x i s t s f o r zR < a n d 1K∨ ≥. ⇒ f i s i n f i n i t e l y d i f f e r e n t i a b l e o n ()0;BR .
2) Since
()() ( ) ( ) 1. . . 1KnKn
nKfz n n n K a z∞−== − −+ ∑()!
!nKn
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()1!!!nKkn
KnnKa a znK∞−=+= + − ∑
Put 0z= ∴()()0!0K
K fK a= +
()()0!K
kfaK =
Replace K by n ∴ ()()0!n
nfan =
Corollary : If the power series
0nn
naz∞
= ∑ has radius of convergence 0R>, then ()nnfz a z=∑ is analytic on ()0;BR .
Theorem : If G is an open connected set and :fG→ i s
differentiable with ()0fz zG′= ∨ ∈, then f is constant.
Proof : Fix a point 0zG∈ and let ()00wf z= .
Let (){}0;Az G f zw=∈ =
T.P.T. AG=.
[i.e. by showing A is both open and closed and A≠∅]
T.P.T. A is closed.
Let zG∈ and {}nz be a sequence in limnnAz z→∞ ∋ = ∴ ()0nfz w=, for each n∈
f is differentiable on G (given) ∴ f i s c o n t i n u o u s o n G. ∴ () ()0lim limnnnnfz f z fz w→∞ →∞= = = zA⇒ ∈ ∴ A c o n t a i n i t s l i m i t p o i n t . ⇒ A i s c l o s e d .
Now, T.P.T. A is open.
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Let ();zB a r∈ and set ()()1gt f t z ta=+ −, 01t≤≤ ∴()()()()() ()()()()11z ft z t a fS Sagt gs tS z St ats tS z St a tS + − ++ − − −+− = × −− +−− ∴() ( )()()() ()()11lim lim
tS tSft z t a fs z S a gt gSzatS tS z St a →→ + −− +− − = ×− −− +− ∴ []'' ( 1 ) ( )gs f s z sa z a=+ −× − ∴ '( ) 0gs=, 01s≤≤()()0,fz zG′ = ∨ ∈∵ ()constant, 0 1gs S⇒ = ≤≤ ⇒ ()()1c o n s t a n t 0gg == ∴ ()()()1c o n s t a n t 0fz g g == = ()0fa w== ⇒ zA∈ ∴ ();zB a R zA∈ ⇒∈ (;)Barc A⇒ ⇒ A i s o p e n a n d ()Az A≠∅ ∈∵
Hence, by the connectedness of G
AG= ∴ f is constant on G.
4.3 SUMMARY
1) If :fG→ i s d i f f e r e n t i a b l e a t a p o i n t 0z i n G, then f is
continuous at 0z.
2)The Increment Theorem : Let :fG→ b e a c o m p l e x v a l u e d
function 0zG∈ and 0r>, ()0,Bz r G∃ ⊂. Then f is differentiable
at 0z i f f ∃ a c o m p l e x n u m b e r α a n d a f u n c t i o n ()()():0 ; 0 , , 0 ;Bs S r h BSη → ∋ ∨ ∈ ()()()00fz h fz h h h+= + α+ η and ()0lim 0hh
→ η=.
3) Chain Rule : Let , G∈ Ω∈ b e o p e n s e t s a n d le t f a n d g b e
differentiable on G and Ω (respectively). Suppose ()fG⊂Ω then gf i s d i f f e r e n t i a b l e o n G and ()()()().gf z gf z fz zG′′= ∨ ∈
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4) If ()()0,,lim
hux hy uxyh → + − e x i s t s t h e n i t i s c a l l e d p a r t i a l
derivative of u w.r.t x as the point (),xy a n d i s d e n o t e d b y
(),uxyx ∂ ∂ or (),xux y .
5) Let ()
0nn
nfz az∞
== ∑ has the radius of convergence 0R> then
i) The function f is infinitely differentiable on ()0;Br and
()() ( 1 ) ( 2 ) . . . ( 1 )k nkn
nkfz n n n n ka z∞−= = −− − + ∑ for zR < and 1K∨ ≥.
ii) If 0n≥ then ()()0!n
nfGn=
6) If G is an open connected set and :fG→ is differentiable with ()0fz zG′= ∨ ∈, then f is constant.
4.4. UNIT END EXERCISES
1) Check for the differentiability of the p ower series
0 ()!nnzfzn∞
= =∑ .
Solution: We know that the series 1!nnzn∞
=∑ converges for all
complex numbers.
/g1178 ()fz′ exist for all z∈ and
11
11 1() ()!( 1 ) ! !nn n
nn nnz z zfz f znn n−− ∞∞ ∞
== =′ = = = = ∑∑ ∑−. () ()fz f z′ = for all z∈.
2) If the series 0()nn naza∞
= − ∑ has the radius of convergence 0R>, then show that 0() ( )nn nfz a z a∞
= = − ∑ is analytic in (; )BaR.
( Hint : Use the fact that f i s i n f i n i t e l y d i f f e r e n t i a b l e o n (; )BaR and 1() 1!n
naf a nn= ∀≥.)
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5
COMPLEX LOGARITHM
Unit Structure
5.0. Objectives
5.1. Introduction
5.2. Logarithmic function
5.3. Branches of Logarithmic Function
5.4. Properties of Logarithmic Function
5.5 Trigonometric and hyperbolic functions
5.6. Summary
5.7. Unit End Exercises
5.0. OBJECTIVES
We are already familiar with a logarithm fu nction,
defined for positive real x. In the same manner one can define a
complex logarithm Log (z) of a complex number z∈. We shall
study the branches of this complex logarithm f unction . The
complex logarithm Log (z) posseses some branches , which we
shall try to investigate. We shall also study the properties of a
complex logarithm in detail.
5.1. INTRODUCTION
With the help of order completeness property of, we
proved in our earlier course that if 0y> a n d 2n≥ i s a n y
integer, then there is a unique positive nu mber x such that
nxy=. x is called thn root of y, since there is a unique positive
number x satisfying this, defining 1nyis justified. We proved that,
for 1a>and x∈, .,xy x yaa a x y+= ∀ ∈and ()yxx yaa= . ():0 ,f→∞defined by ()xfx a= is a bijective function and
it’s inverse is called as the logarithm of y t o t h e b a s e a,
denoted by log ( )ay. We want to discuss these concepts once munotes.in
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again but we consider the logarithm of c omplex numbers
with base e, hence we try to identify the nature o f inverse
of the exponential function of complex vari able z, namely
()zfz e= on some domain D⊆. Here we shall start defining log ( )z for z∈.
5.2 LOGARITHMIC FUNCTION
Definition : For 0z≠, the logarithmic function of a complex
variable z, denoted by logz , is defined as ()log ln arg 2zz i z n = + + π w h e r e []arg ,zθ= ∈− π π o r []0, 2 π
and n∈.
Here, logz is a single valued function.
5.3 BRANCHES OF LOGARITHMIC FUNCTION
Definition : If 0G∉ is an open connected set in and :fG→ is
a continuous function such that(),fzez z G= ∨ ∈, then f is branch
of logarithm.
Theorem: A branch of the logarithm is analytic and its deriv ative is 1z .
Proof: Let ()log ln argfz z z i z= = + b e a b r a n c h o f l o g a r i t h m ,
where 0z≠, ()arg [ , ]zππ=− .
Let ()()()fz u z i v z=+ and zx i y=+ ∴ ()22 1ln tanyui v x y ix− += + + w h e r e 22zx y = + a n d ()1tanyx− θ= ∴ ()22,l nuxy x y = + and ()1tanyvx− =
()22,l nuxy x y = + ∴()22 222,
2ux xxyxxy xy ∂ = =∂ + + and ()22,uyxyyxy ∂ =∂ + ∴ ()1tanyvx− = munotes.in
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∴
()22222
21
1vy y
x x yx yxxx ∂ − − = × = ∂ + + 22yxy− = + a n d
22vxyxy ∂ − =∂ +
Therefore C-R equations are satisfied
() ()logfz zzz∂∂= ∂ ∂ uvixx ∂ ∂ =+ ∂∂22 22xyi
xy xy− =+ + + 21.zzzzz z== =
Theorem : Let 0G∉ b e a n o p e n co n n e c t e d s e t in a n d s u p p o se
that :fG→ is analytic. Then f is a branch of logarithm if f
()1,fzz′= zG∨ ∈ and ()faea= for atleast one aG∈.
Proof : Suppose f is a branch of a logarithm. ∴ ()fzez z G= ∨ ∈ (1)
Differentiate w.r.t. to z on both sides. ∴ ()().1fzef z ′ = ∴()1()fzfze′=
∴ ()1fz zGz′= ∨ ∈
Clearly, from equation (1), ()faea= for atleast one aG∈.
Conversely,
Suppose ()1fz zGz′= ∨ ∈ and ()faea= for atleast one aG∈.
T.P.T. f is a branch of the logarithm.
Define, ()().fzgz ze− = ( 2 ) ∵g is analytic. ∴g is differentiable. ∴()()fzdgz z edz− ′= () ().fz fzdz dez edz dz− − =+ ()()().fz fzef z z e− − ′ =+ −
() ()()11.fz fzez e f zzz− − ′ = − × = ∵
()()fz fzee− − = − ()0gz′= zG∨ ∈ munotes.in
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()gz⇒ = constant = K (Say) (3)
To find K, put za= in equation (2) and (3) ∴ ()().faga ae− = and ()ga K=. ⇒ () ()11..fa faKa e a eaa− − = = = ∵
1= ⇒ 1K =
Put 1K= in equation (3), we get ()1gz=
Put ()1gz= in equation (2), we get ()1.fzze− = ()fzezz G= ∨ ∈ ⇒ f is branch of logarithm (by definition)
A single valued function (is branch of logarithm) []()log ln arg 0 and arg ,zz i z z z = + ≠ θ= ∈− π π i s c o n t i n u o u s
function in the region or a Domain {};0 ,0Dz x i y y x= =+ ∈ = ≤ []()log ln arg 0 and arg ,zz i z z z = + ≠ θ= ∈− π π ∴logz is not defined at the point 0z=.
Theorem : Prove that logz is not continuous on the negative real
axis.
Proof : Let 000zx=< be any point on the negative real axis.
For zx i y=+ with 0, 0xy< <,
we have,
000lim arg lim arg( )
zz xx
yzx i yπ→→
>=+ = = π
For zx i y=+ with 0x< and 0y<,
We have
000lim arg lim arg( )
zz xx
yzx i yπ→→
<=+ = − ∴ T w o l i m i t s o b t a i n e d a r e d i f f e r e n t .
i.e. argz fails to possess a limit every point of the negati ve real axis. ∴ logz is not continuous along the negative real axis.
Theorem: Let 0G∉ be an open connected set in . If a branch of
the logarithm :f→ i s r e l a t e d b y ()()2gz f z i n=+ π [ f o r s o m e
integer n∈] with :gG→ then g is branch of logarithm.(2008)
Proof : Given that f is a branch of the logarithm. ∴ ()fzez z G= ∨ ∈ munotes.in
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Given, ()()2gz ez i n=+ π ( f o r s o m e intn ∈) ∴ ()()2gz f z i nee+π = ()2.fzineeπ =
()gzez z G= ∨ ∈ (){}2and 1fz inez eπ = =∵ ∴ g is a branch of logarithm.
5.4 PROPERTIES OF LOGARITHM FUNCTION
Theorem:
1) ()12 1 2 log . log log 2zz z z i n = + + π , where 1, 0οr1n= − by
definition.
Proof : log ln argzz i z = + where 0z≠ and []arg ,zθ= ∈− π π. ∴()()()12 12 12log . ln . arg .zz zz i zz = + []12 1 2ln ln arg arg 2zz i z z i n= + + + + π
(∵ ()12 1 2 arg . arg arg 2zz z z i n = + + π ) ()()112 2ln arg ln arg 2ziz z iz i n= + + + + π 12log log 2zz i n= + + π
2) 112
2log log log 2zzzi nz = − + π
3) 1log logzz=−
Proof:11 1log ln arg izz z = +
1ln arg arg argzi z zz= − + = ∵ ln argzi z= − − ln argzi z= − + logz= −
Evaluate :-
1) logi ∵ 0zx i yi x=+ = ⇒ = and 1y= ∴ log ln argzz i z = + (by definition) ∴ log ln argii i i = + ln122iiππ=+=
2) ()log 1i − ∵ 11zix=− ⇒ = and 1y=− ,
2211 2zx y = + = + = munotes.in
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∴ ()()log 1 ln 1 arg 1ii i i−= −+ − ()()11ln 2 tan ii− − = +
()()1ln 2 tan 1i− = − ()ln 24i π = −
3) ()log 1i + ∴ 11 , 1 2zi xy z=+ ⇒ = = ⇒ =
∴ ()log 1 ln 1 arg 1ii i i+= + + + ()()11ln 2 tan1i− = + ()()1ln 2 tan 1i− = + ()ln 24i π = +
4) In unit disk (){}0,1 : 1Bz z = ∈ < p r o v e t h a t p o w e r s e r i e s
11log1n
nznz∞
= = − ∑ where 1log1z − is a branch of the logarithm
1log1z − .
Solution : Let ()
1nnzfzn∞
= =∑ ( 1 )
and ()1log1gzz= − ( 2 ) ⇒ g is differentiable.
()
()()()2
211 1log 11 1 11dgzdz z zz − ′== − − − −()()21
1zz − = −
()11gzz′= − ( 3 )
Given, the power series ()
1nnzfzn∞
= =∑
Here, 111,1nnaann+ = = + ∴ ()
11lim lim 1 limn
nn n nnaRnan →∞ →∞ →∞ +
= = ×+ = ()11nn+
110= = ∞ ∵ ⇒ f is analytic in open disk ()0,1B (using corollary) ⇒ f is differentiable in ()0;1B ∴ ()1111.nn
nnnzfz zn∞∞ − − == ′==∑∑
0nnz∞
==∑ munotes.in
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∴ ()11fzz′= − ( 4 )
0 01is a G.S. and cgs to 1,1n n
n nzz Gz∞ ∞
= = < = − ∑∑∵∵
From equation (3) and (4), we get
() () () ()100fz g z f z g z′′− = ⇒ − = ⇒ ()()fz g z− = constant = K ( 5 )
To find k, put 0z= in equation (1), (2) and (3)
()()00 , 00fg= = ∴ ()()00fgK− = 0K⇒ =
Put 0K= in equation (3), we get
()()0fz g z− = ∴ ()()fz g z= ∴11log1n
nznz∞
= = − ∑ for all ()0;1zB∈
Definition : Given 0∉, the principal value o f bz ( i . e . t h e bth
power of z) is defined by .l o g,bb zze b = ∈
Here, bz is analytic. (∵ logz is analytic)
Consider .l og,bbzze b = ∈
Here, bz is multivalued function. ∵ argz (and hence logz ) is a multiple valued function.
Case I : If b i s a n i n t e g e r t h e n .l ogbb zze = i s a s i n g l e v a l u e d
function.
Proof : Let bK=∈. ∴ ()ln arg 2 .l ogkz iz nbb zze e + + π ==
() ln arg 2.zi zKnkiee + +π =
()logkzei = ()()21,iK nenπ = ∈∵
logbze =
.l o gbbzze = is a single valued function.
Case II : If pbq= (real rational) then bz has produces exactly q
values. munotes.in
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Case III : If b is an irrational number or imaginary number then bz
is infinite valued function.
Example 1 : Find the principal value of ii.
Solution : .l o g,bb zze b = ∈ (by definition) ∴ .l ogiiiie =ln argii i ie + =2 2 0.2ii ieeπ+π ==
2iieπ− =
Example 2 : Find all the values of 2ii− .
Solution : .l og bb zze b = ∈ ( b y d e f i n i t i o n ) ∴ 22 . l o gii iie− − =()2l n a r g 2ii ii ne− + + π=20 22ii n
e π − + + π=
2
e−
=4
2nii π+π ()214ine− π +=()()11 4 nen−− π+= ∈
()41neπ +=
Here, the principal value of 2ii− is eπ .(All values are not found.)
Example 3 : Find the value of 2i.
Solution :22 l o giie =()2l n a r g 2ii i ne + + π =()2l n 1 a r g 2iine+ + π =
2014 2.nieeπ+ π = ieπ= ()41,nienπ = ∈∵ cos sini= π+ π10=− +
21i = −
Example 4 : Find all the values of ()()11ii + + .
5.5 TRIGNOMETRIC AND HYPERBOLIC FUNCTIONS
Trigonometric Function : The Complex trigonometric functions
sin and cos are defined by
sin2!iz izeez− − = a n d cos2zi zeez−− + = (2008) munotes.in
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Similarly,
()2tan , seciz iziz iziz izeezz
ee ie e− − − − = = + +
()2cosec , cotiz iziz iz iz izie eizzee ee−
− − + = = − −
Note :
1) 22sin cos 1zz+=
2) ()sin cosdzzdz = a n d ()cos sindzzdz = −
3) ()sin sinzz− = − a n d ()cos coszz− =
4) ()sin sin .cos cos .sinzw z w z w+= +
5) ()cos cos . cos sin .sinzw z w z w + = −
6) sin 2 2 sin .coszz z =
Hyperbolic Function : The complex hyperbolic functions sinh and cosh are defined by
sin2zzeehz− − = and cos2zzeeh− + = .
Similarly, tan , cotzz zzzz zzee eehz hzee ee− − − − − + = =+ −
22sec , coseczz zzhz hzee ee− − = = + −
Note :
1) 22cos sin 1zz− =
2) ()sin cosdhz hzdz =
3) ()cos sindhz hzdz =
4) ()sin sin .cos cos .sinhz w h z h w h w h z + = +
5) ()cos cos . cos sin .sinhz w h z h w h w h z + = +
Relation between Trigonometric and Hyperbolic Funct ion :
1) sin siniz i hz =
Proof : ()()sin2ii z ii zeeizi− − =2!zzee− − =()22zzie ei− −
= ()2zzie e− − = sinih z=
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2) cos cosiz hz =
Proof : cos( )2iz izeez−+ = 2zzee− + =coshz=
3) tan taniz i hz =
4) cos coshiz z =
5) sin sinhiz i z =
6) tan tanhiz i z =
Periodic Function : A function :fG→ is said to be periodic if ∃
a non-zero complex number ()()Tf z T f z z G ∋ + = ∨ ∈.
Here T is a period of the function n.
Periodicity of ze:
Let T be the period of ze ∴ += ∨ ∈zT zee z .
To find T, put 0z= ∴ 021Ti nee eπ == =
Let Ti=α+β ∴ 2ii neeα+βπ = 1eα= and 2ii neeβπ = 0⇒ α= and 2nβ=π ∴ 02Ti i n=α+β=+ π is a period of ze.
OR log1 0T== ( i s n o t p o s s i b l e b y d e f i n i t i o n )
Let Ti=α+β ∴ 1ieα+β= .cos 1eα β= and sin 0β= 0eα⇒ = and 2nβ=π 02 2Ti i n i n⇒ = α+β=+ π= π
2in π is a period of ze ∴ 2zn i zee+ π=
Periodicity of sinz :
Let T be the period of sinz . ∴ ()sin sinzT z+=
Put 0z= munotes.in
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∴ ()sin 0 sin 0T+= sin 0T⇒ = Tn⇒ = π, where 0, 1, 2, ...n= ± ± ∴ () ( )sin 2 sin .cos 2 cos .sin 2 1 sinnzz z z+π= π+ π=− sinz= if n is even. ∴ The period of sinz is 2nπ where n∈.
Periodicity of cosz :
Let T be the period of cos z. ∴ ()cos cos + = zT z
Put 0z= ∴ ()cos 0 cos 0T + = ⇒ cos 1T= ⇒ 2Tn n=π ∨ ∈ ∴ The period of cosz is 2nπ, where n∈.
5.6 SUMMARY
1) For 0z≠, the logarithmic function of a complex variable z,
denoted by logz , is defined as ()log ln arg 2zz i z n = + + π
where []arg ,zθ= ∈− π π or []0, 2 π and n∈.
2) If 0G∉ i s a n o p e n c o n n e c t e d s e t i n a n d :fG→ i s a
continuous function ∋ (),fzez z G= ∨ ∈, then f i s a b r a n c h o f
the logarithm.
3) Given 0c∉, the principal value of bz (i.e. the bth power of z) is
defined by
.l o g,bb zze b = ∈
4) The Hyperbolic functions sinh and cosh are defined by
sin2zzeehz− − = and cos2zzeeh− + = .
5.7 UNIT END EXERCISES
1) Suppose that :fG→ is a branch of the logarithm and nis
any integer . Prove that exp( ( ))nzn f z= for all zG∈. munotes.in
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Solution: Since :fG→ is a branch of the logarithm ( G⊆
is an open connected set .) exp( ( ))zf z∴=for all zG∈.
2exp( ( ).exp( ( )) exp( ( ) ( )) exp(2 ( ))zf z f z f z f z f z∴== + = . exp( ( ))nzn f z∴= for all zG∈. (By induction on power of z.
2) Describe the branches of an analytic function ()fz=√z .
Solution:1() e x p l o g ()2fz z z∴ ==,since
()211 1exp log( ) exp log( ) log( ) exp log( )22 2zz z z z =+ = = . ∴This defines
a n d i t i s a n a l y t i c , w h e r e t h e log( )zis
analytic.
∴Different branches of log( )zyield different branches of z. log( )zhas infinitely many different branches log( ) 2zk iπ+for any
integer kbut there are only two different branches of z.
Since ()11exp log( ) exp log 222zz k iπ =+ whenever k is an even
integer .
3) Find all values of the complex number ii.
Solution: log( ) log( ) arg( ) arg( )ii i i i i iie e e+ −∴ = = = .
Here we know that
35 7arg( ) ..., , , , ,... 2 :222 2 2in nππ π π ππ = − =+ ∈ .
35
222..., , , ,...iie e eππ π−−∴=.
4) Find all values of ()11ii++ (Hint : ()()()11l o g 11iiiie++++= ).
5) Let :fG→ a n d :gG→ b e b r a n c h e s o f az a n d bz
respectively. Show that fg is a branch of abz+ a n d fg is a
branch of abz−.
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Solution: ∴()afz z= and ()bgz z= for all zG∈. ∴:fg G→ d e f i n e d b y () () () .() .ab a bfg z f z g z z z z+== =
() ()abfg z z+= for all zG∈ . ∴abz+ has a branch fg on G.
Similarly, fg is a branch of abz−.
6) Let 12,, . . . ,nzz z be complex numbers such that Re( ) 0kz>and
12Re( ... ) 0k zz z>, for 1kn≤≤. Then show that
12 1log( ... ) log( ) ... log( )knzz z z z=+ + .
Solution : L e t () l o g ()fz z= b e t h e p r i n c i p l e b r a n c h o f t h e
logarithm function ∴() l o g ()fz zee z==.
Take 1...nazz= . Since the arguments of each kz and that of
12...kzz z l i e s b e t w e e n 2n− t o 2π f o r a l l 1kn≤≤. Therefore
12 1 12log( ... ) log ... ( ... )kk kzz z z z iA r g zz z=+ and log( ) log ( )kk kzz i A r g z=+ , for 1kn≤≤. ∴12 1 2(. . . ) ( ) ( ) . . . ( ) 2n nArg z z z Arg z Arg z Arg z kπ=++ ++ w h e r e k i s
any integer . ∴11 2() l o g . . . r g ( . . . )nnfa z z i A z z z=+
[] 12
1log ( ) ( ) ... ( ) 2n
k n
kzi A r g z A r g z A r g z kπ==++ + + +∑
1log ( ) 2n
k kkzi A r g z k iπ= =+ +∑
1log( ) 2n
kkzk iπ= =+∑ ∴() l o g ( ) 12
1fa z ekink k eπ− ===∑
1() l o g ( ) 0n
kk fa z=⇒ − =∑ ∴1() l o g ( )n
kkfa z= =∑ ∴12 1log( ... ) log( ) ... log( )nnzz z z z=+ + .
7) Give the principal branch of 1z− ( H i n t : ()log 1 1log(1 )2zez−= − ) munotes.in
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8) Prove that there is no branch of the logar ithm defined on 0G=−.
(Hint: A s s u m e t h e e x i s t e n c e o f a c o n t i n u o u s f u n c t i o n L(z)
defined on a connected open set G of the complex plane such that
L(z) is a logarithm of z f o r e a c h z i n G, compare L(z) w i t h t h e
Principal branch of log( )z. As α g o e s f r o m 0 t o 2π, since
()iaLe i a= a n d L b e i n g c o n t i n u o u s f u n c t i o n o f α , ∴()22( 1 ) 0iLe i Lππ== =, a contradiction.)
9) Evaluate ii by taking the logarithm in its principal branch.
10) Prove that 2 22sin sin sinhzx y=+
Solution:
()22 2 2sin sin sin cos cos sin sin cosh cos sinhzx i yx i y x i y x y i x y=+ = + = +()22 2 2 2 2 2 2
22 2 2 2 2 2 2sin cosh cos sinh (sin cosh cos sinh )
(sin cosh cos sinh )(sin cosh cos sinh )sin cosh cos sinh sin (1 sinh ) (1 sin )sinhsin sin sinh sinh sinh sin sin sinhxy i x yxy i x y
xy i x y xy i x yxy x y x y x yxx y y y xx y=+ +
=+ −
=+= + + −
=+ + − =+
11) Find the principal value of 2ii−
***** munotes.in
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6
ANALYTIC FUNCTIONS
Unit Structure
6.0. Objectives
6.1. Introduction
6.2. Analytic Functions
6.3. Cauchy Riemann equations
6.4. Harmonic Functions
6.5. The Functions ,sin( ),cos( )zezzetc.
6.6. Summary
6.7. Unit End Exercises
6.0. OBJECTIVES
In this unit we shall characterise the differ entiability of a
complex valued function in terms of it’s po wer series
expansion , in this case the function is sai d to be an analytic
function about some point 0z∈. An analytic function ()fzsatisfies some properties, among these one im portant
property is to satisfy Cauchy-Riemann equations . F u r t h e r w e
shall also see the term by term differentiation of a power series
function, provided that such term by term dif ferentiation is
possible. We shall also study the inverse func tion theorem then
we shall define a class of functions called as harmonic
functions .We shall also discuss the different iability of a
complex valued functions like ,sin( ),cos( )zezzetc.
6.1 INTRODUCTION
Given a function of the complex variable z, we wish to
examine if f is a differentiable function of z or not. As we
saw in the case real valued functions, we look for existence
of the limit 0() ( )limnfz h fzn→+− which should exist regardless munotes.in
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of the manner in which h approaches 0 through complex values.
An immediate consequence is that the partia l derivatives of f
, considered as a function of two real va riables x a n d (( ) ( ) ( ,)yfz fx i y fx y=+ = m u s t s a t i s f y t h e C a u c h y R i e m a n n
equations. Let us define the derivative of a function of
complex variable z at the point 0 zz=∈.
Let zx i y G=+ ∈ a n d :fG→ b e d e fi n e d by , ()()()fz u z i v z=+ ,
where u and v are real valued function
OR ()()(),,,fx y u x y i v x y = + .
Definition : If ()()0,,lim
hux hy uxyh → + − e x i s t s t h e n i t i s c a l l e d
partial derivative of u w.r.t x as the point (),xy and is denoted by
(),uxyx ∂ ∂ or (),xux y .
6.2 ANALYTIC FUNCTIONS
A function f is said to be analytic (or holomorphic or regular )
at a point 0zz= i f f is differentiable at every point of some nbd.
of 0z.
Definition : A function :fG→ i s a n a l y t i c i f f is continuous
differentiable on G.
A function f is analytic on a closed set S if f is differentiable at every
point of some open set containing S.
Theorem : Let ()
0nn
nfz az∞
== ∑ have the radius of converges 0R>
then
1) The function f is infinitely differentiable on ()0;Br and
()() ( 1 ) . . . . . . . ( 1 )K nKn
nKfz n n n Ka z∞−== −− + ∑ for zR < and 1K∨ ≥.
2) If 0n≥ then (0)!n
nfan= munotes.in
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Proof : 1) For zR <, we will write
() () ()
0n
nn n
nfz az S z R z∞
== = +∑ w h e r e ()
0nnn
nSz a z∞
== ∑ a n d
()
1KnK
KnRz az∞
=+= ∑
Put ()11n
ngz n az∞− == ∑ ()limnnSz→∞′ =
Fix a point 0z in ()0;BR
(Choose 0 0,rz r R> ∋ < < and zr R < < ()0zz≠
We will prove that ()()00fz g z′ =
Let 0δ> be arbitrary ()()0;0 ,Bz B r∋ δ⊂
Let ()0,zB z∈ δ then ()()()()()()()()00 0
0
0 00nn n nSz Rz Sz Rz fz fzzz gz lzz zz −− + − − = − −−
()()() ()()()0 000
0 0nn nn
nnSz Sz Rz RzSz Szzz zz − − ′′ = − ++−− ∴()()()()()() () ()0000 0 000nn
nnfz fz S z S zgz S z S z gzzz zz − − ′ − ≤ − + − −−()()0
0nnRz Rz
zz − + −….(1)
Let 0ε>, be given
Now, ()()()0000 11 nnKKK
KnRz Rzaz zzz zz∞
=+ − = −− − ∑
()12 2 100 0 0
0 11...KK K K
K
Knaz z z z z z z zzz∞−− − − =+= − + + + + −∑ ()()012 2 100 0
0 1...nn KK K K
K
KnRz Rzaz z z z z zzz∞−− − −=+ − ≤ + + + + −∑
11 11...KK K
K
Knar r r∞−− −=+≤ + + + ∑ 11.KK
KnaK r∞−=+= ∑ munotes.in
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∵The derived series 11nn
nna z∞−= ∑ is convergent at zr=. ∴The power series 11.KK
KaK r∞−
= ∑ converges for rR<.
For the above 0,ε> ∃ a n i n t e g e r 1N ∋
111..3K
K
KnaK r n N∞−
=+ε < ∨ ≥∑ . (∵ by Cauchy criteria)
Thus, ()()0
03nnRz Rz
zz − ε <− 1nN∨ ≥ ( 2 ) ∵()()00limnnSz g z→∞′ =
For the above 0ε>, ∃ an integer 2N s.t. ()()003nSz g zε′ − < ( 3 )
Choose {}12max ,NN N = ∴ F o r t h i s n, we can find 0δ> s.t.
()()()0
0
03nn
nSz SzSzzz − ε ′ − <− ( 4 )
whenever 0 0 zz< − <δ.
From equation (1), (2), (3), (4) we get ()()()0
0
0333fz fzgzzz − ε ε ε − < + +−=ε whenever 0 0 zz< − <δ ⇒ f i s d i f f e r e n t i a b l e a t ()0.;zBR∈ ∵ z i s a r b i t r a r y . ∴ f i s d i f f e r e n t i a b l e o n ()0;BR
A repeated application of this argument shows that the heigher
derivatives (),, . . . , . . .Kff f′′ ′ exists, so that
()() ( ) ( ) 1. . . 1KnKn
nKfz n n n K a z∞−== − −+ ∑ e x i s t s f o r zR < a n d 1K∨ ≥. ⇒ f i s i n f i n i t e l y d i f f e r e n t i a b l e o n ()0;BR . munotes.in
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2) Since
()() ( ) ( ) 1. . . 1KnKn
nKfz n n n K a z∞−== − −+ ∑()!
!nKn
nKnaznK∞−== − ∑
()1!!!nKkn
KnnKa a znK∞−=+= + − ∑
Put 0z= ∴()()0!0K
K fK a= +
()()0!K
kfaK =
Replace K by n ∴ ()()0!n
nfan =
Corollary : If the power series
0nn
naz∞
= ∑ has radius of cgs. 0R>,
then ()nnfz a z=∑ is analytic on ()0;BR .
Theorem : If G is an open connected set and :fG→ i s
differentiable with ()0fz zG′= ∨ ∈, then f is constant.
Proof : Fix a point 0zG∈ and let ()00wf z= .
Let (){}0;Az G f zw=∈ =
T.P.T. AG=.
[i.e. by showing A is both open and closed and A≠∅]
T.P.T. A is closed.
Let zG∈ and {}nz be a sequence in limnnAz z→∞ ∋ = ∴ ()0nfz w=, for each n∈
f is differentiable on G (given) ∴ f i s c o n t i n u o u s o n G. ∴ () ()0lim limnnnnfz f z fz w→∞ →∞= = = zA⇒ ∈ ∴ A c o n t a i n i t s l i m i t p o i n t . ⇒ A i s c l o s e d .
Now, T.P.T. A is open.
Fix aA∈, since G is open. ∴ ()0;rB a r G∃ > ∋ ⊂ munotes.in
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Let ();zB a r∈ and set ()()1gt f t z ta=+ −, 01t≤≤ ∴()()()()() ()()()()11z ft z t a fS Sagt gs tS z St ats tS z St a tS + − ++ − − − +− = × −− +−− ∴() ( )()()() ()()11lim lim
tS tSft z t a fs z S a gt gSzatS tS z St a →→ + −− +− − = ×− −− +− ∴ ()()1gS f S z S a z a′′=+ −× − ∴ ()0gS′=, 01S≤≤ ()()0,fz zG′ = ∨ ∈∵ ()constant, 0 1gs S⇒ = ≤≤ ⇒ ()()1c o n s t a n t 0gg == ∴ ()()()1c o n s t a n t 0fz g g == = ()0fa w== ⇒ zA∈ ∴ ();zB a R zA∈ ⇒ ∈
(;)zb a r A⇒ ∈⊆ ⇒ A i s o p e n .
Hence, by the connectedness of G.
AG=
• f is constant on G.
6.3 CAUCHY RIEMANN EQUATIONS (C-R Eq.)
Theorem : Let u and v be real valued function defined on the domain
G⊂ and suppose that u and v have continuous partial d erivatives
then :fG→defined by ()()()fz u z i v z=+ is analytic iff u and v
satisfy Cauchy Riemann equation. i.e. uvxy∂∂= ∂ ∂ and uvyx ∂ ∂ =−∂∂
(2006, 2007, 2008, 2009)
Proof : Let zx i y G=+ ∈ and zx i y∆=∆+∆.
Given, :fG→ i s d e f i n e d b y ()()()fu z i v z=+ O R ()()(),,,fx y u x y i v x y = + ∵ f i s a n a l y t i c o n G. ∴ f i s d i f f e r e n t i a b l e a t zG∈. ∴ ()()()0fz z fzfzz + ∆− ′→∆ ( a u n i q u e l i m i t )
as 0z∆→in any manner in .
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Now, ()()()()()(),, , ,ux xy y i vx xy y uxy i vxyfz z fz
zx i y+∆ +∆++ ∆ +∆− + + ∆− =∆∆ +∆ ()()()(),, .,ux xy y uxy vx xy y vxyixi y xi y + ∆ +∆− + ∆ + ∆− =+∆+∆∆ +∆ … … … … … … … ( I )
Suppose 0z→, along the real axis (x-axis) ∴ zx∆=∆ and 0y∆= ∴ ()()()()00,,lim lim
zxfz z fz u x x y u x yzx ∆→ ∆→ + ∆− + ∆ − = ∆∆
()(){},, ivx xy vxy
x + ∆ − + ∆ ∴ () () (),,uvfz x y i x yxx ∂ ∂′= + ∂ ∂ … … … … … … … ( I I )
Suppose 0z∆→, along the imaginary axis (y-axis). ∴ zi y∆=∆ and 0x∆= ()()()()00,,lim lim
zyfz z fz u x y y u x yzi y ∆→ ∆→ + ∆− + ∆− = ∆∆ ()(){},,ivxy y vxyiy + ∆− +∆
()1uvfziy y ∂ ∂ ′= + ∂∂uviyy ∂ ∂ =− +∂∂
21 iiiii i = = = − − ∵
() (),,uvix y x yyy ∂ ∂ =− + ∂∂ … … … … … … . . ( I I I )
From equation (II) and (III)
uv u viixx y y ∂ ∂ ∂ ∂ + = − +∂∂ ∂ ∂
Equating Real and imaginary part on both sides.
i.e. and uv u vxy y x ∂ ∂ ∂ ∂ = = − ∂∂ ∂ ∂
which are Cauchy Riemann equations. munotes.in
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Conversely,
Let zG∈ ∵G is open ()0, ,rB z r G⇒ ∃ > ∋ ⊂
Let ()0;zx i y B r∆=∆+∆∈ .
Given, u and v have continuous partial derivatives. ∴,, , , ,uux vuvxyxy′ ∂ ∂ ∂ ∂ ∂∂∂∂ are continuous on G. ∴The expression ()(),,ux x y y uxy+∆ + ∆− can be written as (by
definition of partial derivative.) ()()()()(),, . , , ,xyux x y y uxy xu xy y u xy x y+∆ + ∆− = ∆ + ∆ + ∅∆ ∆
where, ()0,lim 0xyxy
xi y ∆→ ∆→ ∅∆ ∆ =∆+∆ O R ()0lim 0zz
z ∆→ ∅∆ =∆
…………………(IV)
Similarly, ()()()()(),, , , ,xyvx x y y vxy xv xy y v xy x y+∆ + ∆− = ∆ + ∆ + ψ∆ ∆
where ()0,lim 0xiyxy
xi y ∆+∆→ ψ∆ ∆ =∆+∆ … … … … … . . ( V ) ∴ ()()()(),,fz z fz u x xy y u x y+∆− =+ ∆ + ∆− ()(),,ivx xy y vxy++ ∆ +∆− f r o m e q u a t i o n ( I )
() () ., ,xy xyxu yu x y i xv yv x y=∆ + ∆ + ∅∆ ∆+∆ + ∆ + ψ∆ ∆
f r o m e q u a t i o n ( I V ) a n d ( V )
()()() (),,xx yyxu i v yu i v x y i x y=∆ ++ ∆ ++ ∅∆ ∆+ψ∆ ∆
By Cauchy-Riemann equations
uvxy ∂ ∂ =∂∂ and uvyx ∂ − ∂ =∂ ∂
i . e . xyuv= and 2yx xuv i v=− = ()21i = − ∵ ∴ () ( ) ()()2
xx x xfz z fz x u i v y iv i u i+∆− =∆ + + ∆ + + ∅+ψ
()()xx x xxu i v i yi v u i=∆ ++ ∆ ++ ∅+ψ ()()xxxi yu i v i=∆+∆ + + ∅+ψ ∴()()()xyfz z fzxi y iui vzz z + ∆− ∆+∆ ∅+ψ =+ +∆∆ ∆ munotes.in
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{uv iiz x i yxx x i y ∂ ∂ ∅+ψ =+ + ∆=∆+∆∂∂ ∆+∆ ∵where
0lim 0xiyi
xi y ∆+∆→ ∅+ψ = ∆+∆ } ⇒ f i s d i f f e r e n t i a b l e a t z a n d
()()()
0lim
zfz z fzfzz ∆→ + ∆− ′ =∆uvixx ∂ ∂ =+ ∂∂ ⇒ f′ is continuous (ux ∂ ∂∵ and vx ∂ ∂ are continuous) ⇒ f i s c o n t i n u o u s l y d i f f e r e n t i a b l e . ⇒ f is analytic.
Note: If f(z) is analytic then it can be differentiated d irectly
Example :
(1) Prove that the function ()zfz e= is analytic in . Also find
its derivative.
Solution : Let ()()()fz u z i v z=+ and zx i y=+
Given that, ()zfz e=
().c o s s i nxi y x i y xui ve ee e yi y++= = = +
Equating real and imaginary parts on both sides.
T.P.T. f is analytic.
By previous theorem, we see that in order to prove and is analytic
we have to verify that u and v are satisfy Cauchy-Riemann
equations.
i.e. uvxy ∂ ∂ =∂∂ and uvyx ∂ − ∂ =∂∂
Now, ∵cosxue y= and sinxve y= .cos , sinxx uvey e yxy ∂ ∂ = = ∂∂ uvxy ∂ ∂ ⇒ =∂∂and uvyx ∂ − ∂ =∂∂ ⇒ u a n d v satisfies the C-R equation. ⇒ f i s a n a l y t i c . sin sin2eeih−θ θ + θ== θ munotes.in
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and cos cos2eeih−θ θ + θ== θ
()uvfz ixx ∂ ∂ ′=+ ∂∂cos sinxxey i e y= + ()cos sinxey i y=+ .xi yee= xi ye+=ze=
(Or ()fz∵ is analytic it can be differentiated directly i.e. ()zfz e′=)
2) Show that the function ()sinfz w z== is analy tic and also find dwdz.
Solution : Let zx i y=+ and ui v w+=
Given that sinwz= ()sinui v xi y+= +sin cos cos .sinxi yx i y= + + sin . cos cos .sinxh y i x h y= +
Comparing real and imaginary parts ∴sin . cosux h y= , cos .sinvx h y= ⇒ u a n d v are real valued function of x and y. ∴ cos .cos , sin .sinuvxh y x h yxx ∂ ∂ = = − ∂∂
and sin .sin , cos .cosuvhy x x hyyy ∂ ∂ = = ∂∂ ∴ uvxy ∂ ∂ =∂∂ and uvyx ∂ − ∂ =∂∂ ∴ C a u c h y - R i e m a n n e q u a t i o n s a r e s a t i s f i e d . ∴ ()sinfz z= is an analytic function. ∴ ()wu vfz izx x ∂ ∂ ∂ ′== + ∂∂ ∂ cos cos sin .sinxh y i x h y=+ −
(Or ()fz∵ i s a n a l y t i c i t c a n b e d i f f e r e n t i a t e d d i r e c t l y
i.e.() s i nfz z′=)
3) Using the Cauchy- Riemann equations , v erify that
222xy i x y++ is not analytic.
Solution : /g1179 ()22,2Pxy x y i x y=++
22 , 22yx y xPy i x P x i y P i P⇒ =−+ = + ⇒ ≠ (),Pxy⇒ is analytic. munotes.in
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4) Using Cauchy-Riemann equations , verify th at 222xy i x y+ −
is not analytic.
Solution : ()22,2Pxy x y i x y=+ −∵ 22 , 22 2 2yx xPy i x Px i y i P i x y⇒ = − = −⇒ =+ ()22 , 22 2 2 ,yx x y xPy i x Px i y i P i x y P i P P x y⇒ = − = − ⇒ =+ ⇒ ≠⇒
is not analytic.
(5) Give an example of function which is continuous everywhere but
not analytic
Solution: Let ()fz x y i y=+
,ux y vy∴==. Since u and v are polynomials , they are
continuous everywhere.
Now,,, 0 , 1,uuyyyxxyx x
uv u v
xy y x∂∂∂∂== = =∂∂∂∂
∂∂ ∂ ∂∴≠ ≠ −∂∂ ∂ ∂
Therefore f(z) is not analytic
6.4 HARMONIC FUNCTIONS
I f G is an open subset of , then the function :UG→ (i.e.
Real valued function of complex variable) is harmon ic if, it has
continuous second order partial derivatives and
22
220uv
xy ∂ ∂ +=∂∂ ( T h i s i s c a l l e d L a p l a c e ’ s e q u a t i o n )
e.g. (),. c o sxuxy e y = is harmonic function ?
.cos sinxx uuey e yxy ∂ ∂ = = − ∂∂
22
22.c o s c o sxx uuey eyxy ∂ ∂ = = − ∂∂ ∴ 22
22cos cos 0xx uuey ey
xy ∂ ∂ += − =∂∂ ∴ Above function is Harmonic function.
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Proposition : Let f be a analytic function in a region and ()()()fz u z i v z=+ . If u and v have continuous second partial
derivatives then u, v are harmonic function.
OR
1) If :fG→ d e f i n e d b y ()()()fz u z i v z=+ i s a n a l y t i c t h e n , Reuf= and Imvf= are harmonic functions.
Proof : Given that ()()()fz u z i v z=+ is analytic. ∴ C a u c h y R i e m a n n e q u a t i o n s a r e s a t i s f i e d .
i.e. uvxy ∂ ∂ =∂∂ ( I )
and uvyx ∂ − ∂ =∂ ∂ ( I I )
Differentiate equation (I) partially w.r.t. x and ( II) w.r.t. y. ∴ 22
2.uvxyx ∂ ∂ = ∂ ∂ ∂ and 222.uvxyy ∂ − ∂ = ∂ ∂ ∂ ∴ 22 2 2
220..uv v v
xy xy xy ∂ ∂ ∂ ∂+= − = ∂ ∂ ∂ ∂ ∂∂ ∴ ν is harmonic function.
Differentiate equation (I) and (II) partially w.r.t . y and x
respectively. ∴ 222.uvxyy ∂ ∂ = ∂ ∂ ∂ and 222.uvxyy∂− ∂ = ∂ ∂ ∂
Consider, 22 2 2
220..vv u u
xy xy xy ∂ ∂ ∂ ∂+= − = ∂ ∂ ∂ ∂ ∂∂ ∴ v is harmonic function.
Definition : If :fG→ i s a n a l y t i c a n d ()()()fz u z i v z=+ t h e n Reuf= a n d Imvf= a r e h a r m o n i c c o n j u g a t e i . e . u and v are
harmonic conjugate then u and v are harmonic function and u, v are
satisfied C – R equations.
Example : If :fG→ i s a n a l y t i c a n d ()()()fz u z i v z=+ t h e n
prove that harmonic function u satisfies the partial differential
equations 20.u
zz∂= ∂ ∂ .
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Solution : Given f is analytic.
Let zx i y zx i y=+ ⇒ =−.
Here, (),uu x y= where 2zzx + = and 2zzy−−= . ∴ (),uuxyzz ∂ ∂= ∂ ∂ ..ux uyxz yz ∂ ∂ ∂ ∂ = + ∂∂ ∂ ∂11
22uuxi y∂ ∂= + ∂ ∂
211
21uu i iiixy i i ∂ ∂= − = = = − ∂ ∂ −
(),uuxyzz ∂ ∂= ∂ ∂ ..ux uyxyzz ∂ ∂ ∂ ∂ = + ∂∂ ∂ ∂ 11
22uuxi y ∂ ∂= − ∂ ∂
1
2uuixy∂ ∂ =+ ∂ ∂
OR
1
2ixyz∂ ∂ ∂=+ ∂ ∂ ∂
2
.uuzzz z∂∂ ∂ =∂ ∂ ∂ ∂ 11.22uuiixy xy ∂∂ ∂ ∂=+ − ∂ ∂ ∂ ∂
22 2
221
4.uu uiyx xy ∂ ∂ ∂= + + ∂ ∂ ∂∂2
.uiyx∂− ∂ ∂ 22221
4uuxy ∂ ∂ =+ ∂∂=0
20.u
zz∂ = ∂ ∂
Example: Prove that the function ()32 2 2,3 3 3 2uxy x x y x y = − + − +
is harmonic. Find its harmonic conjugate and corres ponding analytic
function ()()()fz u z i v z=+ .
Solution : Given function ()32 2 2,3 3 3 2uxy x x y x y = − + − + ∴ 22336uxyxx ∂ = − +∂ and 66uxy yy ∂ =− −∂
2
266uxx ∂ =+∂ and 2
266uxy ∂ =− −∂ ∴ 22
2266 66 0uuxx
xy ∂ ∂ += + −− =∂ ∂ ∴ u i s a h a r m o n i c f u n c t i o n .
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To find v, we use Cauchy Riemann equation.
uvxy ∂ ∂ =∂∂ and uvyx ∂ − ∂ =∂∂ ∵ 22 22336 336uvxyx xyxxy ∂ ∂ = − + ⇒ = − +∂∂
B y i n t e g r a t i n g ()22336vx y x y∂= − + ∂∫∫
()32 3363yvx y x y x = − ++ ∅ ()2336xy y x y x= −++ ∅
(1)
where ()x∅ i s a n a r b i t r a r y f u n c t i o n o f x . T o f i n d ()x∅, we use
another equation of Cauchy-Riemann.
uvyx ∂ −∂ =∂∂ ∴()2366 3 6xy y x y y xy xx∂− −=− −++ ∅ ∂ ()66xxy yx ∂∅ = − ++∂()66xy y x′=− −− ∅ ∴ ()0x′∅ =
Integrating, we get ()xc∅ =, where c is constant. ∴ 2336vx y y x y c= −++
i.e. the required harmonic conjugate. ∴ Analytic function () (, ) (, )fz u x y i v x y=+ ()32 2 2 2 333 3 2 3 6xx yxy i x y yx y c=− + − ++ −++
Put xz= and 0y=
3() 3 2fz z z c=++ +
(Alternate method to find harmonic conjugate using Milne
Thompson method)
Given function ()32 2 2,3 3 3 2uxy x x y x y = − + − + ∴let 1(, )xyφ=22336uxyxx ∂ = − +∂ and 2(, )xyφ=66uxy yy ∂ =− −∂
Now, ()2
12 , () (, 0 ) (, 0 ) ( 3 6) ( 0 ) 0fz z i z z p u t t i n g x iz yz φφ = ′ − =+ +==
∴23 2() ( 3 6) 3fz z z d z z z=+= +∫
Put z = x+iy , munotes.in
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3232 2 2 3 32 2 2() ( ) 3 ( )
33 3 6 3fz x i y x i yxx y i x y i y ixx y i y i∴ =+ + +
=+ + + + + +
Separating real and imaginary parts we get v2336xy y x y= −+
Example : If :fG→ d e f i n e d b y ()()()()fz u z u z i v z==+ i s
analytic and ()cos sinxuve y y−= − t h e n f i n d t h e f u n c t i o n ()fz i n
terms of z.
Solution : ()
()fz u i vif z ui v=+=− (1 ) ( ) ( )ifz u i v u i v u v i u v∴+= + + −=−++()Ui V s a y=+
1
2(cos sin ) ( , )(s i n c o s) ( ,)x
x
x
yUe y y x yUe y y x yφ
φ∴ = − =
= −− =
12(1 ) ( ) ( , 0) ( , 0)(1 ) ( ) ( ) (1 )
()xy
zz z
zif z U i U z zifz e i ed z i e d zfz e cφφ′∴+= − = −∴+= + = +
∴ =+∫∫
Proposition: Suppose that f is analytic in a region G . I f ()fz=constant.
Proof : Let zx i y G=+ ∈ and ()()()fz u z i v z=+ .
Given that f is analytic. ⇒ Cauchy-Riemann equations are satisfied.
uvxy ∂ ∂−∂ ∂ and uvyx ∂ − ∂ =∂∂
Here, we given that ()constant fz K = = ( s a y ) .
Let 0k≠ [If 0k=, it is obvious that ()0fz=]. ∴ ui v K + = ∴ 22uv K+ =
∴22 2uv K+= … … . ( I )
Differentiate equation (I) partially w.r.t. x 220uvuvxx ∂ ∂ + =∂∂ ∴ 0uvuvxx ∂ ∂ + =∂ ∂ ∴ 0uuuvxy ∂ ∂ − =∂ ∂ uvyx ∂ ∂ = −∂∂∵ … … … … . ( I I ) munotes.in
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Again, differentiate equation (I) partially w.r.t y
220uvuvyy∂∂ + = ∂ ∂ ⇒0uvuvyy ∂ ∂ + =∂ ∂
0uuuvyx∂∂ + = ∂ ∂ {}uvxy∂∂=∂∂∵ ………….(III)
Multiplying equation (II) by u and equation (III) by v and add
20uuuu vxy ∂ ∂ − = ∂∂
20uuvu vxy∂ ∂ + + = ∂ ∂
()220dxu uv ∂+ = ⇒ 20uKx ∂ =∂ 0ux ∂ ⇒ =∂
Multiply equation II by v and III by u and subtract.
20uuuv vxy ∂ ∂ − = ∂∂
20uuuv uxy∂ ∂ − + = ∂ ∂
()220u uvdy∂− + =
20uKy ∂ ⇒ − =∂
0u
y ∂ =∂
Using Cauchy-Riemann equation
0uvxy ∂ ∂ ==∂∂ a n d 0uvyx ∂ − ∂ ==∂∂
0vvxy ∂ ∂ ⇒ ==∂∂ ∴ f is analytic at z. ∴ f is differentiable at z and
()yvfz ixx ∂ ∂ ′=+ ∂∂00i=+ 0= ∴ ()0,fz zG′= ∈ ()fz⇒ = constant.
Theorem : Suppose that f is analytic in a domain (region) D t hen
a) If ()0, ,fz z D=∀∈ f is constant.
b) If any one of ,R e ,I m ,ff f are f is constant in D (2008) munotes.in
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Proof : Let zx i yD=+ ∈ and ()()()fz u z i v z=+
Given that f is analytic in D. ∴ C a u c h y - R i e m a n n ( C . R ) e q u a t i o n s a r e s a t i s f i e d .
i.e. uv u vandxy y x ∂ ∂ ∂ − ∂ = =∂∂ ∂ ∂ ( 1 )
a) If ()0,fz zD′= ∨ ∈ t h e n f i s c o n s t a n t . ( A l r e a d y d o n e , l a s t
proposition)
b) i) Let f = constant.
∴ 22uv K+= (say)
22uv K⇒ + = (2)
w h e r e (),uxy a n d (),vxy a r e r e a l v a l u e d f u n c t i o n . L e t 0K≠ [if 0K= then nothing to prove.]
D i f f e r e n t i a t e e q u a t i o n ( 2 ) w . r . t . x
220uvuvxx ∂ ∂ + =∂∂
∴ 0uvuvxx ∂ ∂ + =∂∂
∴ 0uvuvxy ∂ ∂ + =∂∂ uvyx ∂ − ∂ =∂∂∵ ( 3 )
A g a i n d i f f e r e n t i a t e e q u a t i o n ( 2 ) w . r . t y
220uvuvyy ∂ ∂ + =∂∂
0uvuvyy ∂ ∂ + =∂∂
0uuvuxy ∂ ∂ + =∂∂ uvxy ∂ ∂ =∂∂∵ ( 4 )
Multiplying equation (2) by u and (4) by v and adding
20uuuu vxy ∂ ∂ − = ∂∂
20uuvu vxy∂ ∂ + + = ∂ ∂
()220dxu uv ∂+ = ⇒ 20uKx ∂ =∂ 0ux ∂ ⇒ =∂
Multiply equation (3) by v and (4) by u and subtracting munotes.in
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20uuuv vxy ∂ ∂ − = ∂∂
20uuuv uxy∂ ∂ − + = ∂ ∂
()220u uvdy∂− + =
20uKy ∂ ⇒ − =∂ ∴ 0u
y ∂ =∂
From equation (1), 0vx ∂ =∂ and 0v
y ∂ =∂
()00 0uvfz i ixx ∂ ∂ ′=+ = + = ∂∂ ∵ z is arbitrary. ∴ ()0fz zD′= ∨ ∈ ∴ b y p a r t ( a )
f is constant.
ii) Let Re constantfK = = (say)
∴(),uxy K = (say)
0ux ∂ =∂ and 0u
y ∂ =∂
∴from equation (1)
0vx ∂ =∂ and 0v
y ∂ =∂
∵f is analytic.
f is differentiable and ()uvfz ixx ∂ ∂ ′=+ ∂∂
()0. 0 0fz i′=+ =
∴()0fz′= ()fz⇒ = constant by (a)
iii) Let Imf = constant = K
∴ (),vxy K = (say)
∴0vx ∂ =∂ and 0v
y ∂ =∂
∵f is analytic.
∴f is differentiate and ()vyfz ixy ∂ ∂ ′=+ ∂∂ munotes.in
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B y C a u c h y R i e m a n n e q u a t i o n 0uv
xy ∂ ∂ ==∂∂
∴()0. 0 0fz i′=+ = ()fz⇒ = constant
b y ( a )
iv) Let ()argfz = constant
()()1arg tanvfz Ku− = θ== ( s a y )
1tanvu− = K
tan cottanuvKu v KvK = ⇒ = =
∴cot 0uv K− =
P u t cut K C =
0uv c⇒ − = unless .0vc =
B u t uc v− is a real part of ()()1ic f z+
()()1ic f z⇒ + = constant by (a)
()(){}constantconstant 1 constant1fz i cic == + =+∵
⇒ f is constant.
Example 1 : Prove that the function ()fz z= i s n o t d i f f e r e n t i a b l e
anywhere in the complex plane.
Solution : We know that, f is differentiable at z if
()()()fz z fzfzz + ∆− ′→∆ (a unique limit)
as 0z∆→ in any manner in -plane. ∴ ()fz z z z z z+∆=+ ∆=+ ∆ ∴ ()()()0lim
zfz z fzfzz ∆→ + ∆− ′= ∆
00lim lim
zzzz zz zzz ∆→ ∆→ +∆ + ∆− = = ∆∆ 0lim
zzz∆→ ∆ = ∆0lim
zxi yxi y∆→ ∆− ∆ = ∆+∆ …………(1)
Let 0z∆→ along the Real axis. ∴ zx∆=∆ and 0y∆= munotes.in
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∴ ()
0lim 1xxfzx ∆→ ∆ ′= =∆
Let 0z∆→ along the imaginary axis. ∴ zi y∆=∆ and 0x∆= ∴ ()
0lim 1yiyfziy ∆→ ∆ ′= = = − ∆ ∴ T w o l i m i t s o b t a i n e d a r e d i f f e r e n t .
i.e. limit is not unique. ∴ G i v e n f u n c t i o n ()fz z= i s n o t d i f f e r e n t i a b l e i n t o c h e c k 0z∆→. xy=, we get 11ii − +.
2) Show that the function ()2fz z= i s d i f f e r e n t i a b l e o n l y a t t h e
origin. (2006 )
Solution : Let 22()fz x y=+.
Since, 22xy+ i s c o n t i n u o u s e v e r y w h e r e , f(z) i s c o n t i n u o u s
everywhere.
022
00 00
0000 0 0 00 0 00 0
00 00
0 0
000() ( )() l i m l i m
() () () ()lim lim
lim lim
limzz z
z z
z z
zzz z fz z fzfzzzzz zz z z zz zz z zzz
zzzz z z zzzz z z
zz
zzz zzδ
δ δ
δ δ
δδ δ
δδ
δδ δδ
δδ
δδ δ δ δδ δ δ
δδ
δδδ→ →
→ →
→ →
→+ − + −′==
++ − ++ −==
++ ++==
=+ +
(i) When zδ is real: Then yδ=0 a n d zzxδδδ== . As 0, 0zxδδ→→ ∴ 00 0 0 00 000() l i m l i m
xxzfz z z z z z x z zzδδδδδδ→→′=+ + = + + = +
(ii) When zδ i s i m a g i n a r y : T h e n 0xδ=and ,.zi yz i yδδ δ δ== −
As 0, 0zyδδ→→ ∴ 00 0 0 0 0 000() l i m l i m ( )
zxzi zfz z z z z z i y z zzzδδδδδδδδ→→−′=+ + = + + = −+
Since the two limits are different along two differ ent paths except at
z=0, 0()fz′does not exist anywhere except at z=0
Hence, f(z) is not differentiable anywhere except at z=0
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3) Let ()()33 33
220, 0 i.e. 000 , 0 i . e . 0xyi xy
xy zfz xy
xy z −++ ≠ ≠ ≠= +
= = =
Prove that C.R. equations are satisfied at the orig in but ()0f′ does
not exist i.e. ()fz is not differentiable there.
Solution : Let zx i y=+ and ()()()fx u z i v z=+ ∴ ()()()fz u z i v z=+
()()33 33
22,000xyi xy
zx y
xy
z −++ = ≠= +
= ∴ ()()33 33
220,00xyi xy
zuxy xy
z −++ ≠ = +
= ∴ ()()33 33
220,00xyi xy
zvxy xy
z + + + ≠ = +
= ∴ ()()()0,,l i m
hux h y uxyuxyxh → + − ∂ = ∂
()()()0,0 0 ,00, 0 lim
huh u u
xh → − ∂ = ∂
3
2
0000
0lim lim
hhhhhhh→→ − − −= = = 1
Similarly,
()()()00, 0, 00, 0 lim
huuu u
yh → − ∂ = ∂3
2
0000
0lim lim
hhhhhhh→→ − − − −= = 1= − ∴ ()3
2
0000
0, 0 lim lim
hhhvhhxh h→→ + − ∂ = = ∂1= munotes.in
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∴()()()0,,,l i m
hvxy h vxyvxyyh → + − ∂ = ∂3322000
0lim lim
hhhhh
hh→→ + += = 1=
uvxy ∂ ∂ =∂∂ and uvyx ∂ − ∂ =∂∂ ∴ C a u c h y - R i e m a n n e q u a t i o n s a r e s a t i s f i e d a t t h e o r i g in. ∴ ()()()0lim
zfz z fzfzz ∆→ + ∆− ′= ∴ ()()()000l i m0 zfz ffz ∆→ − ′= −
where 0z→, in any manner in Complex Plane.
Along the x-axis, 0y= and zx= ∴()() ()33
2
00000l i m l i m
hxxi x
fx fxfxxx →→ + − − ′= = 3
0lim
xx
→ = ()31ix+ 1i=+
Along the y-axis, 0x= and zi y= ∴()() ()332
0000l i m l i m
iy iyyi yfy f yfiy iy →→ −+ − ′= = ()211ii iiiii + + == = +
Let 0z→ along the line yx=, xy=, and ()1zx i xx i=+ = +.
()
()()()() 1000l i m1 xifx ffxi +→ − ′= +3
0.2lim
xix
→ =
22x
()0lim11xixixi→= + + 1ii= +
Hence, ()0f′ is not unique. ∴ The given ()fz is not differentiable at origin.
4) If :fG→ d e f i n e d b y ()()()fz u z i v z=+ i s a n a l y t i c a n d
suppose that (),uvψ=ψ be any continuously differentiable function.
Prove that ()2 22 22fzxy uv ∂ψ ∂ψ ∂ψ ∂ψ ′ += + ∂∂ ∂∂ .
Solution : Let zx i y=+ and ()()(),, ,fx y u x y i v x y = +
Given, ()fz is analytic. ∴ C a u c h y R i e m a n n e q u a t i o n s a r e s a t i s f i e d . munotes.in
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∴ uvxy ∂ ∂ =∂∂ and uvyx ∂ − ∂ =∂∂ (1)
Now, (),uvψ=ψ where (),uu x y= and (),vv x y= ∴ x ∂ψ ∂ and y ∂ψ ∂ exist.
(),uvxx ∂ψ ∂= ψ ∂ ∂
..uvux vx ∂ψ ∂ ∂ψ ∂= + ∂∂ ∂ ∂ ( 2 )
Similarly,
..uvyu yv y ∂ψ ∂ψ ∂ ∂ψ ∂= + ∂∂ ∂∂ ∂
By C.R. equations
..vuyu x v x ∂ψ − ∂ψ ∂ ∂ψ ∂ = + ∂∂ ∂ ∂ ∂ ( 3 )
Squaring and adding equations (2) and (3)
2 22 2
.. 2 . . .uu v u vxyu xv xu x v x ∂ψ ∂ψ ∂ ∂ ∂ψ ∂ ∂ψ ∂ ∂ψ ∂ += + + ∂∂∂ ∂∂ ∂∂ ∂ ∂ ∂
22
.. 2 . . .vu u uux vx uxvx ∂ψ ∂ ∂ψ ∂ ∂ψ ∂ ∂ψ ∂ + + − ∂∂ ∂ ∂ ∂∂∂∂
222 2 2uvuv x u v x ∂ψ ∂ψ ∂ ∂ψ ∂ψ ∂ =+ ++ ∂∂ ∂ ∂ ∂ ∂
22 2 2uv
uv x x ∂ψ ∂ψ ∂ ∂ =+ + ∂∂ ∂ ∂ ( * ) ∵ f is analytic at z. f⇒ is differentiable at z.
()uvfz ixx ∂ ∂ ′=+ ∂∂ and ()uvfzxx ∂ ∂ ′= −∂∂
(*) 22uv uviiuv x x x x ∂ψ ∂ψ ∂ ∂ ∂ ∂ ⇒ + + − ∂∂ ∂ ∂ ∂ ∂
()2 22 22fzxy uv ∂ψ ∂ψ ∂ψ ∂ψ += + ∂∂ ∂∂
5) Prove that the function Re z is differentiable anywhere in .
Solution : ()()fz zfzz + ∆ ′ →∆ ( a u n i q u e l i m i t )
as 0z∆→ in any manner in plane. munotes.in
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Here ()Refz Z= ∴ ()()Refz z z∆+∆= + ∆
6) Prove that Cauchy Riemann equation can be writte n polar Co-
ordinates as 1ud vrr d ∂ = ∂ θ and 1vurr ∂ − ∂ = ∂∂ θ.
Solution : Let ()()(),, ,fr u r i v r θ= θ+ θ
Let cosxr= θ and sinyr= θ 22rx y⇒ = + and ()1tanyx− θ=
Now, (),uu r= θ where r and θ are functions of x and y. ∴ ..uu ruxr x x ∂ ∂ ∂ ∂ ∂θ = + ∂∂ ∂∂ θ ∂ ∵ 22rx y=+ and ()1tanyx− θ= ∴ 2 r
x ∂ =∂2x
22r
xy =
+ cos
r θ cos= θ
and
()2
2
21
1yxy
x yx x− ∂θ − = ×− = ∂ + 22
2xy
x + 2sinr
r − θ =
sinr − θ = ∴ sin.c o suu u
xr r ∂ ∂ ∂ − θ = θ+∂∂ ∂ θ
( 1 )
Similarly, ..uu ruyr y y ∂ ∂ ∂ ∂ ∂θ = + ∂∂ ∂∂ θ ∂ 1.sin cos .uurr ∂ ∂ = θ+ θ ∂ ∂θ
( 2 )
Similarly, sincos . .vv vxr r ∂ ∂ θ ∂ = θ − ∂∂ ∂ θ
( 3 )
and cossin . .vv vyr r ∂ ∂ θ ∂ = θ + ∂∂ ∂ θ
( 4 )
By Cauchy – Riemann Equations ∂ ∂ =∂∂uvxy and uvyx ∂ − ∂ =∂∂⇒ 0uv
xy ∂ ∂ − =∂∂
Consider,
()11cos sin sin cosuvu u v v
xy r r r r ∂ ∂ ∂ ∂ ∂ ∂ − = θ− θ− θ− θ∂∂∂ ∂θ ∂ ∂θ munotes.in
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11cos sin 0uv vu
rr rr ∂ ∂ ∂ ∂ θ− −θ + = ∂ ∂θ ∂ ∂θ
( 5 )
and 0uv
yx ∂ ∂ + =∂∂
1s i nsin cos . cos . 0uu v vrr r r ∂ ∂ ∂ θ ∂ θ+ θ + θ − =∂ ∂θ ∂ ∂θ
11sin cos . 0uv u v
rr r r ∂ ∂ ∂ ∂ θ − + θ + −= ∂ ∂θ ∂θ ∂
( 6 )
Multiplying equation (3) cos×θ and equation (6) sin×θ and adding.
2 11cos sin . cosuv vu
rr rr ∂ ∂ ∂ ∂ θ− − θ θ + ∂ ∂θ ∂ ∂θ 0 =
2 11sin sin . cosuv vu
rr rr ∂ ∂ ∂ ∂ + θ − − θ θ + ∂ ∂θ ∂ ∂θ 0 =
∴()22 1sin cos 0uv
rr ∂ ∂ θ+ θ − =∂ ∂θ ∴ 10uvrr ∂ ∂ − =∂ ∂θ ∴ 1uvrr ∂ ∂ = ∂ ∂θ
Multiplying equation (3) sin×θ a n d e q u a t i o n ( 6 ) cos×θ a n d
subtracting.
1sin .cosuv
rr ∂ ∂ θ θ − ∂ ∂θ 2 1sin 0vu
rr ∂ ∂ − θ + = ∂ ∂θ
1sin . cosuv
rr ∂ ∂ −θ θ − ∂ ∂θ 2 1cos 0vu
rr ∂ ∂ + θ + = ∂ ∂θ
− −
()22 1sin cos 0vu
rr ∂ ∂ − θ+ θ + = ∂ ∂θ ∴()22 1sin cos 0vu
rr ∂ ∂ − θ+ θ + =∂θ ∂ ∴ 10vurr ∂ ∂ − − =∂ ∂θ ∴ 1vurr ∂ ∂ = ∂ ∂θ
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6.5 THE FUNCTIONS ,sin( ),cos( )zezz etc
Exponential Function : The exponential function in Complex
Plane denoted by ze, is defined by as
23
01. . .!2 ! 3 !nz
nzz zezn∞
= = = + + + + ∑
1) The exponential function ()zfz e=is analytic in the whole
Complex Plane and()()fz f z′= with ()01f=.
Solution : We have, ()
0!nz
nzfz en∞
= ′== ∑ . ∴ ()111!1 !nnaann+ = ⇒ = + ∴ ()1n+1 !1lim lim!1n
nn naRan →∞ →∞ + = = ×
()lim 1nn
→∞= +
∵ T h e r a d i u s o f c o n v e r g e n c e o f t h e g i v e n p o w e r s e r i e s is R=∞. ∴The power series converges for all z and convergence is uniform
for each compact subset of . ∴By using corollary.
[If ()nnfz az= ∑ has a radius of convergence 0R> t h e n f i s
analytic in ()0,BR ]
()zfz e= is analytic in whole Complex Plane.
Note: Similarly, 21 2
00sin ,cos(2 1)! (2 )!nnnnzzzznn+ ∞∞
====+∑∑ are analytic in
whole Complex Plane.
Definition :
Entire Function : If the function f is analytic everywhere in whole
Complex Plane ( e x c e p t a t ∞) is called an Entire function o r
integral function . e.g. ,c o s ,s i nzez z .
6.6 SUMMARY
1) A function f is said to be analytic (or holomorphic or regular ) at a
point 0zz=, if f is differentiable at every point of some nbd of 0z. munotes.in
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2) If the power series
0nn
naz∞
= ∑ h a s r a d i u s o f c g s . 0R>, then
()nnfz a z=∑ is analytic on ()0;BR .
3) Let u and v be real valued function defined on the domain G⊂
and suppose that u and v have continuous partial derivatives then
:fG→defined by ()()()fz u z i v z=+ is analytic iff u and v
satisfy Cauchy Riemann equation. i.e. uvxy∂∂= ∂ ∂ and uvyx ∂ ∂ =−∂∂.
4) 23
01. . .!2 ! 3 !nz
nzz zezn∞
= = = + + + + ∑
6.7 UNIT END EXERCISES
1) Give an example of a function which is not diff erentiable at the
origin but the partial derivatives exist and s atisfy the Cauchy-
Riemann equations there .
Solution: Consider :f→ defined by
22()(, ) (, ) ( 0 , 0 )xy x iyfx y x y
xy+= ≠
+ 0( , ) ( 0 , 0 )xy = =. (0,0) 0xf=similarly (0,0) 0yf=
But 0( , ) ( 0 , 0 )22() ( 0 )lim limzx yfz f x yzxy→→−=
+does not exist. For, on
the line ya x=2() ( 0 )1fz f
zaα−=
+∵for 0z≠.
The limit depends
on real number a.
2) Check at what points does the function 2()fz z= i s
differentiable.
Solution : 2()fz z= . Let zx i y=+.
22()fz x y⇒ =+. 2, 2xyfx fy∴= =
f⇒ has continuous partial derivatives for all
f∴is differentiable at z∈, provided that yxfi f=.
22 2 20 ( , ) ( 0 , 0 )yi x y i x x y⇒ = ⇒ − =⇒ =. f∴is differentiable only at the origin (0,0) munotes.in
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3) Show that 22()fz x i y=+ i s d i f f e r e n t i a b l e a t a l l p o i n t s o n t h e
line yx=.
Solution: 22() . 2, 2xyfz x i y f x f i y=+ ∴= =∵ ,xyff⇒ exist and are continuous functions of zx i y=+ 22yxfi y i fx i⇒ == = iff 0xy−=.
This is possible iff xy=.
B y p r o p o s i t i o n f i s d i f f e r e n t i a b l e a t a l l p o i n t s o n t h e l i n e yx=.
4) Suppose f is analytic in a region and at every point , either 0f= or 20f=. Show that f is a constant function.
(Hint: Consider The derivative of 2()fz)
5) Find all analytic functions fu i v=+with 22(, )uxy x y= −
Solution: fu i v=+∵ is analytic.
f∵ satisfies Cauchy- Riemann equations .
.2 ,, 2
2, 2.xy x yx y xyxuv v u v y vvx vy ⇒ =⇒ = = − −=−⇒ = = (, ) 2 ,vxy x y c∴ =+where c is any real constant.
22() 2fx y i x y i c⇒ = − ++
6) If f is a analytic in a region and if f is constant there , then
show that f is constant.
Solution: If 0f=, the proof is immediate , otherwise assume
that 0f≠. Let 220fu i v u v=+ ⇒ + ≠.
T a k i n g t h e p a r t i a l d e r i v a t i v e s w . r . t . x a n d y, we see that 0, 0xx yyuu vv uu vv∴ += +=.
Making use of the Cauchy- Riemann equations , we get , 0, 0xy xyuu vu vu uu∴− = + =
22() 0 0xx y uv u uv⇒ += ⇒ ==, similarly 0yxuv==. f⇒is constant.
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7) Show that (sin( )) cos( )dzzdz= . (Hint: Use
that 1sin( ) ( )2iz izze ei−= − )
8) Find a power series representation for cos( )z.
Solution: 01cos( ) ( ),2!niz iz z
nzze e en− ∞
= =+ = ∑∵
00() ( ),!!nniz iz
nniz izeenn∞∞
==−= =∑∑ 201( 1 )()2( 2 ) !nniz iz
nzeen∞−
=−∴ += ∑
22 4 6
0(1 )cos( ) 1 ...(2 )! 2! 4! 6!nn
nzz z zzn∞
=−∴ == −+ −+∑
9) Show that 22log( )xy+ is harmonic in 0−.
Solution: Let 22(, ) l o g ( )uxy x y=+ . 22 2222,xyxyuuxy xy⇒ = =++ 0− 22 2222 2 22 222 22,() ()xx yyyx xyuuxy xy−−⇒ = =
++ ,xx yyuu⇒ exist and are continuous functions of z on 0−,
also 0xx yyuu+=∵. (, )uxy⇒is harmonic in 0− .
10) For the function f(z) defined by
()2()0zfz i f zz= ≠
= 00if z=
Prove that C-R eq. are not satisfied at the origin , but the function
f(z) is not differentiable at the origin(2009)
11) Find the holomorphic function f(z) w h o s e r e a l p a r t i s 2xy+2x (2008)
12) Find the analytic function ()()()fz u r , i v r ,θθ=+ w h o s e r e a l
part is 22rc o sθ . munotes.in
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13) For the function, ()fz defined by ()()2000z
,z fzz
,z≠==
.
Prove that the Cauchy Riemann equations are satisf ied at ()00, but
the function is not differentiable at ()00, .
14) If :fG→C is analytic, prove that () ()222222Re 2 'fz f zxy∂∂ += ∂∂
15) Construct an analytic function()()()zu zi v zf = + , whose real part
is cos x cosh y . express the result as a function of z.
16) Construct an analytic function ()()()fz u z i v z = + , whose real
part is (c o s s i n)xex yy y − .
E x p r e s s t h e r e s u l t a s a f u n c t i o n o f z.
*****
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7
COMPLEX INTEGRATION
Unit Structure
7.0 Objectives
7.1 Introduction
7.2 Complex Line integrals
7.3 Integration along piecewise smooth path , The Closed
Curve Theorem
7.4 Summary
7.5 Unit End Exercises
7.0 OBJECTIVES
T h r o u g h t h i s u n i t w e s h a l l s t u d y t h e c o n c e pt of complex
integration, an integration of the form ()fz d zγ∫ t a k e n o v e r a
piecewise smooth path γ f u r t h e r w e s h a l l d e r i v e c e r t a i n
properties of this integral. We would like to know further that
what can be the integral of an entire func tion along a
boundary of a rectangle in a complex plane, the answer is
given in form of a closed curve theorem.
7.1 INTRODUCTION
We have to recall theorem on differentiability of a power
series that states that a power series
0()nn
nfz a z∞
==∑ converges for zC R. Then ()fz′ exists and 11()nn
nfz n a z∞−=′ = ∑ on the open disc zC R . T h e r e f o r e a n e v e r y w h e r e c o n v e r g e n t p o w e r s eries
represents an entire function. Our main goal i n this unit is to
study the converse of this result namely th at every entire
function can be expanded as an everywhere convergent power
series. This result has a consequence that ever y entire function
is infinitely differentiable. We shall also a r r i v e a t t h e s e
results by discussing integrals. Let us st art by defining a
Line integral.
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7.2 COMPLEX INTEGRATION
Definition : Trace of a curve :
If []:,xa b → i s a c u r v e , t h e n t h e s e t (){}xt i s c a l l e d t h e
trace x and is denoted by {}(){}:xx t a t b= ≤≤
The trace of x is always a compact set.
Definition : Contour : A contour is a piecewise smooth curve.
Definition : A complex valued function f is said to be continuous on
a smooth curve []:,xa b → i f , ()()()()()fz fx t u t i v t== + i s
continuous.
7.3 INTEGRATION ALONG A PIECEWISE SMOOTH
PATH, THE CLOSED CURVE THEOREM
Definition : Complex Line Integral :
Suppose f i s c o m p l e x v a l u e d , c o n t i n u o u s a n d d e f i n e d o n o p e n set
G⊂ a n d t h a t []:,xa b → i s a p i e c e w i s e s m o o t h c u r v e w i t h {}xG ⊂.
Then, the expression
() ( ) ( ) ( ) ( )1 1
1j b n
j xa ifz d z fx t xt d t fx t xt d t+−
=′′== ∑ ∫∫ ∫
where 01 1 ...nat t t t b−=< < << = is called the complex line integral
of f over x.
This curve x is called path of integration of this integral.
Connection between Real and Complex line integral :
I f ()()()fz u z i v z=+ then the complex line integral ()xfz d z ∫
can be expressed as
()xx xfz d z u d x v d y iu d y v d x = − + + ∫∫ ∫
Theorem : Let x b e s u c h t h a t ()()(),xt xt i yt a t b=+ ≤≤ i s a
smooth curve and suppose that f and g are continuous function on
open set G⊂ containing {}x then ; munotes.in
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i) ()()xxfz d z fz d zα = α ∫∫ where α is a complex constant.
ii) ()()()()xx xfz g z d z fz d z g z d z+ = + ∫∫ ∫
Example 1 : Evaluate 1
xdzz ∫, where () [],0 , 2itxt e t = ∈ π.
Solution : By the definition of complex line integral.
() ( ) ( )b
xafz d z fx t xt d t′ = ∫∫
Here, () ()1,, 0 , 2itfz x t e a bz= = = = π ∴ () ()1,ititfx t x t i ee ′= =
11
itxdzze = ∫2
0.itieπ
∫[]2200dt i dt i tππ = = ∫2i= π
Definition : Rectifiable Curve : A curve is rectifiable if it has finite
length.
Note : Every piecewise smooth curve is Rectifiable.
Definition : If x is s.t. () .xt a t b ≤≤is rectifiable, then its length ()Lx
us defined by () ( )b
aLx x t d t′= ∫
Example 1 : Find the length of the curve () []4, 0 , 2itxt e t= ∈ π.
Solution : Length of ()2
04itxL x i ed tπ
== ∫ 2
04itied tπ
= ∫
2
00244dtdtπ π = = ∫∫8= π
Example 2 : Find the length of the curve ()()[]1,0 , 4xt it t=+ ∈ .
Solution : ∵ ()()1xt it=+ ∴ ()()1xt i′=+ munotes.in
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∴ L e n g t h o f
() ( )4
01xL x id t== + ∫4
22
011dt= + ∫4
02dt= ∫ []402t= 42=
Definition : Opposite Curve :
If []:,xa b → is a given curve then, the opposite curve x− to x is
defined as ()()[];,xt xa b t t ab− =+ − ∈
Example : Let () [];0 ,itxt e t= ∈ π.
1 -1 1 -1x -x
Fig. 7.1
Definition : Let []1:,xa b →and []2:,xa b → b e t w o s m o o t h
curves such that ()()11 2 2xb x a= . Then we define the path []121 122 :, xxa bba+ + −→ as follows.
()()()[]12 1 1 1 if ,xxt x t ta b+ = ∈
OR ()()()[]12 2 12 1 1 22 if ,xxt x t b a tb b ba = −+ ∈ + −∪
The path 12xx+ i s c a l l e d t h e s u m o f t w o c u r v e s 1x o r 2x o r t h e
union of two curves 1x and 2x.
Example 1 : Let ()[],0 , 3xt t t= ∈ and ()2 3, 0 , 32itxt e tπ= ∈ . munotes.in
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x12
3 0x
Fig 7.2
Theorem : Let x be s.t. 11() () ()xt x t i y t=+ be a smooth curve and
suppose that f is a continuous function on an open set G containing {}x. Then,
i) ()()xxfz d z fz d z = − ∫∫
ii) () ()
xxfz d z s z d z ≤ ∫∫
iii) If
[]()(),Max
ta bMf x t∈ = and ()LL x= ( L e n g t h o f x )
t h e n ()
xfz d z M L ≤ ∫.
(This property is called standard estimate for the integral.)
iv) If 1x a n d 2x a r e s m o o t h c u r v e s i n G t h e n , ()()()12 1 2xx x xfz d z fz d z fz d z+ = + ∫∫ ∫, where 12xx+ a r e s u m
of 2 curves.
Proof : i) By definition of opposite curve
()()xt xa b t− =+ −
By definition of complex line integral.
() ( ) ( )b
xafz d z f x t xa b t d t− ′ = − − + − ∫∫
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Put abtu d t d u+−= ⇒ = − , When ,tau b= ⇒
,tbua= =
() () ()( ) .a
xbfz d z fx u xu d u− ′ = − − ∫∫ () ()b
afx u x u d u′= ∫().xfz d z= − ∫
ii) If ()0xfz d z =∫, then there is nothing to prove.
Let ()0xfz d z ≠∫
Put iue− θ= , where () arg
xfz d zθ= ∫ ∴ 1u =
() () ()i
xx xfz d z e fz d z u fz d zθ = = ∫∫ ∫ ( 1 ) ∵ Rezz =
() () Re
xxfz d z fz d z = ∫∫ () Re
xuf z d z= ∫ from (1)
()Re
xuf z d z= ∫()xre u f z dz= ∫
()xuf z d z≤ ∫ (∵Rezz ≤ )
().
xuf z d z= ∫
() ()
xxfz d z fz d z ≤ ∫∫ (∵1u =)
iii) Given that,
[]()(),Max
ta bMf x t∈ = and ()LL x=
B y u s i n g p a r t ( i i ) ,
() ()
xxfz d z fz d z ≤ ∫∫ munotes.in
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() ()() ()b
xafz d z fx t xt d t′ = ∫∫ ()b
aMx t d t′≤ ∫()ML x=
()
xfz d z M L ≤ ∫
iv) Let []11 1:,xa b G → and []22 2:,xa b G → with ()()11 2 2xb x a= . ∴ W e d e f i n e []12 1 122 :xxa bba G+ + −→
() ( )()[]()11 112 if ,
if ,21 2 1 1 2 2xt t abxxtxtb a t b b b a ∈ + = −+ ∈ + − ∴ 12xx+ h a s d e r i v a t i v e ()1xt′ i n []11,ab a n d ()12tb a + i n []11 2 2,bb b a + −
() ( ) ( )( ) ( )122
12 12
12 1bb a
xx afz d z f x x t x x t d t+−
+′ = + + ∫∫
() () ( ) ( )123 1
11 21 2 21 2
11bb b b
abfxt xt d t xt b a xt b a d t++
′′= + −+ −+ ∫∫
() ( ) ( )123
21 2 21 2
11bb b
xbfz d z fxt b a xt b a d t++
′= + −+ −+ ∫∫
Put 12tb a u d td u−+= ⇒ = , then 1tb=, 2ua=, 12 2tb b a=+ −, 2ub=.
() () ()2
22
12b
xafz d z fx u x u d u′= + ∫∫
()()12xxfz d z fz d z= + ∫∫
Note : i) If x i s p i e c e w i s e s m o o t h t h e n , t h e r e i s a p a r t i t i o n
01 ...nPat t t b÷= << < = of [],ab s.t. the restriction kx of curve x to []1,kktt− is smooth for 1kn≤≤.
12 ...nxx x x=+ + + ∴ ()()()()()12 1 2...
xx x x xx x n nfz d z fz d z fz d z fz d z fz d z++ = = + + + ∫∫ ∫ ∫ ∫
ii) () () Re Re
xxfz d z fz d z
≠ ∫∫
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Example 1 : Let ()1fz= and ()x t it=, []0,1t∈ .
() ( ) ( )1
01. .b
xafz d t fx t xt d t i d t i′ = = = ∫∫ ∫
()Re Re 0xfz d t i
= =∫ ( 1 )
and ()1
0Re 1. 1xfz d z i d t = =∫∫ ( 2 )
From equation (1) and (2)
() () Re Re
xxfz d z fz d z
≠ ∫∫
Change of Parameter :
Let []:,xa b → a n d []:,cdσ → b e t w o s m o o t h c u r v e .
Then the curve σ i s e q u i v a l e n t t o c u r v e x i f , t h e r e i s a f u n c t i o n [][]:, ,cd ab∅ → which is contain non-decreasing and with ()ca∅ =
and ()db∅ = s.t. xσ= ∅.
H e r e , w e c a l l t h e f u n c t i o n ∅ a c h a n g e o f p a r a m e t e r . T h i s
new curve x∅ is called the Reparametrization of the curve x.
Theorem : Let []:,xa b → b e a s m o o t h c u r v e a n d s u p p o s e t h a t [][]:, ,cd ab∅ → i s a c o n t i n u o u s n o n - d e c r e a s i n g f u n c t i o n w i t h ()ca∅ = and ()db∅ =. If f is continuous on {}x then
()()xxfz d z fz d z∅ = ∫∫
Proof : Given that []:,xa b → i s a s m o o t h c u r v e a n d [][]:, ,cd ab∅ → i s c o n t i n u o u s n o n - d e c r e a s i n g f u n c t i o n w i t h ()ca∅ = and ()db∅ =.
By hypothesis, there is a change of parameter [][],,cd ab∅ → s.t. ()()()xs x s∅ = ∅
and () ( ) ( ) ( )xs x s s′′′∅ = ∅ ∅ f o r [],sc d∈ ∴ () () () () ( )d
xcfz d z fx s x s d s∅′ = ∅ ∅ ∫∫ munotes.in
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()() ()() ()d
cfx s x s s d s ′′= ∅ ∅ ∅ ∫
Put ()ts=∅ ⇒ ()dt s′=∅
when sc=, ()tc a=∅ = and when sd=, ()td b=∅ =
() ( ) () ()b
xafz d z fx t xt d t∅′ = ∫∫ ∴ ()()xxfz d z fz d z∅ = ∫∫
Fundamental theorem of calculus :
I f f is continuous on [],ab and ()()Fx fx′= in [],ab then () () ()b
sfx d x F b F a= − ∫ .
Primitive or Antiderivative of a function :
A function :fG→ is said to be primitive or antiderivative of f
in G if, F is analytic in G and ()()Fz fz′= in G.
Theorem : Let G be an open set in and suppose that :fG→ is
a continuous function with primitive :FG→. If []:,xa b G → is a
smooth curve, then ()()()xfz d z Fx b Fx a = −∫.
I n p a r t i c u l a r , i f x i s c l o s e d t h e n ()0xfz d z =∫.
Proof : Given that, :fG→ i s a c o n t i n u o u s f u n c t i o n w i t h
primitive :fG→. ∴ ()()Fz fz′= ∴ () ( ) () ()b
xafz d z fx t xt d t′ = ∫∫
() ( )b
aFx t d t′= ∫ (∵ ()()()()Fx t fx t′ =
()()baFx t= ( b y u s i n g f u n d a m e n t a l )
()()Fx b Fx a= −
If x is closed, then ()()xa xb= .
()()()0xfz d z Fx a Fx b = − = ∫ munotes.in
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x
xx
cc
αβ
βx
α
();0nxα=Example : Evaluate 2
xzd z ∫ where () []2,0 ,itxt t t =+ ∈ ππ.
Solution : Given integral is 2
xzd z ∫.
Here, ()2fz z=
()33zFz = is primitive of f.
0,ab= = π.
By previous theorem,
() ( ) ()baxfz d z Fx t=∫
220xitzd z ftπ = +π∫3201.3ittπ =+ π 32103i π = π+ − π()313i= π+ π
()3313i π = +
Definition : The index of curve or winding number : If x i s a
closed rectifiable curve in a n d i f {}xα∉ , then,
()1,2xdzxiz η α= π −α ∫ is called the index of x w.r.t the point α.
It is also called the winding number about α.
();1xη α= ();2xη α= ();1xcη = −
();0xη β= ();1xη β=
Fig. 7.3
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x
αr() 2nx;α=
Fig 7.4
Theorem : If []:0 , 1x → is a closed rectifiable curve and {}xα∉
then ();xη α is an integer.
Proof : Define []:0 , 1g → by,
()()
()[]
00,1txs d sgt txs′ = ∈ −α ∫ ∴ ()00g=, g is continuous on []0,1 and
()1
xdzgz= −α ∫ ( * )
( p u t ()xs z= ()xs d s d z′⇒ =)
To prove that ()12gi n=π for some integer η. ∴ f r o m e q u a t i o n ( 1 ) ,
()()()[],0 , 1xtgt txt′′= ∈ −α
Now, ()()()()() ()()()()()gt gt gt dex t e x t x t e g tdt− − − ′′ −α =+ −α −
()() ()()gtex t x t− ′ = −− α()
()xt
xt −α ()() ()gtex t x t− ′′ = − ().0gte−= 0= ∴ ()()().gtex t− −α = constant = K (say)
OR ()().gtxt K eα−= (2)
To find K, put 0t=
()()0 00gxK e K e K−α= = = {}01e =∵ ∴ ()0 Kx= −α
() []{}:: , 0 , 4itxxt zz r e t = −α= ∈ πit itzr e r e r −α = = =
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Putting the value of K in (2), we get
() ( ) ()()0.gtxt x e−α= −α
Put 1t=,
()()()1 1
0g xex −α =−α 1= (∵x is closed ()()10xx= ) ∴ ()1 2 gineeπ = (∵21ineπ =)
()12gi n=π
For some integer n substitute the above value in (*), we get
2
xdzinz π = −α ∫ ∴ ()1;2xdznn xiz= α= π −α ∫
Component of a Metric space:
A subset D of a metric space X is a component of X if D is a
maximal connected subset of X i.e. D is connected and there is no
connected subset of X that properly contains D.
D1
23
4DD
D{}Gx=
Fig 7.5
Note: If G is open then component of G also open.
Simply and multiply connected domains :
Definition : A domain D is said to be simply connected if any
simple closed curve which lies in D can be shrunk t o a point without
leaving domain D.
Definition : A domain which is not simply connected is said to b e a
multiply connected domain.
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Green’s Theorem :
L e t (),Mx y and (),Nxy be continuous and have continuous
partial derivatives in a domain Ω a n d o n i t s b o u n d a r y x, Green’s
theorem states that,
xNMMd x Nd y d x d yxyΩ ∂ ∂ + = − ∂∂∫∫ ∫
Cauchy Theorem :
Let G be an open set in and suppose that []:,xa b G → is a
smooth curve. If f is analytic with f′ continuous inside and on a
simple closed curve x then, ()0xfz d z =∫. ΩxGf
Fig 7.6
Proof : Let zx i y=+, ∴ ()()()fz u z i v z=+
and Int xΩ= ∴ ()()()()xxfz d z u z i v z d x i d y = + + ∫∫ ()xxudx vdy i vdx udy= − ++ ∫∫ (1)
G i v e n t h a t , f is analytic in Ω and on its boundary x. ∴ f is continuous in Ω and hence u and v are continuous.
Also, given that, f′ i s c o n t i n u o u s i n s i d e a n d o n a s i m p l e c l o s e d
curve x. ∴ P a r t i a l d e r i v a t i v e s o f u and v are also continuous in Ω and on
its boundary x. ∴ B y G r e e n ’ s t h e o r e m
xvuud x vd y d xd yxyΩ− ∂ ∂ − = − ∂∂∫∫ ∫ munotes.in
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and ,xuvvd x ud y d x d yxyΩ∂∂ + = − ∂ ∂ ∫∫ ∫
Substituting above values in equations (1), we get
()
xvu uvfz d z d x d y i d x d yxy x yΩΩ −∂ ∂ ∂ ∂ = − + − ∂∂ ∂ ∂ ∫∫ ∫ ∫ ∫
Given that, f is analytic. ∴ C a u c h y R i e m a n n e q u a t i o n s a r e s a t i s f i e d .
uvxy ∂ −∂ =∂∂ and vvyx ∂ − ∂ =∂∂ ∴ ()
xuu vvfz d z d x d y i d x d yyy yyΩΩ ∂ ∂ ∂ ∂ = − + − ∂∂ ∂∂ ∫∫ ∫ ∫ ∫
()0xfz d z =∫
Note : In Cauchy’s theorem, Cauchy assumed the continuity of
derived function ()fz′. It was Goursat who first proved that this
condition can removed from the hypothesis in the th eorem. The
revised form of the theorem is known as Cauchy- Gou rsat theorem
which we shall study in the next chapter.
8.4 SUMMARY
1) () ( ) ( ) ( ) ( )1 1
1j b n
j xa ifz d z fx t xt d t fx t xt d t+−
=′′== ∑ ∫∫ ∫
()xx xfz d z u d x v d y iu d y v d x = − + + ∫∫ ∫
2) If []:,xa b → is a given curve then, the opposite curve x− to x
is defined as
()()[];,xt xa b t t ab− =+ − ∈
3) If x i s p i e c e w i s e s m o o t h t h e n , t h e r e i s a p a r t i t i o n
01 ...nPat t t b÷= << < = of [],ab s.t. the restriction kx of curve x to []1,kktt− is smooth for
12 2() () ()
xx xfz d z fz d z fz d z h
−∴ =− =<∫∫ ∫εδ δ
0
00 0()( ) ( ) ( ) ( ) ( ) ( ) ( )z zh z zh zh
zz z z zFz h F z f z fw d w fw d w fw d w fw d w fw d w+ + +
′ +− = − =+ = ∫∫∫∫∫ munotes.in
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()()()()()12 1 2...
xx x x xx xn nfz d z fz d z fz d z fz d z fz d z++ = = + + + ∫∫ ∫ ∫ ∫0() ( ) 1 ( )lim ( ) ( )
11() ( ) (() ( ) )11 10( ) ( ) .zh
h
z
zh zh zh
zz z
zh zh
z zFz h Fz f zfz fw d w hhh hfw d w fz d w fw fzd whh
fw fz d w d w hhh h+
→
++ +
+ ++−== −
= − = −
=≤− <= =∫
∫∫ ∫
∫∫εε ε
1kn≤≤.
12 ...nxx x x=+ + +
4) L e t []:,xa b → a n d []:,cdσ → b e t w o s m oo t h c u r v e . Th e n
the curve σ i s e q u i v a l e n t t o c u r v e x i f , t h e r e i s a f u n c t i o n [][]:, ,cd ab∅ → which is contain non-decreasing and with ()ca∅ =
and ()db∅ = s.t. xσ= ∅.
5) Fundamental theorem of calculus :
If f i s c o n t i n u o u s o n [],ab a n d ()()Fx fx′= i n [],ab t h e n () () ()b
afx d x F b F a= − ∫ .
6) Primitive or Antiderivative of a function :
A function :fΚ is said to be Primitive or Antiderivative
of f in G if, F is analytic in G and ()()Fz fz′= in G.
7) The index of curve or winding number :
If x i s a c l o s e d r e c t i f i a b l e c u r v e i n a n d i f {}xα∉ , then,
()1,2xdzxiz η α= π −α ∫ is called the index of x w.r.t. the point α.
It is also called the winding number about α.
8) A domain D is said to be simply connected if any s imple closed
curve which lies in D can be shrunk to a point with out leaving
domain D.
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8.5 UNIT END EXERCISES
1) Suppose 22()fz x i y=+ w h e r e zx i y=+. Then evaluate ()efz d z∫ , where
:( ) , 0 1cz t t i t t=+ ≤≤.
Solution: Consider :( ) , 0 1cz t t i t t=+ ≤≤. Then () 1zt i=+, and
12 2
02() ( ) ( 1 )3cifz d z t i t i d t=++ =∫∫ .
2) Find the integral of the function
taken over a circle
of radius
.
Solution: 22221()xi yfzzxyxy== −
++∵ :) ) c o s ( ) s i n ( ) , 0 2, 0cz t R t i R t t Rπ=+ ≤≤ >∵. Then
()2
0cos( ) sin( )() s i n ( ) c o s ( ) 2cti tfz d z R t i R t iRRππ = −− += ∫∫
3) Let c b e a n y s m o o t h c u r v e i n . L e t () 1fz=. Then find ()cfz d z∫ .
4) Let cbe the Unit circle and suppose that 1f≪ o n c. Then
prove that () 2cfz d zπ∫≪.
(Hint: 1, 2MLπ==. Apply ML formula.)
5) Let cbe any closed curve not passing through the o rigin ,
then show that
210: 0 , 1 ,k
ccdz z dz k kz== ≠− ∫∫ is any integer.
(Hint: 21() ()gz F zz′== ∵ w h e r e 1()Fzz=− a n d ()Fzis analytic
everywhere except at the origin .)
6) Evaluate ()czi d z−∫ , where cis the parabolic segment
2:( ) ,1 1cz t t i t t=+ −≤≤
Solution : L e t ()fz z i=−. Then fis the derivative of an analytic
function 2
()2zFz i z= −. munotes.in
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∴By proposition , ( ) ( ( )) ( ( ))cfz d z F z b F z a= − ∫ ∴22
11() 022iiczzzi d z i z i z
−++ − = −− − = ∫
() 0czi d z− =∫∵.
7) Find 1
z
yzd z∫ . Where (a) γ is the upper half of the Unit circle
from +1 to -1.
(b) γ is the lower half of the Unit circle from +1 to -1.
(Hint: (a) Let () c o s () s i n () , 0tt i t tγπ=+ ≤≤
(b) Let () c o s () s i n () , 2tt i t tγπ π=−− ≤ ≤
8) Let () 2itteγ= , for tππ−≤≤. Find 21(1 )zd zγ−− ∫ .
9) Prove the following integration by parts f ormula . Let fand gbe analytic in G, let γ be a rectifiable curve from ato bin G.
Then show that () () () ()fg f b g b f a g a f gγγ′′ = −− ∫∫ .
10) Evaluate the integral ()sinzze z z d z−+ ∫ γ w h e r e γ i s t h e
circle 2z= .
*****
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8
CAUCHY THEOREM
Unit Structure
8.0. Objectives
8.1. Introduction
8.2. Cauchy Theorem for an Open Star shape d Domain
8.3. Cauchy Integral Formula
8.4. Summary
8.5. Unit End Exercises
8.0. OBJECTIVES
Our main goal in this unit is to show that a function
analytic in a disc can be represented as a power series. We
shall prove Cauchy’s theorem for an open star- shaped domain
and Cauchy integral formula for an analytic function
in a
disc.
8.1. INTRODUCTION
We have seen that a function is analytic o n a closed
curve cbut 0cf≠∫ . For example consider the function
:0f−→ d e f i n e d a s 1()fzz=. In this example
111() 2zzfz d z d z izπ==== ∫∫ . Whereas , the closed curve theorem
states that if fis analytic throughout a disc, the integral
around any closed curve is 0. We shall try to find the most
general type of domain in which the closed curve theorem is
valid. We should note that 1()fzz= is analytic in the punctured
plane 0−. We shall see that the existence of a hole a t 0z=
allows us to construct an example above , for which the
integral is non-vanishing. The property of a domain, which
assures that it has no holes is called simple connectedness.
The formal definition is as follows.
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Definition : A domain D is said to be simply connected if any
simple closed curve which lies in D can be shrunk t o a point without
leaving domain D.
Definition : Singular point :
A point at which the function f i s n o t a n a l y t i c i s s a i d t o b e a
singular point or singularity of the function f.
e.g. ()23zfxz= −
H e r e , f i s n o t d e f i n e d a t 3z= a n d h e n c e n o t a n a l y t i c a t 3z=, therefore 3z= is singular point.
8.2 CAUCHY THEOREM FOR AN OPEN STAR
SHAPED DOMAIN
Theorem : Let G be star like w.r.t. point 0z and suppose that f is
analytic in G. Then there exists an analytic functi on F in G s.t. ()()Fz fz′= in G.
In particular,()0xfz d z =∫, for every closed, piecewise smooth curve
x in G. (2008)
Proof : Given that, G is a starlike w.r.t point 0z and f analytic in G. ∴ B y d e f i n i t i o n , []0,zz G zG ⊂ ∨ ∈
Fix a point z in G and define
() ()
[]()
,00z
zz zFz f d f d = ξ ξ= ξ ξ∫∫
z0zz + h
Fig 8.1
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Choose h∈ with 0h > s.t. ()();Bz h G ⊂ and [],zz h G + ⊂
Since G is starlike w.r.t. point 0z. ∴ T h e t r i a n g l e []01,,zz zh∆= + is contained in G. ∴ f is analytic inside and on the boundary of the tri angle ∆. ∴ B y C a u c h y G o u r s a t t h e o r e m ,
() () ()00() ()0z zz h
zz z hfz d z d
fd fd fdδδ
+
+= ξξ ∫∫ ξξ+ ξξ+ ξξ =∫∫ ∫△△
() () ( )0zhzFz f d Fz h++ ξ ξ− +=∫ ∴ () ( ) ( )zhzFz h Fz f d++− = ξ ξ∫ ∴ () ( ) ( ) ( ) ( )zh zhzzFz h Fz h fz f d f zd+++−− = ξ ξ − ξ∫∫
() ( )() () ()1zhzFz h Fzfz f fz dhh+ + − − = ξ− ξ ∫
() ( )() () ()1zhzFz h Ezfz f fz dhh+ + − − ≤ ξ− ξ ∫ (1)
Given that, F is analytic in G. ∴f is differentiable in G. ∴f is continuous at a point zG∈.
For a given 0, 0ε> ∃ δ> s.t.()()ff z ξ− <ε whenever z ξ− <δ
Choose zh ξ− = < ε ∴ F r o m e q u a t i o n ( 1 ) ,
() ( ).zhzFz h Fzdhhh h+ + − εε < ξ = ∫
By definition ()()()
0lim
hFz h Fzfzh → + − =
()()Fz fz′= in ();Bz h
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Since z is fixed but arbitrary. ∴ ()()Fz fz′= in G. (2) ⇒The derivative F′ exists and is continuous at every point z in G. ∴ F is analytic in G.
From equation (2),
F is primitive of f. ∴ By using the theorem ()0xfz d z =∫ for every closed, piecewise smooth curve x in G.
In Cauchy’s theorem, Cauchy assumed the continuity of derived
function ()fz′. It was Goursat who first prove that this conditi on
can removed from the hypothesis in the theorem. Th e revised form
of the theorem is known as Cauchy-Goursat theorem.
Cauchy-Goursat Theorem : (Cauchy Triangular Theorem ) :
Let f be analytic in an open set GD⊂. Let 123,,zz z be points in
G. Assume that the triangle ∆ with vertices 123,,zz z is continuous
in G then ()0xfz d z =∫ where ∂∆ is the boundary of a triangle ∆.
( 2 0 0 7 , 2 0 0 8 , 2 0 0 9 )
Proof : Given that, the triangle ∆ w i t h v e r t i c e s 123,,zz z i s
continued in G. Let 123,,mm m b e m i d p o i n t s o f l i n e s e g m e n t [][]12 23,,,zz z z , []31,zz respectively.
T h e n w e g e t 4 s m a l l e s t t r i a n g l e s 1234,,,∆ ∆ ∆ ∆.
z
123
m
z
zm
m12
3
123
4
Fig 8.2 munotes.in
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∴()()11 2 233 1zmz m z m zfz d z fz d z∂∆ = ∫∫
()()() ()
() ()31 1 3 12 2 1
23 3 2 3112 23mz m m mzmm
mzmm mm
mm m mfz d z fz d zfz d z fz d zfz d z fz d z= + + + + + ∫∫
∫∫
∫∫ ()()()()1234fz d z fz d z fz d z fz d z∂∆ ∂∆ ∂∆ ∂∆= + + + ∫∫∫∫
()4
1K Kfz d z=∂∆= ∑∫
Among this 4 triangles, there is one triangle, call it ∆, s.t.
() ()
11, 2, 3, 4Kfz d z fz d z K
∂∆∂∆ ≥ = ∫∫ ∴ () ()
14fz d z fz d z∂∆ ∂∆ ≤ ∫∫
Let ()L∂∆ be the perimeter of a triangle ∆, then
() () 112LL∂∆= ∂∆ and diam. 112∆= diam. ∆ (diameter of triangle
means the length of its largest side)
Now, perform the same process on the triangle 1∆ getting a triangle 2∆ with analogus properties.
(i) 012∆=∆⊃ ∆⊃ ∆
(ii) () ()2
24fz d z fz d z∂∆ ∂∆ ≤ ∫∫
(iii) () () 2 212LL ∂∆ = ∂∆
(iv) 2 212diam diam ∆= ∆
Continue the process and at the nth stage, we get
(i) 012 ...n∆=∆⊃ ∆⊃ ∆⊃ ⊃∆ munotes.in
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(ii) () () 4n
nfz d z fz d z∂∆ ∂∆ ≤ ∫∫
(iii) () ()12n nLL ∂∆ = ∂∆
(iv) 12n ndiam diam ∆= ∆
A m e t r i c s p a c e × i s c o m p l e t e i f f f o r a n y s e q u e n c e {}nF o f
non-empty closed sets with 123...FF F⊃⊃⊃ a n d 10,nnndiam F F∞= → ∩ consists of a single point.
Since i s com p l e t e a nd {}n∆ i s a se q u e n c e o f n on - em p ty c l o s e d
sets with 01...∆=∆⊃∆ ⊃
and 1lim lim2nnnnndiam diam
→∞ →∞ ∆= ∆
0= {}lim 0 1n
nxi f x
→∞ = <∵
By using cantor’s theorem, 0nn∞= ∆∩ consists of a single point say 0z.
In particular 0zG∈∆ .
G i v e n t h a t , f is analytic on G. ∴f is differentiable at a point 0zG∈ for every 0, 0ε> ∃ δ> s.t.
()()()0
0
0fz fzfzzz − ′ − < ε− where 0 0 zz< − <ε
By increment theorem,
()()()()()()00 00fz fz z z f z z z z′ =+ − + −η w i t h ()zη i s
continuous and ()z η <ε for 0zz − <δ. ()()()()()()00 0 0
nn n nfz d z fz d z z z f z d z z z z d z∂∆ ∂∆ ∂∆ ∂∆′ = + − + − η ∫∫ ∫ ∫
()()0 00
nzz z d z∂∆=++ − η ∫ (∵()0xfz =∫ i f x i s
closed cure and here n∂∆ is closed.)
() () 0
nnfz d z z z z d z∂∆ ∂∆ ≤ − η ∫∫ munotes.in
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Choose η s.t. 12n ndiam diam ∆= ∆<δ . ∵ ()00;nn zB z∈∆ ⇒ ∆⊂ δ
δn∆
Fig 8.3 ∴ () n
nnfz d z d i a m d z∂∆ ∂∆ < ε ∆ ∫∫
(∵ ()2 η <ε and 0nzz d i a m − < ∆)
()nndiam L=ε ∆ ∂∆
(∵()2n
nLd ∂∆∂∆ = ∫)
()11..22ndiam L =ε ∆ ∂∆
(∵12n ndiam diam ∆= ∆ and () ()12n nLL ∂∆ = ∂∆ ∴ () () 4.n
nfz d z d i a m L
∂∆ < ε ∆ ∂∆∫ ∴ () ().
nfz d z d i a m L
∂∆ < ε ∆ ∂∆∫ ( b y ( i i ) , ∵() () 4n
nfz d z fz d z∂∆ ∂∆ ≤ ∫∫)
Since 0ε> is arbitrary. ∴ ()0fz d z
∂∆ =∫
munotes.in
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Check your progress:
1) State and prove Cauchy Goursat Theorem for a cl osed
quadrilateral.
2) State and prove Cauchy Goursat Theorem for a clo sed Rectangle.
Theorem : Let G be an open set in . Let f be analytic in G except
possibly at a point 0zG∈. Assume that f is continuous in G and that
the triangle ∆ w i t h v e r t e x a t 0z i s c o n t a i n e d i n G . T h e n () 0fz d z
∂∆=∫, where ∂∆ is a boundary of the triangle ∆.
Proof : Let []012,,zz z∆= .
Let 1ξbe a point on the line segment []01,zz and 2ξ be a point on
the line segment []02,zz .
z
12
3z
z12
3
1ξ2ξ
Fig 8.4
Consider the subtriangles, [][][]12 0 121 2 2 31 1 2,, , ,, , , ,zz z z∆=ξ ξ ∆= ξ ∆=ξ ξ ∴()()201 122zz zfz d z fz d z∂∆ ξ ξ ξ = ∫∫()()()2012 1 121 1221zz z zfz d z fz d z fz d zξξ ξ ξ ξ ξ ξ ξ= + + ∫∫∫ ()()()123fz d z fz d z fz d z∂∆ ∂∆ ∂∆= + + ∫∫∫()10. . .fz d z
∂∆= + +∫
(by Cauchy Goursat theorem) munotes.in
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∴ () ()
1fz d z fz d z∂∆ ∂∆ ≤ ∫∫
Put ()maxMf zz∆= ∈∂ ∴ () ()11fz d z M d z M L
∂∆ ∂∆ ≤ = ∂∆∫∫
As 1ξ and 2ξ tend to 0z, perimeter of the triangle 1∆ tends to zero
i.e. ()10L∂∆ →. ∴ ()0fz d z
∂∆ =∫
2ξ1ξ0z1
Fig 8.5
Theorem : Let G be an open set in . Let f be analytic {}G α
for some Gα∈. If f is continuous on G, then ()0fz d z
∂∆ =∫, where ∂∆ is a boundary of the triangle ∆ contained in G.
Star Shaped Domains :
Definition : A set G i n i s s a i d t o b e c o n v e x i f , g i v e n a n y t w o
points z and w in G, the line segment [],zw lies entirely in G.
Fig 8.6 munotes.in
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Definition : A set G i s s a i d t o b e s t a r l i k e ( o r s t a r s h a p e d ) w . r . t .
points Gα∈ if for any point ()z≠α in G, the line segment [],zα
lies entirely in G.
α α
αz
The above set is starlike w.r.t. αααα α
Fig 8.7
Punctured disk is not starlike w.r.t .αααα
Note : Every starlike set is not convex but every convex s et is
starlike.
Question : If f is analytic in a simply connected domain D, then
()2
1 ∫z
zfz d z i s i n d e p e n d e n t o f p a th i n D, joining any two points 1z
and 2z in D.
Solution : Let 1x and []2:,xa b G → be two smooth paths in G such
that ()()121xa x a z== and ()()122xb x b z==. ()()()11 2 2 12,, ,xt x t tt a b≠ −
z
12
zY
Xx12xσD
Fig 8.8 Open sets, closed, half planes if >>>>>>>>Re 0, Im 0zz etc. are
munotes.in
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Form a simple closed curve σ which m oves from 1z to 2z via 1x
and 2z to 1z via 2x−. ∴ f is analytic inside and on a simple closed curve σ.
By Cauchy Goursat theorem,
()0fz d z
σ =∫ ∴ ()()120xxfz d z fz d z
− + =∫∫ ∴ ()()12xxfz d z fz d z− = − ∫∫ ∴ () () ()2
121z
xx zfz d z fz d z fz d z = = ∫∫∫
()2
1z
zfz d z ∫ is independent of path.
Cauchy Deformation Theorem :
Statement : If f is analytic in c domain bounded by two simpl e
closed curves 1x a n d 2x ( w h e r e 2x is inside 1x) and on these
curves, then ()()12xxfz d z fz d z = ∫∫ w h e r e 1x a n d 2x a r e b o t h
traversed in anticlockwise direction.
Proof : Join two curves 1x and 2x by lines AB and CD. Denote, 1xℓ = lower section of 1x from A to D.
A
BC
D1ux1x2ux2x2xℓ1xℓ
Fig 8.9 munotes.in
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1xu = the upper section of 1x from D to A. 2xℓ = the lower section of 2x from B to C. 2xu = the upper section of 2x from C to B.
Form a simple closed curve 1σ by transversing from A to B
then from B to C by 2xu, then from C to D and finally from C to D
and finally from D back to A by 1xu. ∴ f is analytic inside and on the simple closed curve 1σ. ∴ B y C a u c h y - G o u r s a t t h e o r e m .
()10fz d z
σ =∫ ∴()()()()210AB x u CD x ufz d z fz d z fz d z fz d z
− + + + =∫∫∫ ∫ (1)
From a simple closed curve 2σ by transversing from A to D by 1xℓ
then from D to C, then from C to B by 2x− ℓ and finally from B back
to A. ∴ f is analytic inside and on the simple closed curve 2σ. ∴ B y C a u c h y - G o u r s a t t h e o r e m ,
()20fz d z
σ =∫ ∴()()()()120xD Cx B Afz d z fz d z fz d z fz d z
− + + + =∫∫ ∫ ∫ℓℓ (2)
Adding equations (1) and (2), we get
() () () ()
11 220xu x x x ufz d z fz d z fz d z fz d z
+ − + =
∫∫ ∫∫
ℓℓ
() ()
xxfz d z fz d z− = − ∫∫∵ ()()120xxfz d z fz d z − =∫∫ ∴ ()()12xxfz d z fz d z = ∫∫
Generalization of Cauchy Deformation Theorem :
Statement : If f is analytic in a domain bounded by non-inter secting
simple closed curves 1,, . . . ,nxx x w h e r e 1,, . . . ,nxx x a r e i n s i d e x an d
on this curves, then munotes.in
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()()()()12...
xx x x nfz d z fz d z fz d z fz d z = + + + ∫∫∫ ∫ w h e r e ,
12,, , . . . ,nxx x x are traversed in anticlockwise direction.
a
b1
23
4
n-1
nbaaa
Fig 8.10
8.3 CAUCHY INTEGRAL FORMULA
Statement : Let f b e a n a l y t i c i n a s i m p l y c o n n e c t e d d o m a i n G, GD⊂. If x is a simple closed curve in G and be any point inside x
then, ()() ( ) 22z
xfzdz i f z i fz = α = π =π α −α ∫ w h e r e , x i s t r a v e r s e d i n
anticlockwise direction.
Proof : Given that, f is analytic in a simply connected domain G.
construct a circle Γ with centre at α and radius 0r> so that Γ lies
entirely inside x.
r
z
Γα
Fig 8.11 munotes.in
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∴ T h e f u n c t i o n ()fzz −α is analytic in a domain which is bounded
by two simple closed curves x and Γ and on these curves. ∴ B y C a u c h y d e f o r m a t i o n t h e o r e m ,
()()xfz fzdz dzzzΓ = −α −α ∫∫ ()()()fz f fdzzΓ −α + α = −α ∫
()()()()
xfz fz fdzdz dz fzz zΓΓ −α = + α −α −α −α ∫∫ ∫ (1)
Consider, dzzΓ −α ∫ ∴ E q u a t i o n o f t h e c i r c l e Γ is, zr −α =
or [],0 , 2itze r t=α+ ∈ π
itdz i re dt=
zr −α =
[],0 , 2izrr eθ=+ θ∈ π
[],0 , 2itre t=α+ ∈ π
rαα+θα+ireθ
r
Fig 8.12
∴ 22
002it
itdz iredt i dt iz reππ
Γ = = = π −α ∫∫ ∫ ∴ E q u a t i o n ( 1 ) , b e c o m e s ,
()()()() 2
xfz fz fdz dz i fzzΓ −α = + π α −α −α ∫∫
()()()()11
22xfz fz fdz f dziz i zΓ −α −α = π −α π −α ∫∫ ∴ ()()() ( ) 11
22xfz f fzdz f dziz zΓ −α −α ≤ π −α π −α ∫∫ ( 2 ) ()22 2 1ii i π = π = π = ∵∵
munotes.in
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Given that, f is analytic in G. ∴ f is differentiable in G. ∴ f is continuous in G. ∴ f is continuous at a point Gα∈. ∴ F o r a g i v e n 0ε>, 0∃ δ> such that,
() ( ) zf z f −α <δ ⇒ −α <ε
Choose, r s.t. r < δ ∴ F r o m e q u a t i o n ( 2 )
()()1
22xdzfzdz fiz rΓ ε −α < π −α π ∫∫ {}z −α=γ∵
2dzrΓε= π ∫().2LTrε= π .22rrε= π π ∴ ()()1
2xfzdz fiz −α <ε π −α∫ ∵ ε is arbitrary. ∵ ()() 2
xfzifz = π α −α ∫
Theorem : Let f be analytic in a simply connected domain G.
G⊂. If x is closed rectifiable curve in G and {}Gxα∈ , then ()() ( )1;2xfzdz f xiz = α η α π −α∫.
Proof : Given that, f is analytic in a simply connected domain G.
Define ()Fz as follows,
()()()(),
,fz fzzFz
fx z −α ≠α −α=
′ = α ( 1 )
()()()fz fFzz −α =−α is analytic in {}G α.
(∵()()fz f−α and 1z −α are analytic in {}G α) ∴ ' () ( )lim ( ) lim ( ) ( )zzfz ffz f Fz→α →α−α== α= α−α b y ( 1 ) ∴F is continuous at a point Gα∈ and hence F is continuous in G. ∴ F is analytic in G. munotes.in
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∴ B y C a u c h y - G o u r s a t t h e o r e m ()0xFzd z =∫
()()0xfz fdzz −α =−α ∫
()()
xxfzdzdz fzz = α −α −α ∫∫
()() ( ) ( ) ()1.2 , . . . ;2xfzdz f in x xz i = α π α η α= −α π ∫
()() ( )1.;2xfzdz f xiz = α η α π −α ∫
Example 1 : Use Cauchy integral formula, to evaluate 22
1xzdzz + −∫
where x is a circle 2z =.
Solution : Given that, 22
1xzdzz + −∫
()()2211fz zFzzz + ==−− x12 0
Fig 8.13 ∴ F has singular point at 1z=.
Given equation of circle is 2z =. ∴ C e n t r e i s o r i g i n a n d r a d i u s i s 2 . ∴ T h e s i n g u l a r p o i n t 1z=, lies inside the circle. ∴ W e u s e C a u c h y i n t e g r a l f o r m u l a
()() 2zxfzdz i f zz=α = π −α ∫
2212221zxzdz i zz= + = π +−∫[]()21 2 2 3ii=π+= π6i=π . munotes.in
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2) Evaluate cot
xzd z ∫ where x is a circle 12z= .
Solution: coscotsinxxzzd z d zz = ∫∫
For finding singular point put sin 0z=
()1sin 0zn− ⇒ == π
Singular point z=0, lies inside the circle 12z = .
Hence by Cauchy’s integral formula ()cos2c o s 02sinxzdz i izππ== ∫
3) Evaluate sin( )12czzπ−∫ where C is the unit circle oriented clockwise.
Solution: Let I=sin( )12czzπ−∫ 12z= is a point of singularity and lies inside 1z= ()()01, sin2zf z zπ∴==
()()0 22 s i n2Ii f z iπππ ∴==
Theorem : Let f be analytic in a simply connected domain G,
G⊂ a n d s u p p o s e x i s a s i m p l e c l o s e d c u r v e i n G . I f α i s a n y
point inside x then,
()()()21
2xfzfd zi z ′α= π −α∫ , w h e r e x i s t r a v e r s e d i n
anticlockwise direction.
(Note: To prove the theorem we will need the following t heorem
Boundedness Theorem :Let f be continuous on a com pact set S.
then, f i s b o u n d e d o n S i . e . t h e r e e x i s t s a n u m b e r M
s.t.()fz M z S ≤ ∨ ∈)
Proof : Given that, x i s a s i m p l e c l o s e d c u r v e i n G a n d α i s a ny
point inside x.
By Cauchy-integral formula,
()()1
2fzfd zizα= π −α ∫
and1( )()2( )xfzfh d ziz hαπα+=−+ ∫ munotes.in
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αα+hzx
Fig 8.14 ∴ () ( )() ()
()1
2xxfz fzfh f d z d ziz z h α+−α = − π −α −α+ ∫∫
() ( ) () ( )
() ()1
2xzh f z z f zi zz h −α+ −− α = π −α −α+ ∫
()
() ()1
2xhf zdzi zh z = π −α+ −α ∫ ()()()() ()1
2xfh f f zdzhi zh z α+−α = π −α+ −α ∫
() ( ) () ( )
() ()21
2xzf z z h f zdzi zh z −α − − α + = π −α+ −α ∫
()() ()21
2xhf zdzi zh z = π −α+ −α ∫()() ()2 2xfz hdzi zh z = π −α+ −α ∫
() ( ) ()
()()
() ()221
22xxfz d z h fh f f zdzhi zz h z α+−α − = π π −α −α+ −α∫∫
(1)
Theorem will be proved if L.H.S. of equation (1) te nds to zero as 0h→. Choose {}inf :rz z x r z= −α ∈⇒ ≤ −α .
Choose ()2rhh α− α + = < ∴ ()zh z h −α+ ≥ −α − {}ab a b − ≤ − ∵
22rrr > −= munotes.in
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∵ f is analytic in G. ∴ f is continuous on {}x {}2is compact setx ∵ ∴ B y b o u n d e d n e s s t h e o r e m , 0M∃ > s.t.
() . Vfz M ≤ {}zx∈ ∴ F r o m e q u a t i o n ( 1 ) ,
() ( ) ()
()22. 1
22 .2xxhM d zfh f f zdzr hirz α+−α − ≤ π π −α∫∫
3.
xhMdzr = π∫()3.hMLxr = π
L.H.S. tends to zero as 0h→. ∴ ()()()()201lim2 hxfh f f zdzhi z → α+−α = π −α ∫
∴()()()21
2xfzfx d zi z ′= π −α ∫
Generalization of the above theorem :
Theorem : Let f be analytic in a simply connected domain G,
G⊂ a n d s u p p o s e x i s a s i m p l e c l o s e d c u r v e i n G . I f α i s a n y
point inside x, then
()()()1!
2n
n
xfz nfd zi z+α= π −α ∫
where 0,1,n= 2 , . . . x is traversed in anticlockwise direction.
Prove this theorem by induction on n.
Note : If a function f i s a n a l y t i c a t a p o i n t Gα∈, then its
derivatives of all orders are also analytic at a po int α.
Example 1 :
Use Cauchy integral formula or theorem, to evaluate
() ( )sin cos
12xzzdzzz π+ π − − ∫ where x is a circle 32z = .
Solution : Given integral () ( )sin cos
12xzzdzzz π+ π − − ∫
munotes.in
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By partial fraction,
() ( )sin cos sin cos sin cos
12 1 2xx xzz zz zzdz dzzz z z π+ π π+ π π+ π = + −− − −∫∫∫.
Here, () ()()() ( )12fzFz Fzzz== − − h a s s i n g u l a r p o i n t a t 1z=
and 2z=.
(Note : If the singular points lies inside the circle then we use
Cauchy integral formula and it lies outside the cir cle then we use
Cauchy integral theorem.)
G i v e n e q u a t i o n o f c i r c l e , 32z = i.e. 2294xy += ∴ S i n g u l a r p o i n t 1z= lies inside the circle and 2z= lies outside
the circle. ∴ F o r 1z=, we use Cauchy integral formula
()() 2zxfzdz i f zz=α = π −α∫
[]1sin cos2s i n c o s1zxzzdz i z zz= π+ π = ππ + π− ∫ []2s i n c o si=ππ + π[]20 1i=π −
2i=−π … . ( a ) ∴ F o r 2z=, we use Cauchy integral formula theorem ∴ ()0xFzd z =∫ ∴ sin cos02xzz
z π+ π =− ∫ ( b ) ∴ S u b s t i t u t i n g ( a ) a n d ( b ) i n e q u a t i o n ( 1 ) , w e g e t
() ( )sin cos02 212xzzdz i izz π+ π = −π =− π− − ∫
2) 2
31
2z
xedzz − ∫ w h e r e x i s t h e r e c t a n g l e w i t h v e r t i c e s a t i±
and 1i±.
Solution : Here, ()()312fzFz
z = − has singular point at 12z= of
order 3.
munotes.in
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i
-i(1,1)
1+i
1-i1/20
Fig 8.15
By Generalization of derivative of an analytic func tion.
()()()1!
2n
n
xfz nfd zi z+ α= π −α∫ ∴()()()12
!n
n
xfz idz fn z+ π = α−α∫ ∴32
1z
xedz
zz =
− ∫2iπ 212!f ()22
212zdie z
dz=
=π×
2124z
zie= =π 1224ie×=π 4ie=π
Exercise : Use Cauchy integral formula to evaluate,
i) 2cos1xzz π −∫ w h e r e x i s a r e c t a n g l e w i t h v e r t i c e s a t 2i± a n d 2i− ±.
ii) tan
xzd z ∫ where x is a circle 22zπ π − = .
iii)
()32z
xezdzz − −∫ where x is a circle 3z =.
iv) 21
12xzdzzi + −− ∫ where x is a circle 31z − =.
Singular point is 12Pi=+ a n d h e r e c e n t r e ()3, 0c=
() ()22 11 2,3 , 3 0 , 1 122 4pdcp d i = += − + − = > .
∴ S i n g u l a r p o i n t p lies outside the circle. munotes.in
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v) 223xzz
zi + + −∫ where x is a circle 2zi − =.
Cauchy estimate or Cauchy inequality :
Theorem : If f is analytic in an open disk ();BRα a n d ()(),fz M z B R ≤ ∨ ∈α
then ()!nnnMfR α ≤ 0,1, 2, ...n=
Proof : For 0rR<<,
Construct a circle x with centre at α and radius γ.
By generalization of the theorem on Derivative of a n analytic
function.
()()()1!
2n
n
xfz nfd zi z+α= π −α ∫ 0,1, 2, ...n= ∴ ()()1!
2n
n
xfz nfd zz+ α ≤ π −α ∫ ( 1 )
Given that,
()fz M ≤ zx∨ ∈
For any point z on x, we have zr −α = ∴ F r o m e q u a t i o n ( 1 ) ,
()()1!
2n
n
xnMfd zr+ α ≤ π ∫ 1!
2n
xnMdzr+= π ∫1!.22nnMrr+= ×π π ∴()()!nnnMfr α ≤ ∵ rR< is arbitrary. ∴ A s rR→, we have ()()!nnnMfR α ≤ 0,1, 2, ...n=
Cauchy Integral formula for Multiply connected doma ins:
Theorem: I t f i s a n a l y t i c i n a d o m a i n w h i c h i s b o u n d e d b y t wo
simple closed curves 12xa n d x ( w h e r e 2x l i e s i n s i d e 1x) and on
these curves and if 0z is any point in G. Then
0
00121( ) 1 ( )()22xxfz fzfz d z d ziz z i z z= − π− π −∫∫
Where 12xa n d x are transverse in anticlockwise direction munotes.in
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Proof: Construct a circle x with centre at 0zand radius r, so that x
lies inside 1x ∴ The function 0()fzzz− is analytic in a domain which is bounded by
non-intersecting simple closed curves 12,,xx x where (12xa n d x lies
inside 1x) and on these curves
r
z0 xxx1
2
Fig 8.16
By using Cauchy determination theorem,
00011 2() () ()xx xfz fz fzdz dz dzzz zz zz=+−−−∫∫∫ (I)
By using Cauchy integer formula, 00()2( )xfzdz i f zzz=π −∫
Put this value in equation (I),
0
0012() ()2( )
xxfz fzdz i f z dzzz zz=π +−−∫∫
0
00121( )1 ( )()22xxfz fzfz d z d ziz z i z z∴ = − π− π −∫∫
8.4 SUMMARY
1) Let G be starlike w.r.t. point 0z and suppose that f is analytic in
G. Then there exists an analytic function F in G s.t. ()()Fz fz′=
in G.
In particular, ()0xfz d z =∫, for every closed, piecewise
smooth curve x in G. munotes.in
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2) Cauchy-Goursat Theorem : (Cauchy Triangular Theo rem) :
L e t f b e a n a l y t i c i n a n o p e n s e t GD⊂. Let 123,,zz z b e
points in G. Assume that the triangle ∆ with vertices 123,,zz z is
continuous in G then ()0xfz d z =∫, where ∂∆ is the boundary of a
triangle ∆.
3) Cauchy Deformation Theorem :
Statement : If f i s a n a ly t i c i n c d o m a i n b o u n d e d by t w o s i m p l e
closed curves 1x a n d 2x ( w h e r e 2x i s i n s i d e 1x) and on these
curves, then ()()12xxfz d z fz d z = ∫∫ w h e r e 1x a n d 2x a r e b o t h
transverse in anticlockwise direction.
4) Statement : Let f be analytic in a simply connected domain G, GD⊂. If x is a simple closed curve in G and be any point inside x
then, ()() ( ) 22z
xfzdz i f z i fz = α = π =π α −α ∫
where, x is traversed in anticlockwise direction.
5) Theorem : Let f be analytic in a simply connected domain G,
G⊂ a n d s u p p o s e x i s a s i m p l e c l o s e d c u r v e i n G. If α i s a n y
point inside x then, ()()()21
2xfzfd zi z ′α= π −α∫ , where x is traversed
in anticlockwise direction.
6) Generalization of the above theorem :
Theorem : Let f b e a n a l y t i c i n a s i m p l y c o n n e c t e d d o m a i n G,
G⊂ a n d s u p p o s e x i s a s i m p l e c l o s e d c u r v e i n G. If α i s a n y
point inside x, then
()()()1!
2n
n
xfz nfd zi z+α= π −α ∫ , w h e r e 0,1,n= 2 , . . . x i s
traversed in anticlockwise direction.
8.5 UNIT END EXCERCISES
1) Suppose :fG→ be an analytic function, define :GG∅× →
by () ( )(, )fz fwzwzw−∅ =− if zw≠
()fz′= if zw=. munotes.in
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Prove that ∅ i s a c o n t i n u o u s a n d f o r e a c h f i x e d w, (, )zz w→∅is analytic function of z.
( Hint: Take 00 ,zz h ww k=+ = + and 00zh wk+≠+for any ,hk.
Consider(,) ( 0 , 0 ) 0 0 0 0lim ( , ) ( , )hkzh wk z w→ ∅ ++ = ∅ , similarly (,) ( 0 , 0 ) 0 0 0 0lim ( , ) ( , )hkzh wk z w→ ∅ ++ = ∅ f o r 00zw= a n d hk= o r hk≠.
2) Let δbe a closed rectifiable curve in and {}aγ∉. Then
show that for 2n≥, () 0nza d zγ−− =∫ .
Solution: Use the lemma that, if γ is a rectifiable curve and
ϕ is a function defined and continuous on {γ}.
For each 1m≥, () ( ) ( )m
mFz w wz d wγ−=∅−∫ for {}zγ∉. Then each mF is analytic on {}γ−and 1mmFm F+′ = . Take 1, 1m=∅= o n {}γ. Then
1
1() ( )Fz w z d wγ−= −∫ . ∴12() ()Fz Fz′ = on {}γ−. Here {}aγ∈−. ∴12() ()Fa Fa′ = ∵ 2() 0Fa=, since 1()Fais constant number
independent of a. ∴2() 0za d zγ−− =∫ Inductively () 0nza d zγ−− =∫ for 2n≥.
3) Let fbe analytic on (0,;1)DB=. Suppose () 1fz≤ for 1z≤.
Then show that
(0) 1f′ ≤
( Hint: fis analytic in a simply connected set (0,;1)B .
Let γ10;2boundary B=, then γis a simple closed curve and
α=0 is a point inside γ.
Consider 11( )(0)2fzfd zizγπ′ = ∫ munotes.in
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∴ 221( ) 1( ) 1(0) .2 122 2fz fzfd zi zzγγπππ π′ = ≤≤ =∫∫ .
4) Let () 1itteγ=+for 02tπ≤≤.Find 1nzdzzγ∫−for all positive
integers n.
( Hint : Put 2()()nfzfz zz=. Then ()() / ( 1 )1n
n zdz f z z dzzγγ= − ∫∫−.
Where () ( 1 c o s () , s i n () )tt tγ=+ f o r 02tπ≤≤. Apply the following
Cauchy- Integral formula
1!( )()(;)2 ()n
nnf zfa a d zi zaγηγπ+= ∫−, for 1n≥.
5) Use Cauchy- Integral formula to evaluate
(i) 2cos( )1nzdzz∫−γ, γ is a rectangle with vertices at 2, 2ii−∓∓
(ii) 3(2 )zezdzzγ−∫−, γ is a circle 3z=.
Solution (ii) Let ()zfz e z= −.
By generalisation of Cauchy formula for d erivative of
analytic function
1!( )()(;)2 ()n
nnf zfa a d zi zaγηγπ+= ∫−, for 1n≥.
Here 2a= a n d w e h a v e 3z= a n d 2n=.
2
32(2) ( ;2)2 (2 )zie zfd zi zγηγπ−= ∫− .
2
3(2)(2 )zezdz if
zγπ−= ∫−, since (; 2 ) 1ηγ=.
(2)()zfz e=. /g19665 2(2)zfe=. 23(2 )zezdz i ezγπ−= ∫−.
(iii)
() ()sin
2zdzzzγ π π−−∫ where γ is the cicle 2z=
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6) Evaluate the integral ()2,2 c o s ( 2 )1idzezθγγθ θ= ∫+ f o r 02θπ≤≤.
7) Use Cauchy-integral theorem or formula to evalua te ()()2cos sin
1rzzdzz π + π
+ ∫ w h e r e γis the circle 2z = , taken in
positive sense.
8) Evaluate ()()cos
2zedzzzγ+∫ , where γis a unit circle.
*****
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9
THEOREMS IN COMPLEX ANALYSIS
Unit Structure
9.0 Objectives
9.1 Introduction
9.2 Morera’s Theorem
9.3 Liouville’s Theorem
9.4 Taylor’s Theorem
9.5 Fundamental Theorem of Algebra
9.6 Summary
9.7 Unit End Exercises.
9.0 OBJECTIVES
In this unit we shall prove the important th eorems in complex
analysis.
1) Morera’s theorem
2) Liouville’s theorem
3) Taylor’s theorem
4) Fundamental theorem of Algebra
9.1. INTRODUCTION
Given an entire function f, we saw that fhas a power
series representation as 0kka∞
=∑ , where each (0)!k
kfak= . In fact
being an entire function the kt h o r d e r d e r i v a t i v e ()kfzexists 0k∀≥.
I n t h i s u n i t w e p r o p o s e t o p r o v e s o m e i m p ortant
theorems in complex analysis .Let us start with the Taylor’s
theorem.
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9.2 MORERA’S THEOREM
Note: this is a sort of converse of Cauchy Goursat thm.
Theorem:
If f(z) is continuous in a simply connected domain D a n d ()xfz d z∫=0 where x is rectifiable curve in D , then f(z) is analytic in
D
Proof : Suppose z is any variable point and z0 is a fixed in the region
D. Also suppose x1 and x2 are any two continuous rectifiable curve in
D joining z0 t o z a n d x is the closed continuous rectifiable curve
consisting of x1 and -x2. Then we have
12() () ()xx xfz d z fz d z fz d z−=+ ∫∫∫ and () 0xfz d z=∫ (given)
12 2() () ()xx xfz d z fz d z fz d z−∴ =− = ∫∫ ∫
i.e. the integral along every rectifiable curve in D joining z0 to z is
the same
Now, consider a function F(z) defined by
0() ( )z
zFz fw d w=∫ ……..(1)
As discussed above (1) depends only on the end poin ts z0 and z
If z+h is a point in the neighbourhood of z, then we
have
0() ( )zh
zFz h fw d w+
+= ∫ …..(2)
From(1)and (2), we have
0
00 0() ( ) ( ) ( ) ( ) ( )z zh z zh
zz z zFz h Fz fw d w fw d w fw d w fw d w+ +
+− = − =+ ∫∫∫∫ ()zh
zfw d w+
=∫ …….(3)
Since the integral on the RHS of (3) is path indepe ndent therefore it
may be taken along the straight line joining z t o z+h, so that () ( ) 1 ( )() ( )zh
zFz h Fz f zfz fw d w hhh h++−− = − ∫
11() ( ) (() ( ) )zh zh zh
zz zfw d w fz d w fw fzd whh++ += − = −
∫∫ ∫ …….(4)
The function f(w)is given to be continuous at x the refore for a given ε>0 there exist δ>0 s.t. () ( )fw fzε− < s.t. wzδ−< munotes.in
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Since h is any arbitrary therefore choosing hδ< so that every point
w lying on the line
joining z to z+h satisfies ….. (5)
From (4)and(5), we have
() ( ) 1() ( ) () .zh
zFz h Fzfz fw fz d whh++−−≤ − ∫
11zh
zdw hhhεε ε+<= =∫
Since ε i s s m a l l a n d p o s i t i v e , t h e r e f o r e w e h a v e
() ( )() 0Fz h Fzfzh+−− = or
0() ( )lim ( )hFz h Fzfzh→+−=
Hence () ()Fz fz′=
i.e. F(z) i s d i f f e r e n t i a b l e f o r a l l v a l u e s o f z i n D . T h e r e f o r e F(z) i s
analytic in D. Since the deriviative of an analytic function therefore
f(z) is analytic in D
9.3 LIOUVILLE’S THEOREM
Statement : If f is an entire and bounded function, then f is
constant. (2006, 2008)
Proof : Given that, f is an entire and bounded function. ∴ 0M∃ > s.t. ()fz M z ≤ ∨ ∈ ∵ f is an entire function and hence f is analytic everywhere in
Complex Plane and ();BR= α ( s a y ) ∴ ()()!0,1, 2, ...n
nnMfnR α ≤ =
Put z=α
()()!0,1, 2, ...n
nnMfz nR ≤ =
Put 1n=
()1!MfzR ′ ≤′ ∴ ()0MfzR ′ ≤→ as R→∞. ∴ ()0 fz z′ = ∨ ∈. ⇒ f is constant.
Aliter :
Given that f is an entire and bounded function. Let 1z and 2z be
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Construct a circle x with centre at 1z and radius R so that point 2z
lies inside x.
By Cauchy’s integral formula
()()1
11
2xffz diz ξ = ξ π ξ−∫ and ()()2
21
2xffz dizξ= ξ π ξ− ∫
() ( )() ()
12
211
2xxfffz f z d diz zξξ′− = ξ− ξ π ξ− ξ− ∫∫
()21
211
2xfzz dzz ξ ≤ − ξ π ξ− ξ− ∫ ( 1 )
Choose R, so large that 212Rzz − < , since ξ is any point on the
circle x. ∴ 1zR ξ− =
Now, 21 1 21 2 122RRzz z zz z z R ξ− = ξ− +− ≥ ξ− − − > − =
Given that f is bounded function. ∴ ()fM x ξ ≤ ∨ ξ∈ ∴ F r o m e q u a t i o n ( 1 )
() ( )211
2fz fz − ≤
21
2MzzR −
π . xdR ξ ∫
21
2.
xzzMdR − = ξ π ∫
21
22zz M RR − × π =π 212zz MR − = ∴ () ( )21 2 120fz fz z z MR − ≤ − → as R→∞. ∴ ()()120fz fz − = ∴ ()()12fz fz= for any two points 1z and 2z in . ∴ f is constant.
Note : If f is a non-constant entire function the f is unbounded.
Example : Let ()()()fz u z i v z=+ be an entire function and suppose ()uz M z ≤ ∨ ∈. Prove that f is constant. munotes.in
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Proof : Given that, ()()()fz u z i v z=+ is an entire function.
Define ()()fzgz e = ⇒ g is an entire function. ∴ ()()()()fz u z i v zgz e e + = = ()uze =
()1ieθ =∵
()Mgz e < constant. ⇒ g is bounded.
()()uz M z ≤ ∨ ∈∵
Thus, g is entire and bounded function.
By Liouville’s theorem,
g is constant. ∴ ()fze is constant. ∴ ()()()fz u z i v z=+ is constant. ⇒ u and v are constant.
Note : If () ( )
0nn
nfz a z∞
== −α∑ has radius of convergence. 0R>, then f is analytic in ();BRα .
9.4 TAYLOR’S THEOREM
If f is analytic in a domain G, then for any poi nt (),zB R G∈α ⊂, f
has Taylor series expansion,
0() ( )nn
nfz a z∞
= = −α∑ w h e r e ()()!n
nfan α = .
Proof : Given that f is analytic in G.
For 0rR<<, construct a circle x with centre at α and radius r so
that the point z lies inside x.
By Cauchy integral formula, ()()1
2xffz dizξ= ξ π ξ− ∫ ( 1 )
Now,
()()
()11 1
1zz z= = ξ− ξ−α +α− −α ξ−α − ξ−α munotes.in
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()()
()111z− −α = − ξ−α ξ−α
()()
()()
()()
()1
111. . . . . .nn
nnzz z
r−
− −α −α −α =+ + + + + ξ−α ξ− ξ−α ξ−α
(){}1 21. . .xx x− = + + +∵
()()
()()
()()
() ()
()1
11 11. . . .
1nn
nnzz z
z z−
− −α −α −α =+ + + + −α ξ−α −α ξ−α ξ−α − ξ−α
()
()
()1 21 1
21 2
1 2111 . . . . . .1. . . 1. . .1. . . 1nn n
nn
nnxx x x x x
xx x x xx
xx x x x− − +
−
− − − =+ + + + + + + = + + + + + + + + = + + + + + −∵
()
()()
()()
()1
21 1...nn
nnzz zz−
−α −α −α =+ + + + − ξ−α ξ− ξ−α ξ−α ξ−α ( 2 )
Multiplying equation (2) by ()2fi ξ π and integrating w.r.t. ξ over. ∴()()1
2xffz dizξ= ξ π ξ− ∫()()()()21...22xxfz fddii ξ −α ξ= ξ+ ξ+ π ξ−α π ξ−α∫∫
()()
()() ()
() ( )1
2n
n
n
xxfz fzd diz−
ξ −α ξ+−α ξ+ ξ π ξ−α ξ− ξ−α∫∫
(3)
By generalization of the theorem on derivative of a nalytic function.
()()()()1!
2n
n
xf nfdi+ξα= ξ π ξ−α ∫ 0,1, 2, ...n=
OR ()() ()
()11
!2n
n
xffdni+ α ξ= ξ π ξ−α ∫
Substituting all these values in equation (3), we g et
() ( ) ( ) ()()()()()()21 12...2! 1 !n nnzffz f z f x f An− − −α −α ′′ ′= α+−α + α++ +− munotes.in
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() () ()21
01 2 1 ...nnnaa z a z a z A−
−=+ −α+ −α ++ −α + ()() () ()
() ( )!2n
nnn
xfz faa n d A dni z α −α ξ = = ξ π ξ−α ξ− ∫∵
The theorem will be proved if lim 0nnA
→∞ =
() ()
() ( )2n
n n
xzfAdi z −α ξ= ξπ ξ−α ξ− ∫
()
2n
n n
xf zAdz ξ −α ≤ ξ π ξ−α ξ− ∫ ( 4 )
Choose 0 >9 s.t. z −α =9 ()0r<<9
Since ξ is any point on the circle x.
Now, zz z r ξ− = ξ−α +α− ≥ ξ−α − −α = −9
Given that, f is analytic in G. ∴ f is continuous on {}x {}()is compact setx ∵
By boundness theorem, 0M∃ > s.t. ()fM x ξ ≤ ∨ ξ∈ ∴ F r o m e q u a t i o n ( 4 ) ,
() () 22n n
nn
x xMMAd drr rr ≤ ξ = ξ π π− − ∫∫99
9 9
2M=
π().2n
r r π − 9
9r ∴ ()0n
nMrArr ≤ → − 9
9 as n→∞
lim 0, 0 1 and hence 0 1 0n
nxx r→∞ = ≤<< < ⇒ < < ∵99 ∴ lim 0nnA
→∞ = ∴ G i v e n s e r i e s i s c o n v e r g e n t a n d w e w r i t e
() ( ) ( ) () ( ) ( )2...fz f z f x z f ′′ ′= α+−α + −α α +
() ( )
0nn
nfz a z∞
= = −α ∑ w h e r e ()()!n
nfan α =
Put 0α= in equation (1), we get,
()()00!n
n
nn
nff z a z where an∞
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Example 1 : Expand sinz in a Taylor series about 4z π = .
Solution : By Taylor series
() ( ) ( ) ( )()()()()23...2! 3!zzfz f z f f f −α −α′′ ′′ ′ ′= α+−α α+ α+ α+
Here, ()sinfz z= and 4 π α=
()1sin sin42f π α= α==
() ( )1cos cos cos42fz z f π ′′= ⇒ α= α= =
() ( )1sin sin sin42fz z f π − ′′ ′′=− ⇒ α= − α=− =
() ( )1cos cos cos42fz z f π − ′′′ ′′′=− ⇒ α=− α=− = ∴ S u b s t i t u t i n g a b o v e v a l u e s i n e q u a t i o n ( 1 ) , w e g e t
23
11 1 1 44sin ...42 6 22 2 2zz
zz π π − − π − − = + − + + +
()() ()23
21 441. . .4 26 2z
z ππ − − π= + −− − +
Zeros of an Analytic Function :
Definition : A complex polynomial ()pz o f d e g r e e n i s a n
expression of the form 01 2 ...nnaa z a z a z++ + + , where 01,, . . . ,naa a
are complex constants and 0na≠.
Definition : Let G be an open set and suppose :fG→ is a given
function. A point 0zG∈ i s s a i d t o b e z e r o ( o r r o o t ) o f f i f ()00fz=.
e.g. () ( ) ( )2256 326 3 2fz z z z z z z z= −+= −− += − −. Here roots or
zero of, f are 2z= and 3z=.
Definition : If :fG→ is analytic and α in G satisfies ()0fα=,
then α is a zero of f of order (multiplicity) 1m≥, if ∃ an analytic
function :gG→ s.t. () ( ) ()mfz z g z=−α where ()0gα≠. munotes.in
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Example : Let :fG→ be an analytic function then f has a zero
of order 1m≥ at z=α if () ( ) ()mfz z g z=−α where g is analytic on
G and ()0gα≠. (2006)
Solution : Let () ( ) ()mfz z g z=−α w h e r e g i s a n a l y t i c o n G a n d ()0gα≠ ( 1 ) ∴ B y T a y l o r s e r i e s ,
For any ();zB G∈α γ⊂, g has Taylor series expansion
() ( ) ( ) ( )()()2...2!zgz g z g g −α′′ ′= α+−α α+ α+
() ( ) ( ) ( ) ( )()()2...2!m
m zfz z g z g g+ −α ′′ ′ =−α α +−α α + α+
() ( ) () ( )()()2
1...2!m
mm zzgz g g+
+ −α ′′=−α α+−α α + α+
()() ()()()
()()()()
()
()()121
2!1 ! 2 !...mm m
mm
mzz zfz f fmm m
f+ +
+
+ −α −α −α = α+ α+ ++
α+
Clearly, this is a Taylor series expansion about z=α and
() () ()()()1... 0mff f f−′′α= α= α== α= and ()()0mfα≠. ⇒ f has a zero of order 1m≥ at z=α.
Conversely, assume that f has a zero of order 1m≥ at z=α. ∴ () ()()()1... 0mff f −′α= α== α= ∴ () ( )
0nn
nfz a z∞
== −α∑ ∴() ( ) n
nmfz a z∞
== −α∑
() () ()12
12...mm m
mm maz a z a z++
++= −α + −α + −α +
() () ()2
12...m
mm m za a z a z++= −α + −α+ −α +() ()
0mnnm
nza z∞
+
==−α + α∑ () ( ) ()mfz z g z=−α ()0gα≠
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Note : () ( ) () ( ) ()1mmfz m z g z z g z−′′= −α +−α
()()()()()12 . . . 2 . . . . . .!m mm m gzfzm − − ×× + + =
Each of these terms have ()z−α as one form.
()() ()!0 0 . . .mfm gα= α+++
()() () !mfm gα= α
Definition : A zero of an analytic function f is said to be isolated if
it has a neighbourhood in which there is no other zero of f.
Theorem : Any zero of an analytic function is isolated in the set of
its zeros. (2009)
Proof : Let :fG→ be an analytic function.
Suppose f has a zero of order m at z=α. ∴ () ( ) ()mfz z g z=−α ( 1 )
where g is analytic on G and ()0gα≠.
Let 0ε> be given.
Put ()20g α = ε> ∵ g is analytic on G. ⇒ g is continuous at Gα∈. ⇒ f o r t h e a b o v e 0, 0ε> ∃ δ>
s.t.()()zg z g −α <δ ⇒ −α <ε
When z −α <δ i.e. (),zB∈α δ ∴ () ( ) ( ) ()gz g g gz≥α −α −()()()gg g≥ α − + α 2>ε−ε=ε ∴ ()0gz > ε> ∴ ()()0,gz z B≠ ∨ ∈α δ ∴ ()(){}0, gz z B≠ ∨ ∈α δ−α
and (){}0, zz B −α ≠ ∨ ∈α δ−α
or (){}0, zz B −α ≠ ∨ ∈α δ−α
From equation (1)
() ( ) () () { } 0,mfz z g z z B=−α ≠ ∨ ∈α δ−α ∵ α is arbitrary.
∴ Any zeros of an analytic function is isolated in t he set of its zeros.
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9.4 FUNDAMENTAL THEOREM OF ALGEBRA
Statement: Every non-constant complex polynomial has a root.
( 2 0 0 9 )
OR
If ()pz i s a n o n - c o n s t a n t c o m p l e x p o l y n o m i a l t h e n , t h e r e i s a
complex number and α with ()0pα=
Proof : Given that ()pz i s a n o n - c o n s t a n t c o m p l e x p o l y n o m i a l
T.P.T. ()0pα=.
Assume that this is not true.
Suppose ()0pz z≠ ∨ . ∴ ()pz is an entire function.
Let ()()1fzpz= ( 1 ) ⇒ f is an entire function. ∵ ()pz is non-constant entire function. ⇒ ()pz is unbounded (by contra positive statement of Liov ille’s
theorem) ∴ ()lim
zpz
→∞ = ∞ ∴ F r o m e q u a t i o n ( 1 ) , ()()11lim lim 0zzfzpz →∞ →∞ = = = ∞
Let f be defined on an bounded set E.
If for a given 0ε>, 0R∃ > s.t. ()fz − <εℓ w h e n e v e r zR > and zE∈
Then, we say that ()fz→ℓ as z→∞. ∴ ()lim
zfz
→∞ =ℓ
REBclosed disk
( closed disk is compact set ) (O,R)B(O,R)
Fig 9.1 munotes.in
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∴ F o r g i v e n 0, 0Rε> ∃ > s.t. ()fz < ε whenever zR > ∵ f is an entire function. ∴ f is continuous on ()0;BR . ∴ B y b o u n d e d n e s s t h e o r e m .
f is bounded on ()0;BR . ∴ 0M∃ > s.t. ()()0;fz M z B R ≤ ∨ ∈ ∴ f is entire and bounded on ()0;BR ∴ B y L i o u v i l l e ’ s t h e o r e m , f is constant. ∴ F r o m e q u a t i o n ( 1 ) ,
()()11constantconstantpzfz== =
which contradicts the hypothesis that ()pz is non-constant. ∴ O u r a s s u m p t i o n i s w r o n g . ∴ There is a Complex Number α with ()0pα=.
Exercise : Prove that a complex polynomial
()2
01 2 ...nnpz a a z az az=+ + + + has exactly n roots where 01,, . . . ,naa a are
complex constant and 0na≠. (Use fundamental theorem of Algebra)
Theorem : Suppose that f is analytic in domain G. If fz, the set of
zeros of f in G, has limit point in G, then ()0fz= in G.
Proof : Given that f is analytic in a domain G. (){}:0fzz f z=∈ = and α is a limit point of fz.
Let {}nz be a sequence of zeros of f in G, such that limnnz
→∞ = α ∵ f is analytic on G, f is continuous on G. ∴ () ()lim lim 0nnnnff z f z
→∞ →∞α− = = (){} is a zero of 0nnzf z ⇒ =∵ ∴ ()0fα= ∵ z e r o s o f a n a n a l y t i c f u n c t i o n a r e i s o l a t e d . ∴ e i t h e r ()()0;fz z B= ∨ ∈α δ
OR ()(){}0; — fz B≠ ∨ α δα
x is connected if the only sets of x which are both open and
closed are ∅ and x.
Assume that ()0fz≠ in (){};—Bα δα munotes.in
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∵ α i s a l i m i t p o i n t o f fz ⇒ e v e r y n b d o f α continuously
infinitely many points of fz.
For sufficiently large n, there is a point nz s u c h t h a t ()0nfz= i n (){};—Bα δα ∴ O u r a s s u m p t i o n i s w r o n g .
Hence, we must have ()()0;fz z B= ∨ ∈α δ
δ
α
() 0=fZ
Fig 10.2
Given that, G is a domain. ∴ G is open and connected.
We split the set G into two sets. {}:i s a p o i n t o ffAG z=ξ∈ ξ {}:BG A=ξ∈ ξ∉ , where fz is the set of zeros of f in G. ∴ AB=∅∩ and ABG=∪ ∵ Aξ∈ is a limit point of fz in G. ∴ ()()0;fz z B= ∨ ∈ξ δ ⇒ zA∈ ∴ ();zB zA∈ξ δ⇒ ∈ ∴ (),BAξ δ⊂ ⇒ A is an open set and A≠∅ ()A α∈∵
Let B′ξ∈, then ′ξ is not a limit point of fz. ∴ B y c o n t i n u i t y o f f at ,0′ξ ∃ δ> s.t.
()()0;fz z B′≠ ∨ ∈ξ δ ⇒ zB∈
Thus, (),zB zB′∈ξ δ⇒ ∈ ∴ ();BB′ξ δ⊂ ⇒ B is an open set. ∵ G is connected. munotes.in
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It can not be written as a union of two non-empty d isjoint open sets. ⇒ A=∅ or B=∅
But A≠∅ ()A α∈∵ ∴ B=∅ ∴ AG= () and ABG B = = ∅∪∵ ∴ E a c h p o i n t o f G is a limit point of fz. ∴ ()0fz z G= ∨ ∈
Theorem: Let f and g be analytic in a domain G. If T is a
subset of G having limit point α i n G a n d i f ()()fz g z z T= ∨ ∈, then ()()fz g z z G= ∨ ∈. ()()()(){}., : 0Fz fz gz T z GFz= = ∈ = a n d u s e o f p r e v i o u s
theorem.
Theorem : Let f be analytic in a domain G such that for s ome α
in G and ()()0nfα=, 0,1, 2, ...n= t h e n ()0fz z G= ∨ ∈. ( U s e
Taylor’s theorem)
Exercise : Prove that the function ()zfz z e z= − has a zero of order
2 at origin.
()zfz z e z= − ∴by Maclaurin expansion
()zfz z e z= −
23456
234567
234567
2345
2(1 )1! 2! 3! 4! 5! 6!()1! 2! 3! 4! 5! 6!
1! 2! 3! 4! 5! 6!
1()1! 2! 3! 4! 5! 6!zz z z z zzzzzzzzzzzzzzzzz
zz z z zz=+ + + ++ + −=+ + + + + + −=+++++
=+ + + + + ∴since lowest power of z is 2
9.5 SUMMARY
1) Morera’s Theorem :
Statement : Let G b e a d o m a i n i n a n d l e t :fG→ b e a
continuous function s.t. ()0fz d z
∂∆ =∫ for any triangle ∆ in G then munotes.in
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f is analytic in G. (This is a partial converse of Cauchy – Goursat
theorem.)
2) Liouville’s Theorem :
Statement : If f i s a n e n t i r e a n d b o u n d e d f u n c t i o n , t h e n f i s
constant.
1) Taylor’s Theorem: If f is analytic in a domain G, then for any
point (),zB R G∈α ⊂, f h a s t a y l o r s e r i e s e x p a n s i o n ,
0() ( )nn
nfz a z∞
= = −α∑ where ()()!n
nfan α = .
3) Fundamental Theorem of Algebra :
Statement : Every non-constant complex polynomial has a root.
OR
If ()pz i s a n o n - c o n s t a n t c o m p l e x p o l y n o m i a l t h e n , t h e r e i s a
complex number and α with ()0pα=.
9.6 UNIT END EXERCISES
1) Show that an entire function is infinitely differentiable .
Solution: If fis entire , by Taylors expansion of f, fhas a
power series representation. In
fact (0)kf exist 1k∀≥.
0(0)() ,!k
k
kffz z zk∞
== ∀∈∑
We can see that ()fzhas an infinite radius of convergence . ()fzconverges for all z∈.
By the result that Power series are infinite ly differentiable
within their domain of convergence, ()fzis infinitely
differentiable. 11 111(0) (0)()!( 1 ) !kkkkkkkf ffz z zkk∞∞−−====∑∑− 22 222)1 )( 0 ) ( 0 )()!( 2 ) !kkkkkkkk f ffz z zkk−−∞∞
==−==∑∑−and so on.
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2) Find the power series expansion of 2()fz z= around 2z=.
Solution: ∴ 2 (2)() ( 2 ) ( 2 ) ( 2 ) ( 2 ) . . .2!ffz f f z z′′ ′=+ −+ −+ 2() 4 4 ( 2 ) ( 2 )fz z z =+ −+−
3) Find the power series expansion for ze about any point a.
4) Suppose an entire function fis bounded by M, along zR=.
Show that the coefficients in it’s power se ries expansion about 0z= satisfy kkMcR≤. ( Hint: (0)!k
kfck= ,
by Cauchy formula ∴ !( )(0) 121k
zR zkf zfd z kikπ == ∀≥∫+)
5) Let fbe an entire function , if for some integer 0k≥, there
exist positive constants A and B such that ()kfz A B z≤+ , then fis a polynomial of degree atmost k.
( Hint: Use Liouville’s Theorem)
6) Using Morera’s theorem show that the funct ion fdefined by
0()1ztefz d tt∞=∫+ is analytic in the left half plane :R e ( ) 0Dz<.
Solution: ∴ 0011zt
xte
dt e dttx∞∞<= −∫∫′+ for Re( ) 0zx=<.
This integral is absolutely convergent and 1()fzx≤′
Consider 0()1ztefz d z d t d zt∞
ΓΓ=∫∫ ∫+. Here Γ= The boundary of
some closed rectangle in D.
Since, 01ztedtdzt∞
Γ∫∫+converges hence we can interchange the order
of integration . ∴00() 0 01ztefz d z d t d z d tt∞∞
ΓΓ== =∫∫ ∫ ∫+, since 1ztet+ i s a n a l y t i c i n s i d e
and on a closed curve Γ.
∴ By Morera’s Theorem fis analytic in D. munotes.in
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7) Show that 1
0sin(2 )tdtt∫ is an entire function.( Use Morera’s
Theorem. )
8) Show that α is a zero of multiplicity k if and only if
1() () . . . () 0kPa P a P a−′= = = = but () 0kPa≠.
( Hint : Use the Fundamental theorem of Algeb ra .)
9) Find the order of zero at 0z= of the function ()()zfz z z e z=−.
10) Find the Maclaurin series expansion of ()2sinfz z= .
*****
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10
MAXIMUM AND MINIMUM MODULUS
PRINCIPLE
Unit Structure
10.0 Objectives
10.1 Introduction
10.2 Maximum Modulus Principle, Schwarz Lemma, Open
M a p p i n g T h e o r e m
10.3 Automorphisms of the Unit Disc
10.4 Summary
10.5 Unit End Exercises
10.0 OBJECTIVES
After going through this unit, we shall unders tand
1)Maximum modulus principle and open mapping theorem for
analytic functions . 2) Corollaries on open map ping theorem
and maximum modulus principle. 3) Possible Au tomorphisms
of the Unit disc (0;1)B . 4)Harmonic functions and their
properties.
10.1 INTRODUCTION
In previous sections, we have studied the connections
between everywhere convergent power series and entire
functions. We shall now turn our attention t o the general
relationship between power series and analyt ic functions.
According to a theorem, every power series represents an
analytic function inside it’s circle of conv ergence.
Our first goal is the converse of this theo rem. We then
turn to the questions of analytic functions i n arbitrary open
sets and local behaviour of such functions.
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10.2 THE MAXIMUM MODULUS PRINCIPLE
Definition : Let G b e a n y s u b s e t o f . A complex function f
defined on G is said to have local maximum modulus at a point α in
G if, there exists 0δ> s.t. ();BGα δ⊂ and
()()(,)fz f z Bαα δ≤∀ ∈ .
Similarly, f h a s l o c a l m i n i m u m m o d u l u s a t a p o i n t α i n G, if 0∃ δ> s.t. ();BGα δ⊂ and
()()fz f ≥ α (),zB∨ ∈α δ.
Theorem : Suppose f is analytic in a domain G and there i s a
point α in G s.t. ()()fz f z Gα= ∀∈. Then f is constant i.e. if f attains to maximum modulus in G then f is cons tant.
Proof : Given that, f is analytic in domain G (open and connected
set) ∵ Gα∈ and G is open. ∴ 0r∃ > s.t. ();BrGα ⊂
where ();xBr=∂α = boundary of closed disk ();Br α
αrx
Fig 10.1
∴ B y C a u c h y i n t e g r a l f o r m l a ,
()()1
2xfzfd ziz α= π −α ∫
Here x is the circle zr −α =
[]0, 2itzr e t=α+ ∈ π , itdz i re dt= munotes.in
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∴ ()1
2f
iα=
π()it
itfr e
re α+
2
0.iπ
∫itre dt ()2
01
2itfr e d tπ= α+ π ∫
( 1 )
Given that, ()()fz f z G ≤ α ∨ ∈ ∴ ()()itfr e f α+ ≤ α ∴ F r o m e q u a t i o n ( 1 ) , w e g e t
() ()2
01
2itff r e d tπ α ≤ α+ π ∫ ()2
01
2fd tπ≤ α π ∫
() ()[]220022ffdt tππ α α = = ππ∫
()22f α = × ππ()f= α ∴() () ()2
01
2itff r e d t fπ α ≤ α+ ≤ α π ∫
()2
01
2itfr e d tπ= α+ π ∫ ∴ () ()2
02itff r e d tππ α = α+ ∫
() ()22
00itfd t fr e d tππ α = α+ ∫∫
() ()2
02ff d tπ π α = α ∫∵
() ()2
00itff r e d tπ
α − α+ =∫
Here, the integral ()()itff r e α − α+ i s c o n t i n u o u s a n d n o n -
negative. ∴ ()()[]00 , 2itff r e t α − α+ = ∨ ∈ π ∴ ()() []0, 2itfr e f t α+ = α ∨ ∈ π ∴ ()()fz f z x = α ∨ ∈ munotes.in
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This equation holds on all circles 0zs r −α =δ ≤≤ ∴ ()fz is constant in ();Brα ∴ ()fz is constant in ();Brα
(∵If :fG→ is analytic and ()fz = constant if zG∈ then, f
is constant on G.
By using theorem [Let f and g be analytic in a domain G.
If T i s a s u b s e t o f G h a v i n g l i m i t p o i n t α i n G a n d i f ()()fz g z z T= ∨ ∈ then ()()fz g z z G= ∨ ∈]
f is constant in G.
Maximum Modulus Principle :
Suppose f is analytic in a bounded domain D a nd
continuous on D (Closure of D). Then , f attains its maximum on
the boundary D∂ of D. (2006, 2012)
Proof : If f is constant, then there is nothing to prove. Let f be a
non-constant function.
Given that, f is continuous on D (D is a compact set). ∴ f attains its maximum value at same point in D.
Maximum modulus principal, ∴ f does not attain its maximum in D. ∴ f attains its maximum on the boundary D∂ of D. ()DD D= ∂∪
Minimum Modulus Principle :
Suppose f is a non-constant and analytic function i n a
domain G. If f attains its local minimum G at α, then ()0fα=
or f is constant (2006,2007)
Proof : Given that, f is a non-constant analytic function in a domain
G and f attains its local minimum at a point α in G. ∴ ()()ff z α ≤ ();zB G∨ ∈α δ⊂ ( 1 )
To prove that ()0fα=
Assume that this is not a true.
i.e. ()0fα≠ in some open disk (),BrGα ⊂. munotes.in
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Let ()()1gzfz= ( 2 ) ∴ g is analytic in ();BrGα ⊂. ∴ F r o m e q u a t i o n ( 1 ) ,
() ()11ff z≥ α
OR
()()gz g ≤ α (),zB r∨ ∈α ∴ g has a local maximum modulus at a point α in G. ∴ B y m a x i m u m m o d u l u s p r i n c i p l e ,
G is constant in G. ∴ F r o m e q u a t i o n ( 2 ) ,
()()1constantconstantgfzgz== = ⇒ f is constant in G.
which is contradicts that f is a non-constant function. ∴ O u r a s s u m p t i o n i s w r o n g . ∴ ()0fα=
Theorem : If f is a non-constant analytic function in a bou nded
domain G and ()0fz≠ for any zG∈, then f can not attain its
minimum in G.
Example : Let ()zfz e= and ()23 , 1TB i=+ . Find a point in T at
which f attains its maximum value.
Solution : Given function, ()zfz e=()23 , 1TB i=+
(2, 3)(3, 3)= ∂∪BBB
Fig 10.2 munotes.in
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=zrr
10The boundary of T is the circle.
()23 1zi −+ = ∴ B y m a x i m u m m o d u l u s t h e o r e m , f a t t a i n s i t s m a x i m u m
value on B∂. ∴()zfz e = () ( )23 1iieeθ++= ()3s i n 2c o s.iee+ θ + θ= 2cose= + θ
()1ieθ =∵
2cose=+ θ
We know that, the value of cosθ is maximum when 0θ=. ∴ 21 3zee e+ = = 3zee = is the maximum value of f at a point ()3, 3 or 33i+ in T.
Exercise : Let ()fz z= and ()0;1TB= .
Prove that the function f cannot attain its minimum value on the
boundary of T.
Schwarz’s Lemma
Let {}:1Dz z=∈ < be the unit disk and suppose f is analytic in D
with, (i) f (0) = 0 and (ii) ()1fz f o r z D≤ ∈. Then, ()()01fz z z D a n df ′≤ ∀∈ ≤. Moreover, if ()fz z= for
some 0,z≠ t h e n t h e r e i s a c o n s t a n t C w i t h 1c= s . t . ().fcω=ωD∀ω∈.(2004, 2005, 2008)
Proof: G i v e n t h a t , f i s a n a l y t i c i n D , w i t h ( i ) ()00fa n d= ( i i ) ()1fz≤ for .zD∈ Define :gD→ by
()()()/0,00fz z zgzf z≠= ′= g⇒ is analytic in D
Choose r s.t. 01r<< ∴ on the circle zr=
()()fzgzz=
Fig 10.3
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()1gzr≤− on the circle zr=
……..()()1fz a n d z r ≤ =∵ ∴ by Maximum Modulus theorem,
() ()10;gz z B rr≤ ∀∈
As ()()1, 1 0,1rg z z D B→ ≤ ∀∈ = - - - - - - - - - - - - - ( 2 ) ()1fz∴ ≤ - - - - - - - - - ( b y ( 1 ) ) ()fz z z D⇒ ≤ ∀∈
Again, from equation (1) ()()00 1fg′= ≤ --------- (2) ()01f′∴ ≤
If ()fz z= f o r s o m e 0z≠ then,
()()1fz zgzzz= = = g∴ attains its maximum value of some point z inside D . ∴ by maximum Modules principle, g is constant in D
i.e. ()gz c= where, c is constant and 1c= ()fzcz∴ =
or ()fz c z= zD∀ ∈
or ()fcω=ω D∀ ω∈
Example: ;zα in D, define the M.T. ()1zzzα−αφ =−α
Let f be analytic in D and let ()1,fz t h e n<
(i) ()()() ( ) 1 1fz f z
ff zz−α−α ≤ −α−α OR () () ()fxfz zα φ ≤ φ ,zD∀ α∈
(ii) ()()2211 1fzzfz′
≤
− − zD∀ ∈
Solution:
Fix a pint α in D arbitrarily
Put ()1zzzα−αω= = φ
−α 1zω+α⇒ =+αω munotes.in
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Define ()() ( )()()() ( ) 1ffz ffz gff zα−αφ = ω=−α -------------------- (1)
()()()
()1
1
1fff
fz
ffαω+α−α+αωφ =ω+α−α+αω ∴ g is analytic in D, g(o) = o and ()1gω<
By using Schwarz’s Lemma, ()()1 go a n d g′< ω≤ ω D∀ ω∈ ()()() ( ) 1 1fz f z
ff zz−α−α⇒ ≤ −α−α
OR
()() ()fxfz z z Dαφ ≤ φ ∀ ∈
From equation (1)
()()()() ( ) 1fz f ddgdz dz f f z−αω′ω = −α
()() () ( )
()211
1zz
g
z−α +−α α′ω =−α
() () () () ( )()() () ( )()() ()
() ( )21
1ff z f zf z f ff z f ff zff z′′ ′−α ⋅ + −α α α α
=
−α ()()() ()
() ()2 2221 1
11 1ff zzg
zz ff z ′−α −α −α ′ = −α−α −α
Put z=α
()21
go−α
′
( )221−α()21f−α
=
()()221f
f′ α
−α
()()()
()2
2111f
go
f′ −α α
′∴ = ≤ −α ()()1go′ ≤ ∵ munotes.in
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()()2211 1f
f′α∴ ≤ −α−α ()()221f
f′α1∴ ≤ 1−α−α
Put ()()221,
1 1fzzz Dz fz′α= ≤ ∀ ∈− −
iii) ()()() 1fo zfzfo z+ ≤ +
()g ω ≤ ω ∵ D∀ ω∈
() ()
()()1
1 1fz fo zzfo fz−−α ≤ −α−
P u t oα=
()()() () 1fz fozfofz− ≤ −
If 11aa n d b < <
11 1ab a b abab abab−− + ≤ ≤ − + − ()()() ()()()() ()()1 1fz fo fz fozfo a n d fz fz fo− − ≤ ≤ − − ()()() () 1fz fozfz fo− ≤ −
() () () ()()1fz fo z fz fo∴ − ≤ − ()()()()fz fo z z fz fo∴ − ≤ −− ()()()()fz zfz fo fo z∴ + ≤ +
()()() 1ffo zzfo z+ ≤ +
Counting zero
Definition : A zero of order one is said to be a simple zero.
e.g. Let ()232fz z z= −+
Here, f has simple zeros at 1z= and 2z=. munotes.in
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()23fz z′= −
()11 0f′=− ≠
()21 0f′=− ≠
Prove that all zeros of the function sinzare simple.
To find zeros at sinz put sin 0z= ∴ ()1sin 0 sin 0 2 , 0, 1, 2, ...zz n n− π= ⇒ = = = ± ± ∈
()sinfz z=
()cosfz z′=
() ( ) cos 1 0,nfz n n′= π=− ≠ ∨ ∈ ∴ All zeros of sinz are simple.
Theorem : Let f be analytic in a domain G with zeros
12,, . . . ,mα α α (repeated according to order)
If x is a smooth closed curve in G which does not pass through any ks′α then
()()()
11;2mkk xfzdz n xif z=′ = α π ∑ ∫
Proof : Given that, f i s a n a l y t i c i n a d o m a i n G w i t h z e r o s
12,, . . . ,nα α α (repeated according to order or multiplicities.) ∴ ()()()()12 ...mfz z z z=−α −α −α, where g i s a n a l y t i c a n d ()0, 1, 2, ...,kgk mα≠ =
Taking log on both sides and differentiating w.r.t . z, we get
()()()() 1211 1...
mfz g zfz z z z g z′ ′ =+ + ++ −α −α −α
Multiply this equation by 12i π and integrate w.r.t z over { x } on
both side. ()()()() 111 1 1...22 2 2mxxx xfz gz dz dzdz dzif z i z i z i g z′ ′ = + + + π π −α π −α π ∫∫∫ ∫ ∵ g and g′ are analytic in G. ∴ gg′ is analytic in G and x is a smooth closed curve. ∴ B y C a u c h y t h e o r e m , ()()0xgzdzgz′ =∫ munotes.in
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∴ ()()() ( ) () 121;; . . . ;2mxfzdz n x n x n xif z′ = α+ α++ α π ∫
()
1;mkknx
== α∑
Note : ()()()
11,2mkk xfzdz n xif z=′ = α π ∑ ∫= Number of zeros of f i n s i d e x,
where each zero is counted according to its order.
e.g. for () ( ) ( ) ()2
12fz z z g z−− α −α
()()() ( ) ()12 31;, 3 ,2xfzdz x x xif z′ = η α+η α+η α π ∫103=+ +4= Gα
αα1
3
2x
Fig 10.4
Corollary : Let f, G and x be as in the preceding theorem except
that 123,,, . . . ,nα α α α are the points in G that satisfy the equation ()fz=α then,
()()()
11;2mkk xfxdz xif z=′ = η α π −α ∑ ∫ Number of zeros of ()fz=α inside x.
Example : Evaluate ()()xfzdzfz′ ∫ w h e r e ()()2315zzfzz −=+ a n d x i s
the circle 1.2z = .
Solution : Given function, ()()2315zzfzz −= +.
Here, ()fz has simple zero at 0z= and 1z= is a zero of order 2.
Given, equation of circle, 1.2z = . ∴ Z e r o 0z= and 1z= lies inside x. munotes.in
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0 1 1.2x
Fig 10.5
By theorem, ()()1
2xfzdzif z′ π ∫ = number of zeros of inside x where
each zero is counted according to its order. ∴ ()()112 32xfzdzif z′ =+ = π ∫ ⇒()()6xfzdz ifz′ = π∫
Exercise: Evaluate ()()xfzdzfz′ ∫ where ()() ( )2
3132zz zfzz + − + =+ and
x is the circle 1.5z = .
Note : Let []:0 , 1xG → b e a c l o s e d ( S m o o t h ) c u r v e i n a n d
suppose :fG→ is an analytic function. Then fxσ= is also a
closed curve in w-plane. If α i s a C o m p l e x N u m b e r {}{}()fxα∉ σ = , we write,
()()()011;22t dwdtiw i tσ′ σ ησ α= = π −α π σ− α ∫∫()()()()()01
2fx tx t d ti fx t′′ = π −α ∫
()()112 32xfzdzif z′ = = + = π ∫ ()()6xfzdz ifz′ = π∫
Exercise : Evaluate ()()xfzdzfz′ ∫ where ()() ( )2
3132zz zfzz − + =+ and x
is the circle 1.5z = .
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Note : Let []:0 , 1xG → b e a c l o s e d ( s m o o t h ) c u r v e i n a n d
suppose :fG→ is an analytic function. Then fxσ= is also a
closed curve in w-plane. If α i s a C o m p l e x N u m b e r {}{}()fxα∉ σ = , we write,
()()()11;22xxt dwdtiw i t′ σ ησ α= = π −α π σ− α ∫∫()()()()()01
2fx tx t d ti fx t′′ = π −α ∫()()112 32xfzdzif z′ = = + = π ∫
()
1,mkkx
== η α∑ = numbers at zeros of ()fz−α i n s i d e x w h e r e e a c h
zero is counted according to its order.
x
−planeG
−zplanef
ω−planeσ=fx
Fig 10.6
where 12,, . . . ,nα α α are points in G with ()kfα=α.
x[ 0, 1] Gfσ=fx
Fig 10.7
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Theorem : Suppose that f is analytic in ();BaR and let ()fa−α
has a zero of order m at za= then there is an 0ε> and 0δ> s.t. ξ−α <δ and the equation ()fz−ξ has exactly m simple roots in ();Ba ε.
Proof : Given that, ()fα− α has a zero of order m at za=. ∵ Z e r o s o f a n a n a l y t i c f u n c t i o n a r e i s o l a t e d . ∴ W e c a n c h o o s e 0ε> s.t. 2Rε< .
()fz−α has no solution with 02za< − < ε and ()0fz′≠ if 02za< − < ε.
b b1 2
mR
O bε
ε2
−zplane
Fig 10.8
Let x be the circle, za − =ε
i.e. () []2,0 , 1itxt a e tπ=+ ε ∈ . []():0 , 1 ;xB a R → ∵ f is an analytic on ();BaR ∴ fxσ= is also a closed curve in w-plane. munotes.in
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∉εδασ=fx
ω−plane
Fig 10.9
Now, {}α∉ σ
So, 0∃ δ> s.t.(){};Bαδ ∩ σ = ∅
(It means open disk ();Bα δ does not touch trace of σ. ∴ ();Bα δ is contained in the same component of {}w σ.
For ();Bξ∈ α δ i.e. ξ−α <δ ⇒ ()();;ησ α=ησ ξ ( 1 )
Now, ()1;2dwiwσησ α= π −α ∫ () () .fz d w f zd z′ω = ⇒ = ∵ (; )⇒ησα()()1
2xfzdzif z′ = π −α ∫
= N u m b e r o f z e r o e s o f ()fz−α inside x, where each
zero is counted according to its order.
= T h m ()fz−α has a zero of order m at za=)
Again, ()1;2dwiwσησ ξ= π −ξ ∫
()()()1;2xfzdzif z′ ησ ξ= π −ξ ∫ munotes.in
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∴ ()()()( )1;;2xfzdz n mif z′ = ησ ξ=σ α= π −ξ ∫ B y ( 1 ) a n d ();mησ α= ()fz⇒ =ξ has exactly m-roots in ();Ba ξ ∵ ()0fz′≠ for 0 za< − <ε ⇒ T h e e q u a t i o n ()fz=ξ has exactly m-simple roots in (),Ba ξ.
Definition : If x a n d α a r e M e t r i c s p a c e s a n d :fX Y→ h a s t h e
property that ()0f is open in α whenever U is open in X, then f is
called an open map.
Open Mapping Theorem :
Statement : Suppose G is a domain in , f is a non-constant
analytic function on G. Then for any open set U in G, ()fU is an
open. (2007, 2009)
Proof : Given that, f is a non-constant analytic function on G.
Let aU∈ and ()fa=α ∵ U is open 0⇒ ∃ ε> s.t. ();Ba U ε⊂. ∵ f is non-constant analytic function on G. ∴ b y f u n d a m e n t a l t h e o r e m o f a l g e b r a , ∃ an integer 1m≥. ()fz−α has a zero of order m at za=. ∴ b y u s i n g p r e v i o u s t h e o r e m , f o r t h e a b o v e 0, 0ε> ∃ δ> s . t .
for the above 0, 0ε> ∃ δ> s.t. ξ−α <δ and the equation ()fz=ξ
has exactly m simple roots in ();Bα ε.
T h u s ();B∨ ξ∈ α δ, we an find m points in ();Bα ε which
are mapped to ξ by f. ∴ ()()();;Bf Bα δ⊂ α ε ∴ ()();Bf Uα δ⊂ ()();BU α ε⊂∵ ∴ α is interior point of ()0f .
But α is arbitrary. ∴()fU is open.
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10.3 AUTOMORPHISMS OF THE UNIT DISC
A function :fD D→ i s s a i d t o b e a n A n a l y t i c a u t o m o r p h i s m o r
Automorphism of the Unit disc D if f is bijective and if both 1,ff−
are analytic in D.
Note: Let {}0: 1Dz z≠α∈ = ∈ <
For ,,zi nDα define the Mobius transformation ()1zzzα−αφ =−α
This Mobius transformation is analytic in D and als o in DD U D= ∂, ()zαφ is not analytic at a point 1z=α.
Which lies outside the disk D . ( 1D α∈ ⇒α< ∵)
111and z z⇒ = > =αα
Note: :0Dαφ→ is an analytic automorphism.
1. ()()00andααφ =−α φα =
2. for any point ,zD∈ ()zDαφ∈
o o−α +αD
−planeω−plane
Fig 10.10
i.e. ()1zz Dαφ < ∀∈
Since, ()()()() ()2
11zzzzzα−α −αφ =−α −α
()2zz z = ⋅∵
1zz z zzz z z⋅− α − α +αα= −α −α +αα ()
()22
221zz zzz z+α−α +α= +α− α +α()()2zzzz−− α +α< 2−α +α
….()1, 1zDz α∈ ⇒ < α<∵ munotes.in
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()21zα∴ φ < ()1zα∴ φ < ()zDα⇒ φ∈ α∴ φ maps D onto itself i.e. ()0Dαφ =
(3) If ,Dα∈ so is −α and for any ,zD∈
[]()[]()zz zαα − ααφφ = = φφ i.e. 1−ααφ=φ
Now () ()1zzzz−α α α α −α−α φφ =φφ =φ −α 1
1z
z
z−α+α−α=−α+αα
[] ()()22
221
11zzzzzzz−α α−α−α+α−αφφ == =−α+α− α − α
Similarly, []()zzα− αφφ =
1−α− α∴ φ=φ α⇒ φ m a p s D o n t o D i n a o n e - o n e m a n n e r H e n c e , 1−ααφ φand a r e
automorphisms of the Unit disc.
4. For (),∈∂ φ∈ ∂zzD z D i.e. ()1αφ = ∀∈ ∂zz D
Since, ()1α−αφ =−αzzz ∈∂∵zD ∴ For any point ∈∂zD
[],0 , 2θ= θ∈ πize Fig 12.10
()11θθ
αθ θ− θ−α −α∴ φ = =
−α −αii
i iieez
e ee θ
θ−α= −αi
ie
e … … … … ()1θθ − θ = =∵ii iea n d e e
()1αφ = z () = ∵zz ()α⇒ φ∈ ∂zD
α∴ φ ∂ ∂maps D onto D
i.e. ()αφ∂ =∂DD
combining results (2) and (4), we get
αφ maps D onto D. oD∂1z11.
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5. Mobius transformation ()1α−αφ = −α zzz is analytic in D,
()()() ( )
()211
1α−α +−α α′φ =
−αzz
z
z
()22112−α=
−α
In particular,
()210α′φ =−α ≠o (1 α∈ ⇒α< ∵ D
()2101α′φα = ≠−α )2211 0⇒α< ∴ −α +
Proposition: If 1inverse of αφ is −αφ. Furthermore, αφ maps ∂D onto ∂D,
() ( )20, 0 1αα′ φα = ⇒ φ =−z and ()211α′ φα =−α.
Theorem: Let :fD D→ be a one-one analytic map of D onto itself,
with ()0fα= and suppose that αφ is a one-one analytic map of D
onto itself with analytic inverse αφ. Then, there is a complex no. C
with 1C= s.t. fCα=φ.
Proof: Given that, :fD D→ is an one-one analytic map of D onto
itself with ()foα=.
Put (),11z
−α−αω= = φω α<−α ω
Define,
gf−α= φ
g⇒ i s a o n e - o n e a n a l y t i c m a p o f D o n t o i t s e l f a n d ()()()go f o f o−α= φ = α=
….. () ()1fo a n d−αω+α α= φω =+αω∵
and ()()()1gf f z−αω= φω =< …… ()(z−α = φω∵ a n d
:fD D a n d D→ is unit Disk
() ())1zD f z D f z∴ ∈⇒ ∈⇒ < ∴ by Schwarz’s Lemma,
()gω ≤ω D∀ ω∈
a n d ()1go′≤ munotes.in
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l e t ()gω=ω f o r s o m e oω≠ in D ∴ by second part of Schwarz’s Lemma,
T h e r e i s a c o m p l e x n o . C w i t h 1C= s.t.
().gcω=ω D∀ ω∈ ()fc−α⇒ φω =ω ()()fz c zα ⇒ = φ fcα⇒ = φ
10.4 SUMMARY
1) The Maximum Modulus Principle:
Let G be any subset of . A complex function f defined on G is
said to have local maximum modulus at a point α i n G i f , t h e r e
exists 0δ> s.t. ();BGα δ⊂ and ()()();fz f z B ≤ α ∨ ∈α δ.
Similarly, f h a s l o c a l m i n i m u m m o d u l u s a t a p o i n t α i n G, if 0∃ δ> s.t. ();BGα δ⊂ and ()()fz f ≥ α (),zB∨ ∈α δ.
2) Minimum Modulus Principle : Suppose f i s a n o n - c o n s t a n t
and analytic function in a domain G. If f a t t a i n s i t s l o c a l
minimum G at α, then ()0fα=.
3) Schwarz’s Lemma: L e t {}:1Dz z=∈ < be the unit disk and
suppose f is analytic in D with, (i) f (0) = 0 and (ii) ()1fz f o r z D≤ ∈. Then, ()()01fz z z D a n df ′≤ ∀∈ ≤.
Moreover, if ()fz z= for some 0,z≠ t h e n th e r e i s a c on s ta n t C
with 1c= s.t. ().fcω=ωD∀ω∈.
4) Theorem : If f is a non-constant analytic function in a bounded
domain G and ()0fz≠ for any zG∈, then f cannot attain its
minimum in G.
5) Theorem : Let f b e a n a l y t i c i n a d o m a i n G w i t h z e r o s
12,, . . . ,mα α α (repeated according to order)
If x is a smooth closed curve in G which does not pass through any ks′α then ()()()
11;2mkk xfzdz n xif z=′ = α π ∑ ∫
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6) Open Mapping Theorem :
Statement : Suppose G i s a d o m a i n i n , f i s a n o n - c o n s t a n t
analytic function on G. Then for any open set U in G, ()fU is an
open.
7) A function :fD D→ is said to be an Analytic automorphism or
Automorphism of the unit disc D, if f is bijec tive and if both 1,ff− are analytic in D.
10.5 UNIT END EXERCISES
1) Find the maximum modulus of 2zz−in the disc 1z≤.
Solution: ∴ 2(1 )zz z z−= − ∴The maximum modulus is assumed at the boundary o f the disc 1z≤ That is at 1z=−. ∴2
1max 2zzz≤−=
2) Show that the maximum modulus of ze is always assumed on
the boundary of the compact domain .
Solution: Since zxee= where zx i y=+ ∴ze is maximum at a point in the domain with ma ximal x.
(At a point farthest to the right. )
3) Suppose f,g both are analytic in a compact domain D. Show
that () ()fz g z+ takes it’s maximum on the boundary.
( Hint: Take () , ()iifz A e g z B eαβ−== then put () () ()hz f z gz=+ () () () () ()hz f z gz f z gz⇒ =+ ≤ +
Let 0z be an interior point of a compact domain D. Assume that fg+ takes maximum values inside D, say 00() ()fz g z+ 0000() () ()() () ( ) ( ) ( )() ( )hz f z gzfz g z fz g z h zhz hz∴ =+
≤ + ≤ +=
∴≤ ∴The analytic function ()hzassumes it’s maximum at the
interior point 0z ( not on the boundary), which is not possible . ∴() ()fz g z+ takes it’s maximum on the boundary. munotes.in
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4) Let f be analytic and bounded by 1 in the Unit disc a nd
102f= Estimate 34f.
Solution: Since 102f=, define :g→ as follows:
()()1/122fzgzzz= −− for 12z≠
3142f′= for 12z=
Then g i s an a ly t i c i n 1z≤ Letting 1z→we find that 1g≤ on
the disc . ∴12()212z
fz−≤− ∴ for 34z=, ∴ 3245f≤
5) Show that among all functions , which are a nalytic and
bounded by 1, in the Unit disc, Max 13f′ is assumed, when
103f =.
Solution: Suppose 103f ≠, consider 1()3()11( )3fz f
gzff z−=−
∴ 1
31113wf
fw−=− when 1w= and 1f< in 1z<, ∴ By Maximum-Modulus Theorem 1g< in 1z< .
By direct calculations 21
1 3
3113f
g
f′ ′ =− ∴ 1133gf ′′ >
This is a contradiction.
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6) Show that the automorphisms of the Unit D isc are of the
form () , 1izgz eazθ ααα−= <−.
Solution: Let ()1zgzzαα−=−. Then () 1gz=for 1z=.
Since () 0ggα=⇒is an automorphism of the Unit disc.
Assume that f i s a n a u t o m o r p h i s m o f t h e U n i t D i s c w i t h () 0fα=. ∴1hf o g−= is an automorphism with (0) 0h=. ∴By the lemma that describes automorphisms of t he Unit disc , ∴()ihz e zθ= or ()1izfz ezθαα−=− .
*****
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11
SINGULARITIES
Unit Structure
11.0 Objectives
11.1 Introduction
11.2 Three Types of isolated Singularities
11.3 Laurent’s Theorem
11.4 Classification of Singularities by the Princi pal Part of
Laurent’s Expansion
11.5 Casorati- Weirstrass Theorem
11.6 Summary
11.7 Unit End Exercises
11.0 OBJECTIVES
After going through this unit , you will understand the
concept of continuing an analytic function to another region.
We shall also study three types of singula rities of a function ()fz and the theorems like Casorati-Weirstrass th eorem and the
Laurent’s theorem.
Given a singularity 0z of a function ()fz, we shall try to classify
the singularities by finding the principal part of Laurent series
expansion of a function ()fz.
11.1 INTRODUCTION
We shall recall the uniqueness theorem that s tates that if f is analytic in a region D and {}nzis a sequence of distinct
points such that ()0nfz n= ∀ a n d {}nzconverges to some
0z∈, then fis identically zero in a region D. Suppose we are
given a function f, which is analytic in region D. The
question is that of continuing fanalytically to a region 1D such
that gf=on 1DD∩. By the uniqueness theorem such munotes.in
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continuation of fis uniquely determined. The Schwarz
reflection principle is an example of how, in some cases, an
analytic function can be continued beyond i t’s original domain
of analyticity. In this unit, we shall exami ne the possibilities
of such extensions for functions given by power series.
11.2 THREE TYPES OF ISOLATED SINGULARITIES
Definition :
A point at which the function f i s n o t a n a l y t i c i s s a i d t o b e a
singular point or singularity of the function f.
e.g. ()23zfxz= −
Here, f i s n o t d e f i n e d a t 3z= a n d h e n c e n o t a n a l y t i c a t 3z=,
therefore 3z= is singular point.
Definition : A point at which the function f is analytic is said to be a
Regular point .
Definition : A function f has isolated singular point at 0zz= if ∃
an 0R> s . t . f i s d e f i n e d a n d a n a l y t i c i n 0 0zz R< − < b u t n o t ()0,Bz R .
e.g. 1) ()() ( )213zfzzz= −− 1z=and 3z=are points of singularity
2) ()cotfz z= sincoszz=
Put cos z = 0⇒z=nπ ∴Singular points are ,nnπ ∈.
Definition : Let f be analytic 0 0zz R< − <. Let 0z be an isolated
singular point of f. A point 0zz= i s s a i d t o b e a R e m o v a b l e
singularity o f f, if ∃ a n a n a l y t i c f u n c t i o n ()0 :,gBz R →g s.t. ()()fz g z= for 0 0zz R< − <.(2007)
Or
Definition : If a single valued function ()fz is not defined at a
point 0zz= but ()0lim
zzfz→ exists. Then 0zz= is said to be a
removable singularity of f.
e.g. ()sin,0zfz zz = ≠ munotes.in
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In this case, f i s n o t d e f i n e d a t a p o i n t 0z= but
0sinlim 1zz
z → =
exists. ∴ 0z= is a Removable singularity of f.
OR
Define ():0 ;gB R → s.t.
()sin010zzzgz
z ≠=
= ∴ ()()fz g z= f o r 00zR< − <
29() , 33zfz zz− = ≠−
Here, f is not defined at 3z=.
But, ()()()2
33 333 9lim lim lim 633 zz zzz zfzzz →→ → − + − = = =−− exist. ∴ 3z= is a removable singularity of f.
Definition: A singular point which is not isolated is said to be Non-
isolated singular point.
e.g. ()1fz c o s e cz=1sin( )z=
In the delta nbd at zeros, there are other singular point of f ∴z = 0 is
a non isolated singular point of f.
For, Singular points, Put 10sinz=
1nnz∴ = π ∈
10 znn= →→ ∞π, 012n, ,, . . . .= ± ± .Since 10zn= →πas n→∞
Here, z = 0 is a non-isolated singular point, whereas other s ingular
points are isolated.
Theorem: If f has an isolated singular point at 0z, then 0zz=is a
removable singularity of f iff ()()000lim z z f z
zz − =→
Proof: Let ()()000lim z z f z
zz − =→
T.P.T. z = z 0 is a removable singularity of f.
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∴ there exists R > 0 s.t. f is defined and analytic in (){}00;\Bz R z
but not in ()0;Bz R
Define
000() ( )() ,0zzzz f zhzzz≠− = = ( I ) h∴ is analytic in (){}00;\Bz R z and
0000lim ( ) lim ( ) ( ) 0 ( )zz zzhz z z f z hz→→= − == from (I) h∴ is continuous in ()0;Bz R
T.P.T. h is analytic in ()0;Bz R
i.e. T.P.T. () 0hzd z
∂∆=∫ for every triangle int∆= ∆+∂∆ in ()0;Bz R
There are four cases : ∴ By Morera’s Theorem,
h is analytic in ()0,Bz R
from equation (I), ()00hz= ∴ 0z is a zero (root) of h ∴ f an analytic function ():; . .gBzR s t → ()0() ()hz z z gz=−
w h e r e0()0gz≠ ∴ for 0 0zz R<−< ()()()0hz z z gz=−()0()zz f z=− by (I) () ()fz g z∴ = for 0 0zz R<−< ⇒ 0zz= is a removable singularity of f.
Conversely,
S u p p o s e 0zz=is a removable singularity of f.
T.P.T. ()00lim ( ) 0zzzz f z
→− =
By definition, f an analytic function ()0 :,gBz R →
s.t. () ()fz g z= f o r 0 0zz R<−<
000lim ( ) lim ( ) ( ) 0zz zzfz g z g z
→→== ≠ ()000lim ( ) 0 ( ) 0zzzz f z g z
→∴ − =× = ()00lim ( ) 0zzzz f z
→∴ − = munotes.in
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Definition:
If f has an isolated singular point at 0,zthen 0zz= is a pole of f
if 0lim ( )
zzfz
→=∞i.e. for any M >0, ∃0δ> s.t. ()fz M≥where 0 0zzδ<−<
Definition:
If f h a s a p o l e a t 0zz=and 1m≥ i s t h e s m a l l e s t p o s i t i v e
integer s.t. 0() ( )mzz f z− has a removable singularity at 0zz= then, f
has a pole of order m at 0zz=.
Definition:
A p o l e o f o r d e r o n e i s s a i d t o b e a s i m p l e p o l e .
e.g. 3()4zfzz=−
Here, z = 4 is an isolated singular point at f.
3
44lim ( ) lim4 zzzfzz →→= = ∞− 4z∴ = is a simple pole of f.
24()() ( 1 )zfz
zz z=
−−
Here f has simple pole at z = 2 and z =1 is a pole of order 4.
Essential Singularity:
A n i s o l a t e d S i n g u l a r p o i n t w h i c h i s n e i t h e r a p o l e nor a
removable singularity is said to be Essential singu larity. e.g. 1()zfz e=
Here, f has essential singularity at z = 0. T.P.T. z = 0 neither a pole
nor a removable singularity.
Theorem: If a function f(z) of analytic for all finite value s of z and as
(),kzf z a z→∞ = then f(z) is a polynomial of degree ≤ k.
Proof: Since f(z) is analytic for all finite values of z therefore it can
be expanded by Taylor’s theorem in the form
0()nn
nfz a z∞
==∑ , for zR<, where R is large.
Let max ()fz M=on the circle ()zr r R=< . Then by Cauchy’s
inequality , we have nnMar≤ for all values of n munotes.in
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k
nk
nArAr r n k kr−== →∞ ∀ >≤, since ()fz=kAz when z→∞
knknArArr−== , which tends to zero when r→∞since n>k.
Thus an=0, nk>.
Hence, we have f ( z ) = a 0+ a 1z+ a 2 z2+………. +a k zk, which is a
polynomial of degree k≤.
11.3 LAURENT’S THEOREM
Theorem: If f is analytic in ()21 2;, , 0nn Ga R R R= α > then for any
point z in G, f has unique representation
01() ( )()n
nnnnbnfz a z
z∞∞
=== −α + −α∑∑
where, 111( )
2 ()nn
xfadi+ξ= ξ π ξ−α∫, n
= 0, 1, 2 .. and 121( )
2 ()nn
xfbdi−+ξ= ξ π ξ−α∫, n = 1, 2, ……
and 12,xx a r e c i r c l e s 12,rrξ−α = ξ−α = r e s p e c t i v e l y w i t h 221 1Rrr R<< <
Proof: for a given ,zG∈ choose 12ra n d r s.t. 22 1 1Rr z r R<< −α<<
by using Cauchy integral formula for multiply conne cted domain 121( ) 1( )()22 2xxfffz d diz iξξ= ξ− ξπξ − πξ −∫∫ (1)
Consider, 11( )22xfdiξ ξπξ −∫
For any point 1on xξ
Consider
()()
()11 1
1zz z= = ξ− ξ−α +α− −ξξ−α −ξ−α
()()
()()
()()
() ()
()1
11 11. . . . . .
1nn
nnzz z
z−
− −α −α −α =+ + + + − −αξ−α ξ−α ξ−α ξ−α− ξ−α
… … .23 11111 . . . .11nnxx x x xxx− = + + + + + + −− ∵ munotes.in
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()()
()()
()()
() ()1
21.....nn
nnzz z−−α −α −α= + + + +ξ−αξ−α ξ−α ξ−α ξ−α
Multiply the above equation by ()//2fiξπ and then integrating w.r.t. 1over xξ ()()()()()2
11 111....22xx xff z fdd diz i z iξξ − α ξ∴ ξ= ξ+ ξ+πξ − πξ − α π ξ−α∫∫ ∫
()()()() ()1
11() ()......22nn
nn
xxff zzdii−ξξ −α −α+ ξ+ ππξ−α ξ−α ξ−α∫∫
Given that, ()()1
11
2n
xfan di+ξ= ξπ ξ−α∫ n = 0 , 1 , 2 , 3 … … . . ()()1
01 1
110.... ( )2nnnxfda a z a z Riz−
−ξ∴ ξ=+ −α++ −α +πξ −∫ (2)
W h e r e , ()
() ( )1()
2n
n n
xz fRdi z−α ξ= ξπ ξ−α ξ−∫
T.P.T. lim 0nnR
→∞ =
()12n
n n
xfzdRzξ−α ξ≤ πξ −ξ−α∫ ( 3 )
Choose 0 >9 s.t. z−α= 9equation of the circle 1x is, rξ−α = ,
Now, 1 zz z rξ−=ξ−α+α− ≥ ξ−α − −α =− 9
Given that, f is analytic in G ∴ f is continuous on 1x (Compact set)
By boundedness theorem ∃10M> s.t. ()fMξ≤ 1x∀ ξ∈
Put all the above values in equation (3) we get
() ()11
11 11112 2n n
nnn
xxMMRd drr rr≤ ξ= ξ π −π −∫∫9 9
99
()111122nMrrr= ⋅ ππ− 9
911
110nMras nrr= → →∞−9
9 lim 0nnR
→∞∴ = munotes.in
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∴ from equation (2) ()()()
011
2nn
n xfda ziz∞
=ξ ξ = −απξ −∑ ∫ ( 4 )
Consider,
()()
21
2xfdizξ − ξ πξ −∫
For any point ξ on 2x,
Consider
()
()11 1 1
1zz zzz−== =ξ− −ξ −α +α−ξξ−α−α −−α
()()
()()
()()
()1
11 11. . . . . .
1nn
nn zz zz
z−
− ξ−α ξ−α ξ−α= + + + + − ξα−α −α −α −α −−α
()
()()
()()
() ()1
21.....nn
nn z zz z z−ξ−α ξ−α ξ−α= + + + +−α−α −α −α −α
Solving in the same manner as above, we get, ()() ()121
2nnn xf bdiz z∞
=ξ − ξ= πξ −−α∑ ∫ _ _ _ _ _ _ _ _ _ _ _ _()∗
From equation (1), (4) and ()∗, we get
()
() 01()n nnnnnbfz a z
z∞∞
=== −α + −α∑∑ __________ ()∗∗
Note: (i) equation ()∗∗ can also be written as () ( )nn
nfz a z∞
=−∞= −α∑
for 21Rz R<−α<
1) Where x is the circle rξ−α = with 21Rr R<<
()()11
2n n
xfadi+ξ= ξπ ξ−α∫ 0, 1, 2n= ± ±
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221
Proof: If xis the circle rξ−α = s.t. 21,Rr R<< then both functions ()()()()11nnffand+ −+ξξ
ξ−α ξ−α a r e a n a l y t i c i n 21RR<ξ−α < b y u s i n g
Cauchy De-formation Theorem, ()()()()11
111
22n nn
xxffad dii++ξξ= ξ= ξππ ξ−α ξ−α∫∫ ()()()()11
211
22n nn
xxffbd dii−++ +ξξ= ξ= ξππ ξ−α ξ−α∫∫ Fx1
12=∫∫FFxx2x
Fig 11.1
We observe that, nnba−=
From ()∗∗
() ( ) ( )
01nnnn
nnfz a z a z∞∞−−
=== −α + −α ∑∑
() ()
01nnnn
nnaz az∞− ∞
== −= −α + −α∑∑
() ()nn
nfz a z∞
=−∞= −α∑
Where ()()11
2n n
xfadi+ξ= ξπ ξ−α∫ 0, 1, 2.......n= ± ±
2. The Laurent Series expansion is Unique.
Proof: Suppose that we have another Laurent series expansi on
() ()mm
mfz A z∞
=−∞= −α∑ f o r 21Rz R<−α< ( 1 ) munotes.in
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We prove that equation (1) is identical with
() ()nn
nfz a z∞
=−∞= −α∑ f o r 21Rz R<−α<
Where ()()11
2n n
xfadi+ξ= ξπ ξ−α∫, 0, 1, 2.......n= ± ±
To prove that nmaA=
Let x be the circle rξ−α = with 21Rr R< < ()()11
2n n
xfadi+ξ= ξπ ξ−α∫= ()
()111
2m
m n
mxAdi∞
+
=−∞ ξ−α ξπ ξ−α∑∫i
()1 1
2mn
m
mxAdi∞−−
=−∞= ξ−α ξπ∑∫ ()1 1
2mn
m
m xAdi∞−−
=−∞= ξ−α ξπ∑ ∫
()1 1
2mniim
mAr e i r ei2π ∞ −−θθ=−∞ 0= π∑ ∫ r ξ−α =∵
[]0, 2ireθ⇒ ξ =α+ θ∈ π
() 1
2mn i mn
m
mAr e di2π ∞−− θ
=−∞ 0= θπ∑ ∫ . This integral 2,mn= π =
0mn= ≠
0 122mAri= ππ nmaA= ∴ Laurent series is unique in G.
Note:
1. The Laurent series for given function
is () ( ) ( )
10nnnn
nnfz a z a z−∞ −∞
=− == −α + −α∑∑ 21Rz R<−α<
1) The part ()
1nn
naz∞
=−−α∑ of Laurent series is called the
principal part of f(z) at z=α.
2) The part ()
0nn
naz∞
=−α∑ of Laurent series is called the
Analytic part of f(z) at z=α.
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2. If f has a pole of order m at z=α, then ()
()()mgzfz
z=
−α where, g
is analytic at a point α and ()0gα≠
e.g.
()23()3zfz
z=
− 2()gz z= is analytic at point 3 and 2(3) 3 9 0.g = = ≠
11.4 CLASSIFICATION OF SINGULARITIES BY THE
PRINCIPAL PART OF LAURRENT’S EXPANSION
Corollary: L e t z=α b e a n i s o l a t e d s i n g u l a r i t y o f ()fz a n d l e t
() ()nn
nfz a z∞
=−∞= −α∑ b e i t s L a u r e n t e x p a n s i o n i n ();0 ,nnaRα
0zR<−α< (Punctured disk or deleted nbd of α).
Then,
i) z=α i s a r e m o v a b l e s i n g u l a r i t y o f f i f f 01naf o r n= ≤− i . e .
(Principal part is zero) (2008)
ii) z=α is a pole of order m iff 00mnaa n d a−≠ ≠ for ()1nm≤− +.
i.e (Principal part is finite)
iii) z=α is an essential singularity of f iff 0na≠for infinitely many
negative integers n. i.e. (Principal part is infini te)
Proof: Given, () ()nn
nfz a z∞
=−∞= −α∑
() ()
10()nnnn
nnfz a z a z−∞ ∞
=− == −α + −α∑∑ ( 1 )
i) Let 0na= for 1n≤−
T.P.T. z=α is a removable singularity of f.
From equation (1)
() ( )
0nn
nfz a z∞
== −α∑ ()01naf o r n = ≤∵
() 0
1()nn
nfz a a z∞
==+ −α∑
() 00 0
1lim ( ) lim 0 0n
nzznfz a a z a a∞
→α →α==+ −α =+ = ≠
∑ munotes.in
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()()()()0lim lim lim 0 0zz zzf z z f z a
→α →α →α−α = −α = =ii z∴ =α is a removable singularity of f
Conversely, suppose, z=α is a removable singularity of f. ∴ ∃ an analytic function ():;gB Rα→ s.t.
()()fz g z= i n 0zR<−α< ∵g is analytic in ();BRα ∴ for any point ();zB R∈α
g has Taylor series expansion
0() ( )nn
ngz a z∞
=∴ = −α∑ ∴ t h e L a u r e n t s e r i e s e x p a n s i o n f o r f ( z ) m u s t c o i n c i d e with the
Taylor series expansion for g(z) about z=α
()
0()nn
nfz a z∞
=∴ = −α∑ 01naf o r n⇒ = ≤− (Compare equation (1) and above equation)
E.g. 3sin() , 0zzfz zz−= ≠
357
3.......3! 5! 7!
()zzzzz
fzz−− + + += 241...........3! 5! 7!zz= −+ −
Principal part is zero i.e. 01naf o r n= ≤−.
This is a Laurent series expansion for f(z) but pri ncipal part contain
no negative power of z.
i.e. 01naf o r n= ≤− 0z∴ = is a removable singularity of f.
ii) Given 0( 1 )mnaa n d a f o r n m−≠ = ≤− +
() ( )nn
nfz a z∞
=−∞= −α∑
10() ()nnnn
nnaz az∞∞
=− == −α + −α∑∑
() ()()1
0() . . . . . .nmn m
na afz a zz z∞
− −
==+ + + −α−α −α∑
Multiplying above equation by (),mz−α munotes.in
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() ()1
1
0() . . . . . ( ) ( )m nmm
m n
nzf z a a z z a z∞−
−−
=−α =+ + −α +−α −α∑
() () () ()1
1
0lim ( ) lim ......m mm nm nzznzf z a a z z a z∞−
−−→α →α= −α =+ + −α +−α −α ∑
00. . . .00ma−=+ + + + + 0ma−= ≠
() ()1lim ( ) lim 0m
mzzzf zz a+
−→α →α∴ −α = −α = ∴ the function ()()mzf z−α has a removable singularity at z=α z∴ = α is a pole of order m
Converse, (Exercise)
E.g. 3() ,zefzz= Here f has pole of order 3 at z = 0
23
23
331. . . .2! 3! 1() 1 . . . .2! 3!zzz
zzfz z
zz
++ + + = = + + + +3211 11.......2!. 3! 4!z
zzz= + + + + +
This is a Laurent series expansion for f (z) b u t p r i n c i p a l p a r t o f
Laurent series is finite. ∴ f has a pole of order 3 at z = 0
iii) Combine part (i) and (ii) and by definition es sential singularity,
we see thatz−α i s a n e s s e n t i a l s i n g u l a r i t y o f f i f f 0na≠ f o r
infinitely many negative integers n.
e.g. 1()zfz e=
23411 1 11. . . . . . .2! 3! 4!
Prinicpal partz zzz=+ + + + +
This is a Laurent series expansion for f(z) a n d p r i n c i p a l p a r t o f
Laurent series is infinite. 0z∴ = is an essential singularity of f.
Theorem: Let f be analytic in ()00zR R<−α< >. Then, f has a
pole of order m at z=α i f f t h e r e e x i s t s a n a n a l y t i c f u n c t i o n , ():;gB R α → s . t . ()()
()mgzfz
z= −αwhere ()0gα≠ munotes.in
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(Note: We know that the geometric series
0nnz∞
=∑ c o n v e r g e s f o r 1z< and we write 23
011. . . . . . .1n
nzz z zz∞
=== + + + +−∑
Similarly, the geometric series
0(1 )nnnz∞
=−∑ converges for 1z< and
we write 234
01(1 ) 1 . . . . . .1nn
nzz z z zz∞
== − =−+ −++∑ )
Example:1
Expand 25()6fzzz=+− in a Laurent series valid for (i) 23z<<
(ii) 2z< ( i i i ) 3z> (iv) 02 4z<−<
Solution: Give function
() ()255 1 1()23 2 36fzzz z z zz== = −− + − ++− ( 1 )
1) For 23 ,z<< if 2z< then 21z<
10011 1 2 22 21nnnnnzz z zzz∞∞+== = = = − −∑∑
if 313zzt h e n< < 10011 1(1 ) (1 )33 3 3313nnnnnnnzzz z∞∞+==== − = − + +∑∑
From equation (1), we get
()11002() 1
3nnnnnnnzfz
z∞∞++=== −− ∑∑
32 2
432 2322 2 11......... .....23 33zz
zzz=+ + + + −− + −
32 2
432 2322 2 1 1........ .......333zz
zzzz=+ + + + −+++ (In this case, z = 0
is an essential singularity)
This is the required Laurent series for 23z<<
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ii) For 2z<
23 1 123zzza n d < < ⇒ < <∵
()100 211 1
22 221 2nnnznnzzz∞∞+==− == − = − − − ∑∑
()10011 11( 1 )33 3 3313nnn nnnnzzz z∞∞+ = === − = − + +∑∑ ∴ from equation (1) 1100() (1 )
23nnnnnnnzzfz∞∞++===− −− ∑∑
This is the required Laurent series for 2z< (In this case, z = 0 is
removable singularity)
iii) For 3z>
2332 1 1za n dzz > > ⇒ < <∵
()120011 1 2 22 1 2nnnnn zzz z z∞∞+==== = − − ∑∑
()130011 1 3 33 1nnnnn zzz z z z∞∞+==== = + + ∑∑
from equation (1), 110023()nnnnnnfz
zz∞∞++=== − ∑∑
This is the required Laurent series (In this case z = 0 i s e s s e n t i a l
singularity)
iv) For 02 4z<−<
put z – 2 = u ⇒ z = u + 2 ∴ from equation (1)
() () () ( ) ()55 5()23 22 5 5fzzz zz u u= = =− + −− ++ 04 5 15uu<< < ⇒ <
() 0 551() (1 )5451nn
unufzu∞
=∴ = = −⋅+∑
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Example: 2
Expand 24()23fzzz=+ − in a Laurent series valid for
(i) 13z<< ( i i ) 3z> ( i i i ) 1z<
(iv) 01 4z<−<
Example 3:
Expand 4cos( ) 1()zfzz− = in a Laurent series about z = 0 and name
the singularity .
Solution:41()Coszfzz−=
246
41. . . . . . . . 12! 4! 6!zzzz−+ −+ −=2
211..........4! 6!2!zz−= + −+
Z = 0 i s a p o l e o f o r d e r 2
Example 4: Find the Laurent series of f(z)=() ( )112zz z−− in the
annular region 12z<<. (2012)
Solution: f(z)=() ( )112zz z−−
By partial fraction 112 , 1 & 12zfor zz<< < <
()
() ()11 1
1 2 14 12fzzzzz∴ = −−
−− ()()111111224zzzz−−−−
= −−
2
211 11 11. . . . . 1. . . . . . . .24 2 4zz
zz zz== − ++ + − ++ +
Definition:
A set D in a Metric space X is said to be Dense iff ,DX= where ,D
is closure of D.
OR
Let D be a subset of ()D⊆, we say that D is dense in if, for
any 00andω∈ ∈> ()0, Bn Dω∈ ≠φ
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11.5 CASORATI -WEIERSTRASS THEOREM
If f has an essential singularity at ,z=α t h e n f o r e v e r y 0δ>, ();0 ,nnfaα δ is dense in . (2005, 2008) OR ()();0,nnfa αδ=
Proof: Let ()(){};0, 0 ; \nn Ga z B= αδ =< −α<δ= αδ α
Given that, f has an essential singularity at z=α ∴ f is analytic in G.
T.P.T. f(G) is dense in i.e. (){}fG=
i.e. T.P.T. for given 0,0 , 0 ,zω∈ ∈> δ> ∃
s.t. 0 () za n d f z−α<δ −ω<∈
Fig 11.1
Assume this is not true
i.e. Assume there is 00andω∈ ∈> s.t.
0 ()fz−ω ≥∈ ()2; 0 ,nn Ga∀∈ = αδ
0 ()lim
zfz
z →α−ω∴ = ∞−α
The function ()0fzzωα−− has a pole at z=α 0zz= is a pole of f if 0lim ( )
zzfz
→=∞
Note: If lim 0xx
→α−α= then 2lim 0xx
→α−α = −p l a n e αδ x2−p l a n e −p l a n e 0ωε
ω−p l a n e f (2) munotes.in
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If 1m≥ i s t h e o r d e r o f t h i s p o l e t h e n , ()()0() mfzzz−ω−α−α h a s a
removable singularity at z=α.
1
0lim ( ) 0m
zzf z+
→α∴ −α −ω = ( 1 )
Now, 11() ()mmzf z zf z c c++−α =−α − +
11 1lim ( ) lim ( )mm m
zzzf z zf z c z c++ +
→α →α ∴ −α ≤ −α − +−α
= 0 + 0 - - - - - - - - - b y ( 1 )
1lim ( ) 0m
zzf z+
→α∴ −α = ∴the function () ( )mzf z−α has a removable singularity at z=α ∴f has a pole of order m at z=α which contradicts the hypothesis
that z=α is an essential singularity of f. ∴ Our assumption was wrong
Hence, 0 ()fz−ω<∈ zG∀ ∈ ()fG⇒ is dense in i.e. {}()fG= 0zz∗ = is a removable singularity ()00lim ( ) 0zzzz f z
→⇔ − = ∗∗ If z=α is an essential singularity of f then f has Laurent series
expansion about z=α. ∗∗∗ I f f i s a n a l y t i c i n ();Bαγ t h a n f o r a n y ();zB∈α γ a n d h a s a
Taylor series expansion about z = α.
()1(1 )
1() ( ) () () ()mm mm
mdgz z f z g z z f zdz−−
− = −α ⇒ = −α∵
() () ( )1(1 ) (1 )lim ( ) limmm mm
zzdgg z z f zdz−−−
→α →α⇒ α= = −α
11.6 SUMMARY
1) A point at which the function f i s n o t a n a ly t i c i s s a i d t o b e a
singular point or singularity of the function f.
2) A function f has isolated singular point at 0zz= if ∃ an 0R>
s.t. f is defined and analytic in 0 0zz R< − < but not ()0,Bz R . munotes.in
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3) If f has an isolated singular point at 0z, then 0zz=is a removable
singularity of f iff ()()000lim z z f z
zz − =→
4) If f has a pole at 0zz=and 1m≥ is the smallest positive integer
s.t. 0() ( )mzz f z− has a removable singularity at 0zz= then, f has a
pole of order m at 0zz=
5) Laurent Theorem: If f is analytic in ()21 2;, , 0nn Ga R R R= α >
then for any point z in G, f has unique representat ion
01() ( )()n
nnnnbnfz a z
z∞∞
=== −α + −α∑∑
Where, 111( )
2 ()nn
xfadi+ξ= ξ π ξ−α∫ n = 0 , 1 , 2 , …
,121( )
2 ()nn
xfbdi−+ξ= ξ π ξ−α∫ n = 1 , 2 , 3 , … … … … …
and 12,xx a r e c i r c l e s 12,rrξ−α = ξ−α = r e s p e c t i v e l y w i t h 221 1Rrr R<< <.
6) Let f be analytic in ()00zR R<−α< > T h e n , f h a s a p o l e o f
order m at z=α iff there exists an analytic function, ():;gB R α → s . t .
()()
()mgzfz
z= −α w h e r e ()0gα≠
7) Casorati Weierstrass Theorem:
If f has an essential singularity at ,z=α t h e n f o r e v e r y 0δ>, ();0 ,nnfaα δ is dense in .
11.7 UNIT END EXERCISES
1) Each of the following functions f has an isolated singularity
at 0z=. Determine it’s nature , if it is removab le singularity ,
define (0)f, so that f is analytic at 0z= if it is a pole, find the
singular part; if it is an essential singula rity determine {}():0 fz zδ<< for arbitrarily small values.
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(a) sin( )()zfzz= (b) cos( ) 1()zfzz−=
(c) 21()(1 )zfzzz+=− ( d) 1() s i nfz zz=.
Solution: (a) ∴ 00sin( )lim ( ) lim 1zzzfzz→→==exists. ∴ 0z= is a removable singularity of f .
Define :( 0 ; )gB r→ as sin( )() , 0zgz zz= ≠1, 0z= =.
Then () ()fz g z=for 0zR<
(b) ∴ 00(cos( ) 1)lim ( 0) ( ) lim 0zzzzzf zz→→−− ==
∴ By theorem on removable singularity, 0z= is a removable
singularity of f . ∴Define (0) 0f=.
Define :( 0 ; )gB r→ as cos( ) 1() 0sfz zz−= ≠
00z= =
Then () ()fz g z=for 0zR<
(c) f has a pole at 0z=.
2) Classify the singularities of (a) cot(z) ( b ) 2exp(1/ )1zz−
(c) ()
() ( )2sin12zfz
zz z=
−− ( 2 0 0 5 )
Solution: (b) Let 21exp
()1zfzz=−.
2
111explim ( ) lim ( )1zzzfz fzz→→
== ∞⇒−∵has a pole at 1z=.
Since ()()()AzfzBz= , where 21() e x p , () 1Az Bz zz== −.
(1) 0, (1) 0AB B≠ = ⇒has a zero of order 1 at 1z=/g19665 ()fzhas a pole of order 1 at 1z=.
We know that f i s a n a l y t i c i n a d e l e t e d n e i g h b o u r h o o d (0; ) 0BR− o f 0 a n d f i s n o t a n a l y t i c a t 00zz=⇒ =is a
singularity of f. Since we know that 211exp
lim1zzz−− does not exist . munotes.in
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∴0z= is not a removable singularity of f. ()fzcant be written as 2() ()()()Az AzfzBzz== , where A and
are
analytic at 0z= ∴ 0z= is not a pole of f. ∴0z= is an essential singularity of f.
(c) ()
() ( )2sin12zfz
zz z=
−−
Solution: f(z) has pole at z=0, z=1, z=2 of order 1,1,2 respectively.
3) Find the Laurent series expansion of (a) 221zz+ about 0z=.
(b) 214z− about 0z=.
Solution:
(a)
Let242 2 2 211 1 1()11 ( )fz zzz z z z−== − = −
++ −−220(1 )kkkzz− ∞
= = −−∑121(1 )kkkz∞+
−= −∑
4) Check whether 0z= i s a r e m o v a b l e s i n g u l a r i t y o f sin( )()zfzz= or not.
Solution: sin( )()zfzz= ∴ 24sin( )() 1 . . .3! 5!zz zfzz== −+ −
Here all the coefficients 0kc−=for 0k>. ∴ f(z) has removable singularity at z=0
5) Show that the image of (0;1) 0B− u n d e r t h e f u n c t i o n 1() c o sfz e cz= is dense in the Complex plane. (Hint : 0z= is
essential singularity of 1sinz. Make use of the following
theorem: If f i s a n a l y t i c i n a d e l e t e d n e i g h b o u r h o o d D o f 0z
except for poles at all points of a sequence {}0nzz→. Then ()fD is dense in the Complex plane. munotes.in
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6)Expand 12()zfz z e = in a Laurent series about z = 0 and name the
singularity.
7) Expand
()23()1zefz
z=
− in a Laurent series about z = 1 and name
the singularity. (Hint:
() ()()()()21222 2 233 3()
11 1zzz eeee e efz
zz z−−⋅== =
−− −)
8) Determine the number of zeroes, counting multipl icities, of the
polynomial z4-2z3+9z2+z-1 inside the circle 2z=
9) Expand ()() ( )12zfzzz=−− i n a L a u r e n t s e r i e s v a l i d f o r
(i)02z<< (ii) 1z< (2008)
10) Expand ()
()231zefz
z=
− in a Laurent series about z=1 and name
the singularity. (2007)
11) Expand ()() ( )12zfzzz=−− i n a L a u r e n t s e r i e s v a l i d f o r
(i)12z<< (ii) 2z> ( 2 0 0 7 )
12) Expand ()()()112fzzz−=−− in a Laurent series valid for
( i ) 12z<< , (ii) 2 z<< ∞ .
13) Expand () ( )13s i n2fz zz =−+ i n a L a u r e n t s e r i e s a b o u t 2z=− and name the singularity.
(14) Expand ()2143fzzz = + + in a Laurent series valid for
( i )13z < < , (ii) 3z < .
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12
RESIDUE CALCULUS AND
MEROMORPHIC FUNCTIONS
Unit Structure
12.0 Objectives
12.1 Introduction
12.2 The Residue Theorem and it’s Application
12.3 Evaluation of Standard Types of Integrals by t h e R e s i d u e
Calculus Method
12.4 Argument Principle
12.5 Rouche’s Theorem
12.6 Summary
12.7 Unit End Exercises
12.0 OBJECTIVES
In this unit we shall study the generalisat ion of the
Cauchy closed curve theorem to functions h aving isolated
singularities. We shall prove the Residue theore m and further we
shall use it to evaluate the standard types of integrals like
0 2 1sin( ),, ( )1zxd xdx f z dzxx∞∞
−∞= ∫∫ ∫+etc. . We shall also prove the
Argument Principle and Rouche’s theorem for M eromorphic
functions in the complex plane .
12.1 INTRODUCTION
In this unit, we now seek to generalize the Cauchy closed
curve theorem to functions, which have isolated s ingularities. If is a
circle surrounding a single isolated singularity 0zand 0() ( )kk kfz c z z∞
=−∞ = − ∑ in a deleted neighbourhood of 0z t h a t
contains trace of a circle γ, then 12ff i cγ π−= . Thus the
coefficient 1c− is of special significance in this context. We munotes.in
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shall see some of the applications of th e Residue theorem.
Let us start defining the Meromorphic fu nctions.
Residues: Let f has an isolated singularity at z=α and
Let () ()nn
nfz a z∞
=−∞= −α∑ be its Laurent expansion about z=α Res ();fα= Coefficient of ()1z−−α in Laurent series = 1a−.
12.2 THE RESIDUE THEOREM AND IT’S
APPLICATIONS
Proposition: If f has a pole of order m at z=α a n d () ( ) () ,gz z f z t h e n=−α Res ()()()
()1
;1!mgfm−αα=−
Proof: Given that, f has a pole of order m at z=α z∴ =α is an isolated singularity of f ∴by definition, 0R∃ > s.t. ∴f is analytic in (){}0; \zR o r B R<−α< αα ∵ f has a pole of order m at z=α
()()()() , 0mgzfz w h e r e g
z∴ = α≠−α and g is analytic in ();BRα. ∴ for any ();zB R∈α , g has Taylor expansion about z=α
()
0()nn
ngz a z∞
== −α∑ where ()()!n
nganα= --------- (2)
() () ()1101 1 1 () ( ) . . . . . . . .mmmmm m gz a a z a z a z a z− +− + =+ −α++ −α + −α + −α
() () () ()011
1()() . . . . . . .m
mm maa a gzfzz zz z−−∴ = = + ++−α−α −α −α
()2
12() . . . . . . . . .mm maa z a z++ + + −α+ −α +
This is a Laurent expansion for f(z) about z=α ∴ by definition of residue,
Res ();fα= Coefficient of ()1z−−α in a Laurent series 1ma−=
Res ()()()1;1!mgfm−α=− ----------------------- ()()!n
nganα =∵
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Calculation of Residues:
1) If f has a pole of order mat z=α then,
Res ()()1
11;l i m ( ) ( )1!mm
mzdfz f zm dz−
−→αα= −α− Where m = order
2) If f has a simple pole at z=α, then,
Res ()()();l i m
zfz f z→αα= −α
Example: Determine the residue of ()
() ()2223efz
zz= −−at its
poles
Solution: Given function,
() ()2()23zefz
zz=
−−
H e r e f h a s s i m p l e p o l e a t z = 2 a n d z = 3 i s a p ole of order 2.
For z= 2:
Res ()()();l i m
zfz f z→αα= −α
Res() ( )
() () ()222222;2 lim 2 2 l im223 23zzeef
z →→
= −⋅ =
−− −
=
() ()222223 1ee=
+ − 2e=
For z=3
()()() ( )1
11;l i m1!mm
mzdRes f z f zm dz−
−→αα= −α−
3, 2mα= =
()()()2
231Re ;3 lim 321 ! zdsf z
d →∴ = −−() ()223e
zz−−2
23lim2zdc
z d →=−
() ()22311lim2 2z
zeez z →−=+−−() ()333 321
32 32eee e=−× + = −+− − 0=
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Example: Determine the residues of () c o t ()fz z = at its poles
Solution: Here
()()()()()
()
()12
2
0
0
2
3....
()
00 1
1m
kxxx x
in i
nk k
n x
zfz d z fz d z fz d z fz d zfz d z a r e i r e d
zz z
efz
zπ
θθθ
δ∞
=−∞=++ +
∴ =
<−<< <
=
−∫∫∫ ∫
∑ ∫∫
()coscotsinzfz zz== f∴ has simple pole at ,nπwhere nis an integer
()()()Re ; lim
zsf z fz→αα= −α
() ( )cosRe , lim .sinznzsfn z nz→ππ= −π 00 form
By using L’hospital rule, () s i n c o slimcosznzn z zz→π−− π +1=
Example: Compute the residue of ()4sinzfzz=
at the pole z=0.
(2012)
Solution:
()
()357
44
3
3sin 1.......3! 5! 7!
11.....65 ! 7 !
11 Re ,06zz z zfz zzz
zz
zz
sf c o e f f i c i e n t o fz== −+ −+= −+−
−==
Residue Theorem or Cauchy Residue Theorem
Let f b e a n a l y t i c i n a d o m a i n G e x c e p t f o r t h e i s o l a t e d singular
points 12,, . . . . . . . ..mzz zIf x is a simple closed curve which does not pass
through an of the points kz then
()2R e ( ; )1mfz d z i sf zkk x = π
=∑ ∫2i=π×[sum of residue of f at its pole
inside x ], where x is traversed in anticlockwise direction. (2008)
munotes.in
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Proof: Given that, 123,,. . . . . . . . .mzz z z are isolated singular points in G.
Assume these points lie inside x
Choose positive numbers 12,, . . . . . . .mrr r s o s m a l l t h a t n o t w o c i r c l e s
:kk kxz z r−=intersect1, 2........km= and every circle ()1, 2, ........kxk m − is inside x.
x1 2 3 4
mzrxxxxmr r r r zz z z
m1 23 41 2 34
Fig 12.1
∴ the function f is analytic in a domain which is bounded by non-
intersecting closed curves 1, 2,.........mxx x and on the curves.
By using Cauchy Deformation Theorem ()()()()12....
m xxx xfz d z fz d z fz d z fz d z=++ + ∫∫∫ ∫
() ()
1km
k xxfz d z fz d z=∴ =∑ ∫∫ … … … … . .()1 f∴has an isolated singular point at kzz= f∴ has Laurent expansion about kzz=
() ()nnknfz a z z∞
=−∞∴ = − ∑ 0kkzz r<−<
Any point on the circle = centre + radiusieθ×
Consider,
()()naz znknkkxf z d z x∞
−
=−∞∑∫
=∫ ()nzz d zkk
nx∞−
=−∞∫=∑
Any point on the circle kx is,
ikkzz r eθ=+ []0, 2θ∈ π
i
kdz ir e dθ= θ munotes.in
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()2
0()
kin i
nk k
n xfz d z a r e i r e dπ
θθθ∞
=−∞∴ =∑ ∫∫()2
1 1
0.in n
nk
nai r e dπ ∞+θ +
=−∞= θ∑ ∫
Note: 2
(1 )
02, 1ined nπ
θθπ+== −∫
0, 1n≠−
() ( )11
12k
xkfz d z a i r−+
−∴ = π∫12ia−=π ()2. R ekis f i z=π b y
definition of residue.
Put this value in equation (1) to get.
() ()
12R e;mkk xfz d z i sf z== π ∑ ∫ ()
12R e ,mkkis f z==π ∑
Example 1: Use Residue Theorem to evaluate
() ()22124xz
zz z+
− +∫ where x is the circle 3z=
Solution:- By using Residue Theorem ()2xfz d z i=π∫sum of residues of f at its poles inside x]……..(1)
Here ()
() ()22124zfz
zz z+=
− + f∴ has a simple pole at 0z= and 2zi= and 4z=− is a pole of
order 2.
But, simple poles 0, 2zz= =lies inside x
-4 -3 23 0
Fig 12.2
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For simple pole at z=0 ()()()Re ; lim
zsf z fz→αα= −α
() ( )
() ()2201Re ;0 lim 024zzsf z
zz z →+= −×
− +
02
lim
z→=()221
2+
() () () () () ()222 2010 1lim24 24 0 2 0 4zz
zz zz →++ = =
− + − + − +=132−
Similarly,
2( 2 1 )
100212 ( 2 1 )(2 1) 32zzzz z zzn ieeee e eee z n izn i z
zxπππ−−−++∴ =⇒ += ⇒ =−⇒ =−⇒⇒ =+
⇒ =+ =
∈
()5Re ;272sf = ∴from equation (1)
() ()() ()2
212R e ; 0 R e ; 224xzis f s f
zz z+ = π +− +∫ 15 1 52232 72 8 4 8 9ii−− =π += π + ×× 92 02272 4i−+=π =×1172 4iπ×
×2
() ()2
211 114424xzizz z+ = π− +∫
(2) Use Residue Theorem to evaluate
()3
4sin
xzdzzπ
−∫ w h e r e zis a closed square bounded by 2,x=± a n d 2y=± .
Solution : By using residue Theorem, ()2xfz d z i=π∫[sum of residue of f at its poles inside x]…….(1) munotes.in
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Here ()
()34sinzfz
zπ=
− f∴ has a pole of order 3 at 4zπ⋅=
But, the pole 4zπ= lies inside x
For pole of order 3,m=
()() ( )1
311Re ( ; lim31 !mm
zdsf z f zdz−
−→αα= −α−
f (z)
-21
2 1 -1114=z
-2-12
Fig 12.3
()3 31
311Re ; lim43 1 ! 4 zdsf z
dz−
−→αππ = − − ()34sin 2zπ⋅
−
()2
2
41lim sin2zdzdzπ→= ()
41lim sin2zzπ→= −
1sin24π= −11222−=× 1Re ;422sfπ−∴ =
()
()3
4sin2
xxzfz d z d z
zπ∴ ==
−∫∫1122 2ii−−π = π
(3) coshzedzz∫ where x is circle 3z=
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Solution :- by using residue Theorem ()2xfz d z i=π∫ [sum of residue of f at its poles
Here ()coshzefzz=
Let coshz=0
2( 2 1 )00212 ( 2 1 )(2 1)2zzzz z zzn ieeee e eee z n izn iπππ−−−++∴ =⇒ += ⇒ =−⇒ =−⇒⇒ =+
⇒ =+ f∴has simple poles at 212nzi+= π
But , the simple poles 2ziπ=± lies inside 3z=
For simple pole ()()()Re ; lim
zsf z fz→αα= −α
22Re ; lim22 c o s hziesf i zizπ→ππ = − 00form
()2
212limsinhz
zizi e e
z π→π − + =2cos sin0. 12.1sin2iiiiπ π ++= =π 1=
3
2Re ; lim22 c o s hziesf i zizπ→−ππ − =+
()222210 2limsinhsinh2i
zizi e ee
ziπ−
π→ π +++ ==π ()()()cos sin22sin2i
i−π π+
=π1=
[]21 14coshz
xeiiz=π+= π∫
(4)
() ( ) ( )2
2112 2 5xz
zz+−− +∫ w h e r e x is the circle4z=
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12.3 EVALUATION OF STANDARD TYPES OF
INTEGRALS BY THE RESIDUE CALCULUS
METHOD
* Application of Residue Theorem to evaluate real Integrals.
Type-I
Integral of the type ()2
0cos,sinFdπθθ∫, Where ()2
0cos ,sinFπθθ∫ i s
rational function of cosθ and sinθ
Consider, ()2
0cos ,sinFdπθθ θ∫
Put [],0 , 2izeθ= θ∈ π, .i
idz dzdz i e d dizieθ
θ=−θ ⇒ θ= =
cos2
sin2iiiiee
ee
iθθθθ−+θ=−−θ=12
121122 21122 2z zz z zzz zz z zii i z−
−+++= = =
−+ −= = =
()2 22
011cos , sin ,22xzz d zFd Fzz i zπ++∴θ θθ = ∫∫,Where
()2211,22zz ifz Fzz i z++=
2i=π[sum of residue of f at its pole inside x]
(by residue Theorem)
2ax bx c++ is polynomial than the root are ()242bba cxa−± −=
(1) Use Residue Theorem to evaluate 2
02c o sdπθ+ θ∫
Solution :- the given integral is 2
02c o sdπθ+ θ∫
Put [],0 , 2izeθ= θ∈ π, idz i e dθ=−θ .idz dzddizieθθ= ⇒ θ=
121cos22 2iiee z z zzθ− θ −++ +θ== = munotes.in
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2
2
01
2c o s 122xdd zizz
zπθ∴ = + θ++∫∫ w h e r e x is the unit circle 1z=
11
iz= 21
41
2zz
z++xdz∫ 221
41xdzizz= ++∫
… … … ( 1 )
Put ()2141fzzz=++ f h a s s i m p l e p o l e a t () 2 41 6 4
2z−± −= =()232−±
i . e . 23z=−±
but, the simple pole 232 1 . 7 3=−+ ≈−+ lies inside x
For simple pole, ()()()Re ; lim
zsf z fz→αα= −α ()()2231Re ; 2 3 lim 2 341zsf zzz→−+ −+= −− + ×++ 00form
() 2311lim2422 3 4z z →−+= = +−++ 123= ∴ by using Residue Theorem ()2xfz d z i=π∫ [sum of residue of f at its poles inside x]
22
41dz
zz =
++1
2iπ 33xiπ = ∫
Put this value in equation (1)
2
02
2c o s3diiπθπ = × + θ ∫
2
022c o s3dπθπ∴ = + θ ∫
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(2) use residue Theorem to evaluate 2
0sindababπ
θ >+ θ ∫
Given , 2
0sind
abπθ+ θ∫
Put []0, 2izeθ= θ∈ π 2i
idd zdz ie d dizieθ
θ⇒ = θ ⇒ θ= =
121sin222iez z ziii z−θ −−−θ= = =
2
2 cos12dziz
xxd
ab zabizπθ∴ = + θ−+∫∫ w h e r e x is the unit circle 1z=
dz
iz=21
2
2aiz bz bai z×+ −x∫ 22
2xdzbz aiz b= −∫ … … … . . ( 1 )
Put ()212fzbz aiz b=+ − f∴ has simple pole at ()2224422ai bz aib+= − ± 22222ai b ab−± −=
i.e. f has simple pole at 22baza ib−=−±22aa bib± −=−
Let 221aa bib−−α= − and 22aa bib−+β=−
()111 1 . . . . . . . . . . . .baα β= ⇒ αβ= ⇒ α=< <β∵ ∵
But the pole α lies inside x
For simple pole,
()()()Re ; lim
zsf z fz→αα= −α
() ( ) 11211Re ; lim .
2 zsf zbz aiz b→αα = −α+ −00form munotes.in
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11lim22zbz ia→α= +
1
2b=
()22aa bib−−−1
22aiia = − +
2222ia b a i+ −+
() 1221Re ;
2sfia bα=
−
By using Residue Theorem, ()2xfz d z i=π∫ [sum of residue of f at its poles inside x]
2212
2iia b=π ×
−
()22xfz d zabπ=
−∫
Put this value in equation (1) we get
222 22022cosd
abab abπθπ π = × = + θ−−∫
3. Use residue theorem to evaluate
0cosd
abπθ+ θ∫ a > b !
Hint: First evaluate 2
0cosd
abπθ+ θ∫
Then use the property 2
0cos cosdd
ab abππ
0θθ = 2 + θ + θ∫∫
4. Evaluate
012 c o s
53 c o sdπ+ θ θ+ θ ∫ &
5. Evaluate 2
0cos3
54 c o sdπθ θ−θ∫ &
6. Evaluate
01cosdaaπθ >+ θ ∫
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Type-II
Integral of the type ()fx d x∞−∞∫
Improper Integral where ()() , () , ()()Pxfx w h e r e p xQ xQx= are
polynomial in x.
Example: If ()KMfzR≤ f o r Reizθ= w h e r e K > 1 a n d M a r e
constants.
Then lim ( ) 0RxRfz d z
→∞ =∫ where, Rx is the semi circle are of radius
R as shown in figure.
Proof: Given ()KMfzR≤ for Reizθ=
x x
R -RR
O
Fig 12.4
() ()
xxRRfz d z fz d z∴ ≤ ∫∫ K
xRMdzR≤ ∫
Put Reiizd z R i e dθθ= ⇒ = θ
()i
K
xRMfz d z R i edRπ
θ
0∴ ≤ θ∫∫ KMRR= × π
1() 0K
xRMfz d z a sR
R−π∴ ≤ → →∞∫ lim ( ) 0RxRfz d z
→∞∴ =∫ munotes.in
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Example: E v a l u a t e ()fx d x∞−∞∫ , w h e r e
()() , () ()()Pxfx P xa n d Q xQx= are polynomials in x .
Solution:
Consider ()xfz d z∫ w h e r e x is a closed curve consisting of
large semicircle Rx o f r a d i u s R a n d t h e r e a l a x i s f r o m – R t o R
traversed in the anticlockwise direction.
x
R -RR
O0
θ∈ π,
−plane
Fig 12.5
We choose only those poles of f which lie in the up per half of the
complex plane. ∴ by residue theorem, () 2xfz d z i=π∫ [sum of residues of f at its pole inside x] () () 2R
xRfz d z fx d x i−+= π∫∫ [sum of residues of f at its pole inside x]
on real axis zx= lim ( ) lim ( ) 2R
RRxRRfz d z fx d z i→∞ →∞−∴ + = π ∫∫ [sum of residues of f at its
poles inside x] () 2Of x d x i∞∞+= π∫ [ s u m o f r e s i d u e s o f f a t i t s p o l e s i n s i d e x]
……….. (by previous example
0lim ( )Rxfz d z→∞= ∫ munotes.in
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Note: If ()()()PzfzQz= where, P(z) and Q(z) are polynomial in z such
that,
(i) Q(z) = 0 has no real roots.
(ii) The degree of Q(z) is greater than that of P(z ) by atleast 2, then, () 2xfz d z i=π∫ [Sum of residue of f at its poles inside x]
Example:
(1) Evaluate
() ( )2
223
14xdxxx∞
−∞+
++∫
x
R -RR
Ox
Fig 12.6
Where x i s a c l o s e d c u r v e c o n s i s t i n g o f l a r g e s e m i c i r c l e Rx o f
radius R and real axis form –R to R traversed in th e anticlockwise
direction.
degQ(z)≥degP(z)+2
2() 3Pz z=+ & ()()22() 1 4Qz z z=+ + ∴ this general method is applicable
Consider,
() ( )2
22314z
zz+++∫
Here,
() ( )3
223()14zfz
zz+=++
Here f has simple poles at zi=± and 2zi=± but the poles zi= and 2zi= lies inside x.
For simple pole munotes.in
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Res ()();l i m ( )zfz f z→αα= −α
Res
() ( )
() ( )2
223;l i m14zizfi z i
zz →+ = − × ++()()() () ()()()()22
2233
lim lim44zi ziziz z
ziziz ziz →→− ++
= = − ++ ++
()()()() ()2
21333 1 221 4 2 4 12 ( 3 )4iii iiii−+ + −= = = =−+ − ++
()1Re ;3sf ii =
Res() ( )
() ( )2
22 23;2 lim 214zizfi z i
zz →+= −++()()() () () () ()22
22 22233lim lim12 2 1 2zi zirrzi zzzz i z i zz i→→− ++= =
+ − ++ +
() [] () () ()22
22(2 ) 3 4 3 4 344 141 4 21 2 2ii
i ii ii i+ −+= = =−+ + + +
()1Re ;22sf iPi =
By using Residue Theorem
() ( )2
223214xzdz izz+ = π++∫ [Sum of Residue of f at its pole lies inside x] 11 4 12232 1 2iiii+=π += π 56= π
() ( ) () ( )22
22 2233 5614 14R
xRRzxdz dx
zz xx−++∴ + = π++ ++∫∫
() ( ) () ( )22
22 2233 5lim lim614 14R
RRxRRzxdz dx
zz xx →∞ →∞−++∴ = π++ ++∫∫ ----(1) munotes.in
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T.P.T.
() ( )2
223lim 014 RxRz
zz →∞+ =++∫
() ( )222
22 22 22
03R e31414ii
ii
xRRe dzdz
Re Re zzθθπ
θθ+ θ+ ≤ + +++∫∫Re , [0, 2 ]Rei
iput z
dz i dθ
θ = θ∈ π ⇒ = θ ∵
()() ( )2
22314RR d
RR2π
0+ θ≤ − −∫
22 22
12 1 2 2 22111111ii
izz z z e R e
R Reθθ
θ + ≥ − ⇒ + ≥− ⇒ ≤ −+ ∵
() ( )()() ()3 332
22 2 1402213
14 11R
xRRRRzdz dzz Rπ ++ ≤ θ++ −−∫∫
()() ()32
142211
11R
RRR+ ⋅ π
=
− −0as R→ →∞
() ( )2
223lim 014 RxRzdz
zz →∞+∴ =++∫
Put this value in equation (1)
() ( )2
22350614xdx
xx∞
−∞+ π+ = ++∫
() ( )2
2235
6 14xdx
xx∞
−∞+ = π ++∫
(2) Use Residue theorem to evaluate
()2
2201xdxx∞ +∫
Hint: First calculate
()2
221xdxx∞
−∞ +∫ and then use the property munotes.in
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() ()22
222202
11xxdx dxxx∞∞
−∞= ++∫∫
Solution: Consider
()2
221xzdzz+∫ w h e r e x i s a c l o s e d c u r v e
consisting of large semicircle Rx of Radius R and real axis from –R
to R traversed in the anticlockwise direction. Here
()222() ,1zfz
z=
+
Here, f has pole of order z at zi=± but, the pole zi= lies inside x.
For pole of order 2
Res ()()()2 1;l i m ( )2! zidfi z i fzid z →= − −
()
()()
() ()2222222lim lim
1zi zizi zdzzidzziziz →→−= − =
− ++ ()22lim
zidz
dzzi→=
+
(3) 2
42210 9xxdxxx∞−∞−+
++∫
Type-III: Integral of the type ()sim mx f x dx∞−∞ ∫ or cos( ) ( )mx f x dx∞
−∞ ∫
m > 0 where ()()()PxfxQx=
x
R -RR
Ox
Fig 12.7
Consider ()imz
xef z d z ∫ where x is a closed curve consisting of large
semicircle Rx of radius R and the real axis from –R to R travers ed in
the anticlockwise direction. munotes.in
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255
∴ by residue theorem () 2imz
xef z d z i=π∫ [Sum of residue of f at its pole lies inside x] () () 2R
imz imzRRef z d z ef x d x i−∴ + = π∫∫ [ Sum of residue of f at its pole
lies inside x] lim ( ) lim ( ) 2R
imz imx
RRxRef z d z ef x d x i→∞ →∞−∴ + = π ∫∫[Sum of residue of f
at its pole inside x] --------- (1) lim ( ) 0imz
RxRef z d z
→∞∴ =∫ --------(by next example) ∴ from equation (1),
()0s i n ( ) 2Cosmx i mx f x dx i∞−∞++ = π∫ [ S u m o f r e s i d u e o f i t s p o l e s
inside x]
Example: 1) Use Residue Theorem, to evaluate 2cos34xdxx∞
−∞∫+ .
Solution: Consider 3
24iz
xedzz +∫
x
R -RR
Ox
zθ=π0θ=
Fig 12.8
Where x i s a c l o s e d c u r v e c o n s i s t i n g o f l a r g e s e m i c i r c l e Rx o f
radius R and the real axis from –R to R traversed i n the
anticlockwise direction.
Put 3
2()4izefzz=+
Here, f is a simple poles at 2zi=± but, the pole zz i=lies inside x.
For simple pole, munotes.in
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Res ()()2;l i m ( )zifz f z→α= −α
Res() ( )()
() ()3 3
2222;2 l im 2 l im224iziz
zi zizi e efi z izi ziz →→−= −× = − + +
()233 ( 2 ) 62lim22 2 4iz i i iziee ezi ii i→= = =++
()6Re ;24!esf i− =
• by residue theorem,
• () 2xfz d z i=π∫ [Sum of residue of f at its all poles inside x] 362244iz
xeedz iiz−∴ = π+∫ 362.
2 4iz
xeedz
z−π ∴ = +∫
33622 2 44R iz iz
xRReedz dx ezx−−π∴ + =
++∫∫
33622lim lim2 44R iz ix
RRxRReedz dx ezx−→∞ →∞−π + =
++∫∫ ------------- (1)
T.P.T. 3
2lim 04iz
RxRedz
z →∞ =+∫
Put []Re 0,2izθ = θ∈ π
..idz i R e dθ= θ
22 2
2 22114444i
iRe R
R Reθ
θ+ ≥ − ⇒ ≤ −+
3. R e
3
2 22Re
4 4iii
iz
i
xRei dedz
z Reθθ2π
θ
0 θ ≤
+ +∫∫
()()2
3
2
0sin
4iR ReC o s i dRπ≤ θ+ θ θ−∫()3
2
04RS i nRe
Rπ− θ=
−∫ munotes.in
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…()31iR Cose θ =∵ 220,0 , Sinπ θ ≥ θ∈ π∵
1
0=θySin=θySin2θ=πy
2π π =θx
Fig 12.9
()63
2 22e
4 4Riz
xReRdz dz Rπ−θ2
π
0∴ ≤ ⋅ θ+ −∫∫ ……. 2Sinθ θ≥ π∵ 2Sin−θ⇒ − θ≤ π63RRS i n−θ⇒ − θ ≤ π
()62202e
64R
R
RRπ−θ
π= −−π()6602
22
6 4RRReeR R−π −××πππ= × −− −()3
2134Re
R−−π= −−()3
2 421
31RRe
R−π= −−00as R→ →
3
2lim 04iz
RxRedz
z →∞∴ =+∫
Put this value in equation (1) 623304 4Cos x i Sin xex∞−−∞+ π+ = +∫
Equating Real part from both sides 623
2 4Cos xdx e
x∞
−
−∞π = +∫ munotes.in
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(2)
()22
00Sinmxdx m a Rxx a∞ > ∈+∫
(3)
()2201Cosmxdxx∞ +∫ [ ()zi=± is a pole of order 2 ]
(4) 245Sinxdxxx∞−∞ ++∫
Type-IV: Poles on the Real axis
I n t h i s c a s e , w e c a n n o t u s e r e s i d u e t h e o r e m b e c a u s e the pole z
= 0 lie on real axis and hence on x.
Example: Prove that
0sin2xdxx∞π=∫
Solution: Consider
0izedzz∞
∫
-RrRY
X-rx xRo
Fig 12.10
Where xis a closed curve consisting of
(i) a real axis from r to R
(ii) a large semicircle Rx of radius R.
(iii) a real axis from –R to –r
(iv) a small semicircle rx of radius r. ∵ the pole z = 0 lie outside x, ∴ by Cauchy- Goursat theorem, 0iz
xedzz =∫ munotes.in
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0Rrix iz ix iz
rx RxRreeeedx dz dx dzxz xz−
−∴ + + + =∫∫ ∫ ∫
In 3rd integral x is negative dx⇒ will be dx−, using this negative
sign make limits r to R. 02RRix ix iz iz
rr x xRree e edx dx dz dzxx z− − + + =∫∫ ∫ ∫ 202Rix ix iz iz
rx xRree e eid x d z d zix z z−− + + =−∫∫ ∫ 20R iz iz
rx xRrSinx e eid x d z d zxz z+ + = ∫∫ ∫ 2iz izeeSinz− =∵
Taking limit 0Ra n d r→∞ →
0lim 2 lim lim 0R iz iz
RR rrx R rSinx e ei dx dz dzxz z →∞ →∞ →∞
→ + + =∫∫ ∫ ------------(1)
T.P.T. lim 0iz
RxRedzz →∞ =∫ (Proof as similar as Type III)
Consider iz
xRedzz∫
Put iizr e d zi r e dθθ= ⇒ = θ
0iiz i r ei
i
xreedz ire dz reθ π θ
θ = −× θ ∫∫ - - - - - - - -
xxff = ∫∫∵
Taking limit as 0r→
0
00limiz
xxredz i e d izπ
→ = − θ=−π∫∫ ∴ from equation (1),
020 0Sinxi dx ix∞ + −π=∫ 00Sinxdx ix∞
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2Sinx ixi∞
0π∴ =∫
02Sinx
x∞π∴ = ∫
Meromorphic Function (M.F):
If f is defined and analytic in an open set G ⊂ except for
poles. Then f is a Meromorphic function on G.
()2
.. ()5zeg f zzz =−
Theorem (1) If f has a zero of order m at ,z=α t h e n R e s
();()fzmfz′ α=
Proof: Given that, f has a zero of order m at z=α
∴ ∃ an analytic function ():;gB R α →
s . t . () ( ) ()mfz z gz=−α w h e r e () 0gα≠
D i f f . w . r . t . z
() ()1() () ()mmfz m z g z z g z−′′= −α +−α
()
()() []
() ( )1() ()m
mfz z m g z g zfz zg z−′′−α +
−α ()()() ( )mg z g zzg z′ += −α
Laurent expansion for ()()fzfz′ about z=α ()() ()()(),0fz g z mzRfz m g z′ = + < −α<−α ()()gzgz′ ∵ i s a n a l y t i c i n (),BRα a n d h e n c e i t h a s T a y l o r s e r i e s
expansion about z=α.
∴ By the definition of residue,
Res ()
();fzmfz′α== Coefficient of ()z−α i n t h e L a u r e n t s e r i e s
expansion.
Theorem (2) I f f h a s a p o l e o f o r d e r n a t z=β t h e n R e s ()();fznfz′ β=−
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Proof: Given that, f h as a pole of order n at z=β ∴ ∃ an analytic function ():;gB R β→
s.t.
()()()ngzfz
z=−β w h e r e ()0gβ±
Diff. w.r.t. z
()() ( ) ( ) ()
()12nnnzg z g z n zfz
z−′ −β −− β′=
−β
()
()()() ( )( )
() ()1n
nnzz g z n g z fz
fzzz−′ −β −β − ′=
−β × −β()nz−β ×()gz
()()()() ( )zg z n g zzg z′−β − = −β ()() ()gzngz z′= − −β ()() ()()()0fz g z nzRfz z g z′′− = + < −β<−β ()()gzgz′ ∵ is analytic in ();Rββ a n d h e n c e i t h a s T a y l o r s e r i e s
expansion about z=β ∴ by definition
Res ()
();fznfz′ β = −
12.4 THE ARGUMENT PRINCIPLE
Let f be a Meromorphic function in a domain G and h ave only
finitely many zeros and poles. If x is a simple closed curve in G s.t.
no zeroes and poles of f lie, on ,x then
()
()2ffxfzid z Z Pfz ′ π = − ∫ Where, ,ffZP , denote respectively the
number of zeros and poles of f inside ,x each counted according to
their order or multiplicity. (2004, 2 007)
Proof: Given that, f is a Meromorphic function in domain G .
P u t ()()()fzFzfz′= ⇒ the singular points of F inside xare the zeros and poles of f. ∴ by Residue Theorem, munotes.in
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()()()2xxfzFzd z d z ifz′== π∫∫ [ S u m o f r e s i d u e s o f F a t i t s s i n g u l a r
points inside x. -----------(1)
If jα is a zero of f of order jm, then Res ()();jjfzmfz′ α =
If kB is a pole of f of order kn, then Res ()();kfznfzκ′ Β= −
From equation (1)
()
()()
()()
()2R e ; R e ;jK
xfz fz fzdz i s s Bfz fz fz ′′ ′=π α + ∫ ()()1
2jKiK xfzdz m nif z′ = −π∑∑∫
()
()1
2ffxfzdz Z Pif z′ = − π∫
Example:
(1) Use Argument Principal to evaluate ()()xfzdzfz′ ∫ where ()()() ( )214zfzzz z−= − − and x is the circle 3z=.
Solution: by Argument Principal
()()1
2ffxfzdz Z Pif z′ = −π∫ where, ffZa n d Z are the no. of zeros and
poles of f inside xeach --(1)
Given function,
()()() ( )214zfzzz z−=−−
Here f has simple zeros at z = 2 and z = 0, Z = 1 a nd z = 4 are simple
poles
01x
34 2
Fig 12.11
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Given equation of circle 3z =
But, simple zero z = 2 and simple poles z = 0, 1, l ies inside x. ∴ from equation (1) ()()112 12xfzdzif z′ = − = −π∫()()2xfzifz′⇒ = −π∫
(2) Use the argument principal to evaluate ()()xfzdzfz′ ∫ w h e r e
()()() ()2123zfz
zz z−= − − and x is the circle z=π
Solution:
B y a r g u m e n t p r i n c i p l e
()()1
2ffxfzdz Z Pif z′ = −π∫
Given function ()()() ()2123zfz
zz z−=−−
Here f has simple zero at 10za n d z= = is a poles of order 2 and z =
2, 3 are simple poles.
Given equation at circle, z =π
But simple zero z = 1 and simple poles z = 2,3 and z = 0 is pole of
order 2 lie inside x.
From equation (1), ()()[] [ ]21 42 3xfziifz′ = π − = π−∫6i=−π .
(3) Evaluate cotzzππ=∫ by using argument principle
Solution: cos 1 .coscot
zz zzzzd z d zSin z Sin zππ πππ ππππ π== === ∫∫ ∫ ∴by argument principal
1cot 2pfzzdz i Z Pπππ == − ∫2ffiZ P= − …………. (1)
Here ()fz S i n z= π has simple zeros at
zn nπ=π ∈
zn n∴ = ∈ munotes.in
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0, 1, , 2, 3, .....z∴ = ±± ± a r e s i m p l e z e r o s o f f a n d f h a s n o p o l e s b u t
zeros 0,1, 2,z= ± ± 3 lies inside x i.e. z=π []cot( ) 2 7 0zzi
=ππ= −∫14i=
(4) Evaluate tan
xzd z π ∫ where x is the circle z=π
Take as an Exercise.
12.5 ROUCHE’S THEOREM
Suppose f and g are Meromorphic functions in a nbd. o f ();BR α with no zeros and poles on the circle {};xz z R= −α=. If (),,fg fgzz P P a r e t h e n o . o f z e r o s ( p o l e s ) o f f a n d g c o u n t e d
according to their order and if ()()()2fz g z g z x− < ∀∈ t h e n , ff g gZPZP−= −. ( 2 0 0 5 , 2 0 0 8 , 2 0 0 9 )
Proof: Given that, f and g are Meromorphic functions in a nbd. of ();BR α
Put ()()()fzFzgz= F⇒ is a Meromorphic function in a neighbourhood of ();BRα
Let ()Zx t= be any point on the circle x.
Then, () []Re 0, 2itzx t t== α+ ∈π
1x
αRF
oσ=Fx2π(t)σ
−zplaneω−planeB( ;1)i
Fig 12.12 x ∵ is a simple closed curve in Z-plane and F is analy tic. .Fx∴ σ= is also a closed curve in w-plane ∴ for any []0, 2t∈π munotes.in
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() ()()()()()()11 1fx t
tF x t
gxtσ− = −= − - - - - - - - - - - - - - - - - - ( 1 )
Given that ()()()fz g z g z z x− < ∀∈
()11()fz
gz− <
Put ()zx t= ()()()()11fx t
gx t−<
Put the above value in equation (1), we get ()11tσ− < {}((1,0);1)B σ⊂ ∴ 0 belongs to the unbounded component of {}\wσ ∴ by definition of winding no;
();0 0ησ = (∵ 0 lies outside the curve σ)
1020d
iσω⇒ =πω −∫
Put ()()[]0, 2td t d t t ′ω=σ ⇒ ω=σ ∈π
()
()2
0102tdtitπ′σ =πσ∫ - - - - - - - - - - - - - - - - - - - ( 2 )
Now, ()()()()()fx ttgx tσ=
()()() ()()()()() ()() ()
()()2xtgx t f xt f x t g x t xtt
gxt′′′ ′−⋅′∴ σ =
()
()()()
()()()() ()() ()()
()()()2gxt gx t f x t g x t txtt fx tgx t′′ − ′σ ′= σ
()() ()()
()() ()()() ()() () ()()
()() ()()()2
22gxt f xt g xt gx t f x txtgx t gx t f x t gxt ′ ′= − munotes.in
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()
()()()()()()()()()()fx t g x t txtt fx t g x t′′ ′σ′= − σ
()()()()()()()()()2
0102fx t g x txt d ti fx t g x tπ′′
′ = − π∫ ----------------- (2)
Put ()()zx t d zx t d t′ = ⇒ = ()()()()102xfz g zdzif z g z′′= − π∫ ()()()()11
22xxfz g zdz dzif z ig z′′⇒ = ππ∫∫ ∴ by the argument principle
ff g gZP ZP−= −
Example:
(1) Use Rouche’s theorem to prove that all zeros of the polynomial
7351 2 0zz− += lie between the circles 1z= and 2z= .
Solution:
C o n s i d e r t h e c i r c l e 1:1xz=
Let () ()7351 2 1 2fz z z a n dg z= − + = g⇒ has no zeros inside 1x
For any point 1zx∈
() ()7351 2 1 2fz g z z z− = − +−735zz= −735zz≤ +
()3 715 1 1 5= + = +6=()12 9z< =
() () ()fz g z g z v − <{})zx∈ Foxσ = ()()()1fz g z g z z x∴ − < ∀∈ ∴ by Rouche’s theorem,
fgZZ= ( H e r e , t h e r e a r e n o p o l e s ) f⇒ has no zeros inside 1x
Consider the Circle 2:2xz=
Let () ()73 751 2fz z z a n d g z z= − + = g⇒ has 7 zeros, counting order, inside 2x munotes.in
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For any point 2zx∈
() ()73 751 2fz g z z z z− = − +−351 2z=−+()3 351 2 5 1 2 5 8 1 2 4 0 1 2zz= + = + = ×+= +
()752zg z=< = ()()()2fz g z g z z x∴ − < ∀∈
Hence all zeros of the polynomial 7351 2 0zz− += l i e b e t w e e n t h e
circles 12za n d z= =
(2) Use Rouche’s theorem, to prove that ()znea za e= > has n zeros
(roots) inside the circle 1z=
Solution: Consider the circle :1xz=
Let ()nzfz a z e= −
()nand g z az = F g⇒ has n zeros, counting order, inside .x
For any point ,zx∈
() ()nz nfz g z a z e a z− = −−
zzee==1e=()ag z<= ()()()fz g z g z z x∴ − < ∀∈ ∴ by Rouche’s theorem fgZZ= ⇒ f has n-zeros inside the circle 1z=
(3) Use Rouche’s theorem to prove that every polyno mial of degree
n has n zeros.
12.6 SUMMARY
1) Residues:
Let f has an isolated singularity at z=α and
Let () ()nn
nfz a z∞
=−∞= −α∑ be its Laurent expansion about z=α Res ();fα= Coefficient of ()1z−−α in Laurent series = 1a−.
x1
Fig 12.13
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2) Calculation of Residues:
1) If f has a pole of order mat z=α then,
Res ()()1
11;l i m ( ) ( )1!mm
mzdfz f zm dz−
−→αα= −α− Where m = order
2) If f has a simple pole at z=α, then,
Res ()()();l i m
zfz f z→αα= −α
Let f be analytic in a domain G except for the isolated
singular points 12,, . . . . . . . :mzz zIf x i s a s i m p l e c l o s e d c u r v e w h i c h
does not pass through an of the points kz than.
3) Cauchy – Residue Theorem:
()2R e ( ; )1mfz d z i sf zkk x = π
=∑ ∫
2i=π[sum of residue of f a t i t s p o l e i n s i d e x ]
W h e r e , x is traversed in anticlockwise direction
4) Meromorphic Function (M.F):
I f f i s d e f i n e d a n d a n a l y t i c i n a n o p e n s e t G except
for poles. Then f is a Meromorphic function on G.
5) The Argument Principle:
L e t f b e a M e r o m o r p h i c f u n c t i o n i n a d o m a i n G a n d have
only finitely many zeros and poles.
I f x is a simple closed curve in G s.t. no zeroes and p oles of f
lie, on ,x then ()
()2ffxfzid z Z Pfz ′ π = − ∫ where, ,ffZP , denote
respectively the number of zeros and poles of f ins ide ,x e a c h
counted according to their order or multiplicity.
6) Rouche’s Theorem:
S u p p o s e f a n d g a r e M e r o m o r p h i c f u n c t i o n s i n a n b d . of ();BR α with no zeros and poles on the circle {};xz z R= −α=. If (),,fg fgzz P P a r e t h e n o . o f z e r o s ( p o l e s ) o f f a n d g c o u n t e d
according to their order and if ()()()2fz g z g z x− < ∀∈ t h e n , ff g gZPZP−= − munotes.in
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7) Schwarz’s Lemma
L e t {}:1Dz z=∈ < be the unit disk and suppose f is
analytic in D with, (i) f (0) = 0 and (ii) ()1fz f o r z D≤ ∈
Then, ()()01fz z z D a n df ′≤ ∀∈ ≤
8) A function :fD D→ is said to be an Analytic automorphism or
Automorphism of the Unit disc D, if f is bij ective and if both 1,ff− are analytic in D.
12.7 UNIT END EXCERCISES
1) Show that the Polynomial 10 2() 2 4 1Pz z z=+ + h a s e x a c t l y
two zeroes in 1z<.
Solution: Let 21 0() 4 , () 2 1fz z g z z= = + ∴ () ()fz g z> for every number on the Unit circle. ∴ By Rouche’s theorem, the number of zeroes of ()fg+
inside the curve ()1Z== the number of zeroes of
inside the
curve ()1Z=. ∴ 10 224 1zz++ has exactly two zeroes in the curve 1z<.
(Here the number of zeroes of
inside the curve ()1Z=.
11 211 8222 4zzfzdzif i z ππ==′ == =∫∫ .
2) Suppose that f is entire and ()fz is real if and only if z is
real .Use the Argument Principle to show that f can have atmost
one zero.
( Hint: Consider the image of the circle zR=. Here f maps the
entire upper semicircle ,0zR y=> i n t o e i t h e r t h e u p p e r h a l f
plane or the lower half plane .
Similarly, f maps the entire lower semicircle ,0zR y=> into
either the upper or lower half plane , becau se ()Arg w△ is atmost
π in any upper/ lower half plane ∴ () 2Agf zπ≤△ as z traverses
through the circle zR=.
∴ The Number of zeroes of ()fz in zR≤
11() 2 122Argf z πππ= ≤ =△ . munotes.in
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3) Find the number of zeroes of 1()3zfz e z= − in 1z≤. ( Hint :
Let 1() , ()3zfz zg z e= = ∴ () () , 1fz g z zz> ∀ ≤)
4) Find the number of zeroes of 642() 5 3 1fz z z z= − + − in 1z≤(
Hint : Take 46 2() 5 , () 3 1fz z g z z z= = + −
∴ on 46 21, ( ) 5 5 3 1zf zz z z= = = ≥+ −
Also () ()fz g z= only at i∓. ∴ There are 4 zeroes of f in 1z≤.
5) Show that for each 0R> if n is large enough then
2
() 1 . . .2! !nnzzPz zn=++ + + has no zeroes in zR≤.
(Hint: ()znPz e→ as n→∞.)
6) If f is Meromorphic on G and :fG∞→ is defined by
()fz=∞ if z is pole of f
()fz= otherwise. Show that f is continuous on G.
7) Find 41dxx∞
−∞∫+.
Solution: Here 41izeπ= and 234izeπ=
Represent the poles of 411z+ in the upper half plane. Since
each of these is a simple pole ∴ The residues are given by the values of 31()4fzz′ = at these
poles. ∴Res ()4
411;2 28 1i
ei
zπ
=− + +and
Res 3441;
1ie
z+π()1228i =−−
∴3 4
44 411 22R e ; R e ;211 1i
i dxis e s e
xz zπ
ππ π∞
−∞ =+ = ∫ ++ +
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8) Evaluate sin( )xdxx∞
−∞∫ .
Solution: We know that the function ixex has pole at 0x=. ∴ We modify the integral as follows: sin( ) 1Imixxedx dxxx∞∞
−∞ −∞−= ∫∫ .
11ix iz iz
M
M IM IMee edx dz i dzxz zπ−−−== − ∫∫ ∫∵ ∴1lim 0iz ix
M IMeedz dx izxπ∞
→∞ −∞−=⇒ =∫∫ ∴1s i n ( )Im .ixexdx dxxxππ∞∞
−∞ −∞−=⇒ =∫∫ .
9) Evaluate 031dxx∞∫+
Solution: L e t 31() .1fzz=
+Then f h a s p o l e s a t 31izeπ= a n d 3323 1,iize zeππ== − = ∴Res 31 3log( ) 1 3;92 21izi ize
zππ == − ++ ∴Res 32 3log( );31izize
zππ
== −+ ∴Res 33 3log( ) 1 3;92 21sizS i ize
zππ == − ++ ∴3log( ) 2Re ; 39 1k kzsz
zπ =− ∑ + ∴0 23log( ) 2Re ; 39 11k kdx zsz
xzπ ∞ =− =− ∑ ∫ ++
10) Evaluate 0(1 )dxxx∞∫+.
( Hint: The integral has the form 1
0()axdxpx−
∞∫ with 1012a<= <
Use the formula ()11
2(1 )
01R e ;() ()aa
ia
k kxzed x s zpx pxπ−−
∞ − − =∑ ∫ the
sum on R.H.S. is taken over the zeroes of t he function ()pz. munotes.in
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11) Evaluate 2
02( )dsπθθ∫+∞
Solution: Consider 2
0 212
2c o s ( )41zdd z
izzπθθ == ∫∫+++
2124R e ;3 2 33 41s
zzππ= −= ++
12) Evaluate the integral ()2
221xdxx∞−∞∫
+ .
13) Use Cauchy-Residue theorem to evaluate
2sin1zdzzγ∫
+ where γ is the circle ziπ−= .
14) Use Cauchy-Residue theorem to evaluate20cos54xdxx∞∫ + .
*****
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13
MOBIUS TRANSFORMATION
Unit structure
13.0 Objective
13.1 Introduction
13.2 Conformal Mapping
13.3 Some standard transformation
13.4 Mobius Transformation Or Bilinear Transformation O r
Linear Fractional Transformation
13.5 Summary
13.6 Unit End Exercise
13.0 OBJECTIVE
After going through this unit you shall come t o know about
• Special type of functions called transformation fro m →
• The combination of special function to give rise to a
transformation called Mobius Transformation
• Special properties of Mobius Transformation like fi xed point
and cross ratio.
• Method to find the bilinear transformation using va rious
method.
13.1 INTRODUCTION
There are certain transformation that can be readil y described
in terms of geometry. In this chapter, we are mainl y concerned with
certain geometric interpretations of functions and finding the image
of a given figure under a given bilinear function .
13.2 CONFORMAL MAPPING
A differentiable map 2:fΩ→ is said to be conformal map if
()()()()()det 0 & , , , {0}zz zDf z Df Df αβα β α β ≠∀ ∈ Ω = ∀∈ −
Thus, conformal map is preserves the angle between two intersecting
curves in munotes.in
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Proposition: Let Ω b e a d o m a i n i n a n d :fΩ→ b e a m a p .
Then f is any analytic function with ()0 fz z′≠∀∈ Ω if and only if f
is conformal map with det(Df z) > 0 z∀∈ Ω
Proof: Let f be analytic, ()()zDf f zαα α ′= ∀∈. Then
() ()()()()()()() ()()()Re Re, ,, / { 0 }fz fz
fz fz
fz fzαβ αβ
αβ αβ αβαβ αβ′′
′′ == ∀∈′′
Thus f is conformal map.
Let f(z)=u(z)+iv(z),z∈Ω. By Cauchy Riemann equation, the
Jacobian of f =( u(x,y),v(x,y) ) is xyxxxy xxuuuvvvvu−= .
Hence det(Df x)=u x2+ u y20≥. Now ()xxfz u i v′=+ & ()0fz′≠.
Hence det(Df x)=u x2+ u y2=()2fz′>0.
Conversely, Fix z∈Ω. Since f i s c o n f o r m a l m a p , ()()()()1, 0 , 0,1 1, 0zzDf Df i==
So, ()()1, 0 0,1zzDf Df ⊥ . Let ()()21, 0 ,zDf a b = ∈. Then ()()1, 0 ,zDf b a=± .
Since ()()()det 0, 0,1 ,zzDf Df b a>= − .Let .ai bα=+ T h e n
()zDfβα β= (Verify) and f is complex differentiable.
13.3 SOME STANDARD TRANSFORMATION
(i) Translation : wz c=+ where c is a complex constant.
The transformation wzc=+is simply a translation of the axes
and as such as preserves the shape if the region in z-plane.
Note : in particular translation maps circle in z-plane onto circles in
the w-plane.
(ii) Rotation and Reflection : wc z= w h e r e c i s a c o m p l e x
constant.
Let Re , ,ii iwz r e c eφθ αρ== =
Now, ()Reii i ire e r eφθ α α θρρ+= ⋅ = ∴ R&rρφα θ== + ∴ Thus the transformation maps a point P(r,θ) i n t h e z - p l a n e
onto a point (, )Prρα θ′+ in the w-plane.
i.e. the image is magnified (or diminished) by rρ and rotated
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Note : in particular wc z= maps circle in z-plane onto circles in the
w-plane.
(iii) Inversion: 1wz=
Let Re ,iiwz r eφθ==
1Re
1R&i
irerφ
θφθ∴ =∴== −
i.e. transformatiom 1wz=maps P(r,θ) i n t h e z - p l a n e o n t o a p o i n t (, )Prθ′− in the w-plane.
Note : in particular 1wz= maps circle in z-plane onto circles in the
w-plane.
13.4 MOBIUS TRANSFORMATION OR BILINEAR
TRANSFORMATION OR LINEAR FRACTIONAL
TRANSFORMATION
Definition: A transformation ()az bsz wcz d + ==+, where ,,,abcd are
complex constant and 0ad bc−≠ i s c a l l e d M o b i u s T r a n s f o r m a t i o n
or Bilinear Transformation or Linear fractional tra nsformation.
Note : If 0bdad bc ad bcac −= ⇒ = ⇒ = . ∴ ()()b azaz bawd cz d cza + +== + + ac = = constant.
Thus, 0ad bc−≠ i s a n e c e s s a r y c o n d i t i o n f o r t h e M o b i u s
Transformation : ()az bsz wcz d + ==+
1) If S is a Mobius transformation, then 1S− is the inverse mapping
of S i.e. ()()()()11SS z z S S z− − = = .
Let ()az bSzcz d + =+ w h e r e ,,,abcd a r e c o m p l e x c o n s t a n t a n d 0ad bc−≠. ∴ ()1dz bSzcz a− − = − + munotes.in
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∴ ()() ()()11SS z S S z− − = dz bScz a − = − +
1 dz babcz a dz bcdcz a − =+ × − + − + − +
adz ab −=bcz ab−+
cz a
− +cz a− + ×
cdz bc cdz−−ad+ adz bczbc ad − =− + ()()ad bczad bc − = −z=
Similarly, ()()1SS z z− = .
Consider ()() ()()() ( ) ( )az b az bszcz d c z dλλ λ
λλ λ+ +==++ (0 )λ≠
⇒ The ω−efficients a, b, c, d are not unique.
12 1aa b21 2 1 21 2 1 22 1 1ca a d db b c cb c d++ + 2d
12 1aa b−22 1 1 21 2 2 1 2 1 1ca b c da b c db c d−−−2d=1212 2221 12 12 2 1 12aa dd abc d bbcc abcd − + −
=()()11 2 2 22 1 1 22 2 2ad a d bc bc bc a d− + − = ()()11 2 2 22 1 1 2 2 22ad a d bc bc a d bc−− −
= ()()11 1 1 2 2 22.0ad bc a d bc−− ≠. i.e. the coefficients ,,,abcd
are not unique.
3) Let S be a Mobius transformation on ∞.
()az bSzcz d + =+ where ,,,abcd are complex constant and 0ad bc−≠.
()()()b zazSzd zcz + =
+bazdcz + =
+
()aSc ∞= w h e n 0c≠
=∞ w h e n 0c=
Again, ()
()az b az bSzd cz d czc + + ==+ +
dSc − = ∞ when 0c≠
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4) If S a n d T a r e M o b i u s T r a n s f o r m a t i o n s , t h e n (TS ),
(composition of T and S ) is also a Mobius Transformation.
Let ()11 111az bSzcz d + ξ==+ w h e r e ,,,abcd are complex constant and
11 1 10ad bc−≠ a n d ()2222abTcd ξ+ ω= ξ=ξ+ w h e r e 222 2,,,abcd a r e
complex constant and 22 2 20ad bc−≠.
[]()()TS z TS z=1111az bTcz d +=+
() ( )1122
11 1122111 az bTS z a bcz d az bcdcz d + =+ × + + + +
12 2 1 2 1 21
11aa z a b bcz bd
cz d + ++ =
+11cz d+
× 12 12 1 2 1 2ac z bc cd z dd + + + 12 21 21 2112 12 1 2 1 2aa z ab bcz bdac z bc cd z dd + ++ = + + + ()()() ()12 2 1 21 2112 1 2 12 1 2aa bc z a b bdac cd z bc dd + + + =++ + zxz α+β =+δ
where, 12 2 1 21 21,aa bc ab bdα=+ β=+ , 12 1 2,xa c c d=+ 12 1 2bc ddδ=+
Now, ()()()()12 21 12 12 21 21 12 12xa ab c b cd d a b b da cc dαδ−β =+ + − ++ ()12 12 1212 12 12 2 112 12 12211 2 122 1 211 2aa bc aa dd bbcc bcdd aabcab c d a bcd bc dd=+ + + −
+ + +
0≠
Hence, composition of S and T is also a Mobius transformation.
Proposition : If S is a M.T. then, S is a composition of tran slation
rotation, inversion and Magnification.
Proof : Consider a Mobius Transformation, ()az bSzcz d + =+ w h e r e , ,,,abcd are complex constant and 0ad bc−≠
Case I : When 0c=
()az b a bSz zdd d + == +
If ()1aSz zd = and ()2bSz zd =+ munotes.in
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then, () ( ) () 21 21 2aa bSS SS z S z z S zdd d == = + =
21SS S=
In this case, Mobius transformation is a compositio n of translation,
rotation and Magnification.
Case II : When 0c≠
()ca z bSzcc z d + =+()az bc ad adccz d + + − =+()()()()bc ad cz dacc z d c c z d − + =+ ++
21.bc ad adcz cc − = + +
If () () () 12 3 21 ,,bc addSz z S z Sz zczc − =+ = = and ()4aSz zc =+
() () ()()()4321 4 3 2 1SSSSz SS S S z=432dSSSzc =+431SSdzc= +
43cSScz d= + ()4 22bc ad cScdc − = × + ()4bc adScc z d−= + ()bc ad acc z d c − =+ +
()ab ca dcc z d + − =+()Sz= ∴ 4321SSSSS=
I n t h i s c a s e , M . T . i s a c o m p o s i t i o n o f t r a n s l a t i o n , rotation,
inversion and magnification.
Fixed Points :
Definition : Let G be a subset of ∞ and :fG∞→. Then point
0zG∈ is said to be a fixed point of f if ()00fz z=.
e.g. i) Let ()2fz z=.Here, f has fixed points 0, 1 and ∞.
ii) Let ()1fzz= . Here, f has fixed point 1 and – 1.
iii) Let ()()3fz z i=+ +.Here, f has fixed point ∞.
Example 1 : What are the fixed points of Mobius transformation? munotes.in
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Solution : Consider a M.T., ()az bSzcz d + =+ w h e r e , ,,,abcd a r e
complex constant and 0ad bc−≠.
For fixed points,
Put
()az bSz z zcz d + = ⇒ =+2az b cz dz⇒ + = +()20cz d a z b⇒ + −− =
() ()24. .2da da b czc − −+ −− ⇒ =⇒ () ()242ad da b czc + ± −− = ∴ A . M . T . c a n h a v e a t m o s t 2 f i x e d p o i n t s u n l e s s ()Sz z z= ∨
2) ()21izSzz + =+
Solution : For fixed points
Put ()21izSz z zz + = ⇒ =+22iz z z⇒ + = +()212 0zi z⇒ + − −=()()211 82iiz − − ± −− ⇒ =
Definition : For any three distinct points 234,,zzz in C∞, the cross
ratio of four points 1234,,,zz zz i s d e f i n e d t o b e
()()()() ()13 24123414 23,,,zz z zzz zzzz z z − − =− −.
For 12zz=, ()()()() ()2324
2234
2423,,, 1zzzzzzzzzzzz − − = =− −
For 13zz=, ()()()() ()3324
3234
3423,,, 0zzzzzzzzzzzz − − = =− −
For 14zz=, ()()()() ()4324
4234
4423,,,zzzzzzzzzzzz − − = = ∞− −
Definition : If 1z∞∈ t h e n t h e c r o s s r a t i o ()1234,,,zz zz i s t h e
image of 1z under the unique. Mobius transformation which t akes 2z to 1, 3z to 0 and 4z to ∞. i.e. ()()11 2 3 4,,,Sz z z z z= .
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Note : If M is any Mobius transformation and 234,,www a r e
complex number s.t. ()()211, 0Mw Mw= = a n d ()4 Mw=∞ t h e n ()()234,,,Mz z w w w= .
Proposition : If 234,,zzz a r e d i s t i n c t p o i n t s i n ∞ a n d T i s a n y
Mobius transformation then, ()()1234 1 2 3 4,,, , , ,zz zz T zT zT zT z = f o r
any fix 1z.
Proof : Let ()()1234,,,Sz z z z z= ⇒ S is a M.T. (1)
and ()()231, 0Sz Sz= = and ()4Sz=∞
Given that, T is any M.T.
Put 1MS T− = ∴ () ()()()1
22 21MT z ST T z Sz− == = ∴ () ()()()1
33 30MT z ST T z Sz− == = ∴ () ()()()1
44 4 MT z ST T z Sz− == = ∞ ∴ ()()234,,,Mz z T z T z T z=
Put 1zT z= ∴ ()()11 2 3 4,,,MT z T z T z T z T z=
()()()111 2 3 4,,,ST T z T z T z T z T z− = ()1MS T− =∵
()()11 2 3 4,,,Sz T zT z T z T z= ∴ ()()1234 1 2 3 4,,, , , ,zz zz T zT zT zT z = () is M.T.S ∵
i.e. The cross ratio is invariant under Mobius Tran sition.
Proposition : If 234,,zzz are distinct points in ∞ and 234,,www
are also distinct points in ∞ then, there is a unique M.T. S s.t. ()22,Sz w= ()33Sz w= and ()44Sz w=.
z 2 34 S is a M.T. w2 3 4 z z wwS
()zp l a n e∞− ()wp l a n e∞−
Fig 13.1
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S
∞Mobius
transformation
SProof : Let ()()234,,,Tz z z z z= and ()()234,,,Mz z w w w= . ⇒ T a n d M a r e M o b i u s T r a n s f o r m a t i o n s a n d ()21Tz=, ()30Tz= and ()4Tz=∞, ()()231, 0Mw Mw= = and ()4 Mw=∞.
Put 1SMT− = ∴ () ( ) ( )1122 21Sz M T z M w− − == =
() ( ) ()1133 30Sz M T z M w− − == =
() ( ) ( )1144 4Sz M T z M w− − == ∞=
Let R be another M.T. s.t. ()jjRz w= f o r 2, 3, 4j= ()() () () ()11 122 2 2RS z R S z R w z− − − = = = ()() () () ()11 133 3 3RS z R S z R w z− − − = = = ()() () () ()11 144 4 4RS z R S z R w z− − − = = =
Here, ()1RS− composite map of S and 1R− has 3 fixed points. ∴ 1RS I− = ( I d e n t i f y m a p ) SR⇒ =
Hence, S is the unique transformation.
Proposition : Let 1234,,,zz zz be distinct points in ∞ then the cross
ratio ()1234,,,zz zz i s a r e a l n u m b e r ⇔ a l l f o u r p o i n t s l i e s o n a
circle (or a straight line). (2009) Proof : Let :S∞∞→ be defined by ()()234,,,Sz z z z z= = Real
number.
Fig 13.2
Then, {} (){}1
234:, ,, i s r e a lSz z z z z− = i . e . i m a g e o f ∞ u n d e r
the Mobius Transformation is a circle. munotes.in
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We will prove that the image of ∞ u n d e r t h e M o b i u s
transformation is a circle.
Let ()az bSzcz d + =+ w h e r e ,,,abcd a r e c o m p l e x c o n s t a n t a n d 0ad bc−≠.
If zx=∈ and ()1wS x− = , then ()Sw x=∈. ∵ ()Sw is purely real number. ∴ () ()Sw sw= ∴.
.aw b a w bcw dcw d + + =++ ∴ ()()()()aw b c w d a w b cw d+ + = + + ⇒ 22ac w adw bcw bd ac w awd bcw bd + + + = + + + ()()()()20ac ac u ad bc w bc ad w bd bd− + − + − + − = ( 1 )
Case I : If ac is not real, then 0ac ac−≠.
I f ac is real then ac ac ac==
0ac ac⇒ = =
From equation (1), ()()()()()()20ad bc ad bc bd bd
ww w
ac ac ac ac ac ac − − − + + − =−−−
Put ad bcxac ac −=
− and bd bdac ac − δ=
−
20wx w x w + + −δ=
22 2wx w x w x x + + + = + δ
()wx R w x R+= ⇒ −− = which is the equation of the
circle with centre at (- x, 0) and radius equal to R.
Where, 2Rx= + δ
2
.ad bc bd bdac a c a c a c − − = +
−−0ad bc
ac ac − = ≠−
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Case II : If ac is real then 0ac ac−=
From equation (1) ()()()0ad bc w bc ad w bd bd− + − + − =
Put ad bcα= − & ()ib d b dβ= −
0.2ww i−
−−αα⇒ − − β =
Im ( ) 0.i∴ αω − β =
Im ( ) 0⇒ αω−β =.
or ()Im 01wβ − α=
α ( 2 )
Note : Consider a straight line L in the Complex Pl ane . If a
is any point in L and b is its direction vector then,
abL
Fig 13.3 :I m 0zaLzb − ==
OR
{}:, i . e . za b t t t== + ∈ − ∞<< ∞ ∴ The point w l i e s o n a l i n e d e t e r m i n e d b y e q u a t i o n ( 2 ) f o r
fixed α and β. ∴
Theorem : A Mobius transformation takes circles onto cir cles. :I m 0zaLzb − == munotes.in
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Proof : Let Γ be any circle in ∞ (z-plane). Let S be any Mobius
Transformation. Suppose 234,,zzz a r e d i s t i n c t p o i n t s o n t h e
circle Γ.
S
M.T.is az
23
4z
z 2ω3ω
4ω
ω
zST
′=Γ′ΓΓ
Fig 13.4
Put jjSz w= for 2, 3, 4j= . ⇒ 234,,www are distinct points in ∞ (w-plane). ∴ T h e s e t h r e e d i s t i n c t p o i n t s 234,,www determine a circle in w-
plane.
T.P.T. ()S′Γ=Γ
Since, 234,,zzz are distinct points in ∞ and S in a M.T. ∴ ()()234 2 3 4,,, , , ,zz z z S zS z S z S z = for any point z.
()234,,, . . .Sz w w w= ( 1 ) ∴ I f z∈Γ then the cross ratio ()234,,,zz z z is a real number. ∵ 234,,,zz z z all lie on a circle … ∴ ()234,,,Sz w w w is a real number. (by equation (1)) ⇒ These four points 234,,,Sz w w w lie on a circle ′Γ .
Put Sz w=
As z moves on the circle Γ, then the corresponding point w moves
on the circle ′Γ under a M.T. ‘ S ’. ⇒ ()S′Γ=Γ.
H e n c e , a M o b i u s T r a n s f o r m a t i o n t a k e s c i r c l e s o n t o circles.
13.6 UNIT END EXERCISE
Example : Find a M.T. which maps points 1, 0, 1z=− o n t o t h e
points 1, , 1.wi=− − Also find the image of unit circle 1z = in the -plane under this M.T. munotes.in
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Solution : Given points 23 4,1 , 0 , 1zz z z z= = − = = a n d ww=,
21w=− 3wi=− , 41w=.
y = 0-1Y
Xv
u11 -1ii− 1ω=zplane−planeω−
Fig 13.5
∴ M . T . i s g i v e n b y ()()234 2 3 4,,, , , ,zz z z ww w w =
( P r o p o s i t i o n o n p a g e n o . 8 1 ( 2 ) ) ⇒ ()()() ()()()() ()324 3 2 442 3 4 23zz z z ww w wzz z z ww w w − − − − =− −− −⇒ ()()() ( )()()() ( )01 1 1 111 0 11zw izw i − −− + −− =−− − − −+ ⇒ ()()()()11zi zw zwi− + −=− + ⇒ zw− zi z wi z++ − wi z w=+ −iz− ⇒ zi z wwi+= + ⇒ ()ziSz w izi − ==+ (1)
which is a required bilinear transformation. ∴ F r o m e q u a t i o n ( 1 ) , ziwizi − =+ ⇒ 11wz iw iz wz iz iw += + ⇒ −=− ()1zw i i w⇒ −=−
1iwzwi − ⇒ =−
Given equation of unit circle is 1z =. ∴ 111iwzi w w iwi − = = ⇒ − = = − 221iw w i⇒ − = −
()()()()11iw iw w i w i⇒ −− = −−
()()()()11iw iw w i w i⇒ − −= − − munotes.in
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2
1
11
2
3
331,0 & 0,0 &0wiwiz z
z zzzz
w wwww−= ⇒ = ∴−−
=∞ ===∞ ==∵∵
()()()()11iw iw w i w i⇒ − + = − +
2211iw iw w w iw iw⇒ + −+ = + −+ 22 0iw iw⇒ − = ⇒ 0ww w w−= ⇒ =
Put wui v=+, ∴ ww ui vui v= ⇒ + = + ui vui v⇒ + = − 20iv⇒ = 0v⇒ =
which is the equation of real axis.
1) Find a Mobius transformation , which send 1, ,1i − onto 1, , 1i− respectively.
Solution : (1) 1, ( ) , ( 1) 1.ab a ib abff i i fcd c id cd++ −+ = = − = = −==++ −+. ,ab cd⇒ += −− ai b di c a d and b c+= −⇒ = = −, 00ba dc cd dc d a−=−⇒−− =−⇒ =⇒ =
()1,bcfzcz cz z∴ = = = − since 0c≠.
()1fzz∴ =− is the required bilinear transformation .
2) Find the fixed points of the mapping 1zwz=+.
Solution: L e t 0z b e t h e f i x e d p o i n t o f t h e m a p p i n g
()2 0
00
00,11zzwf z z zzz== ⇒ = ⇒ =++ l e t 00 0 00zx i y x=+ ⇒ =
and 00y=.
00z∴ = is the fixed point of the mapping 1zwz=+.
3) Find the Mobius Transformation bilinear mappin g sending ,, 2ii i− onto 1,0,3∞ respectively. munotes.in
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Solution: Let () ()0az bfz a d b ccz d+= −≠′+ be the required bilinear
mapping . We know that fmaps dc− onto ,dic∞⇒−=−
()az bdi c f zcz ic+∴ = ⇒ =+
Now () () 002ai b ai bfi fi b a ici ic ci++=⇒ == = ⇒ =−+
()()().az iaz aifzcz ic c z i−−∴ = =++ () ()112233 3afi fic = ⇒ ==∵ ac⇒ =.
()()(),az izifzcz i z i−−∴ = =++ here 0a≠
(Alternate method for problems involving infinity)
4) Find the bilinear transformation which maps the p o i n t s ,, 0zi=∞onto the points 0, ,zi= ∞
Solution: We have transformation 12 3 1 2312 3 1 2 3() ( ) ( ) ()() ( ) ( ) ( )www w zz z zwwww zzz z−− − −=−− − −
Since,1z=∞ and 3w=∞, we define N and D of LHS and RHS by 3wand 1z respectively.
2
1
11,0 &0z zzzz=∞ ==∵ and 2
3
33,0 &0w wwww=∞ ==∵ 231
12 3(1 ) ( )() ( 1 )()( )zzww
ww z z−−−−=−−
Put z2=i, z 3=0 and w 1=0, w 2=i ∴1wiwiz z−= ⇒ =−−
5) Find a Mobius transformation , which send 1, ,1i − onto 1, , 1i− respectively.
Solution : (1) 1, ( ) , ( 1) 1.ab a ib abff i i fcd c id cd++ −+ = = − = = −==++ −+. ,ab cd⇒ += −− ai b di c a d and b c+= −⇒ = = −, 00ba dc cd dc d a−=−⇒−− =−⇒ =⇒ =
()1,bcfzcz cz z∴ = = = − since 0c≠.
()1fzz∴ =− is the required bilinear transformation . munotes.in
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6) Find the fixed points of the mapping 1zwz=+.
Solution: Let 0z be the fixed point of the mapping
()2 0
00
00,11zzwf z z zzz== ⇒ = ⇒ =++ l e t 00 0 00zx i y x=+ ⇒ =
and 00y=.
00z∴ = is the fixed point of the mapping 1zwz=+.
7) Find the Mobius Transformation/ bilinear mapping sending ,, 2ii i− onto 1,0,3∞ respectively.
Solution: L et () ()0az bfz a d b ccz d+= −≠′+ be the required bilinear
mapping . We know that fmaps dc− onto ,dic∞⇒−=−
()az bdi c f zcz ic+∴ = ⇒ =+
Now () () 002ai b ai bfi fi b a ici ic ci++=⇒ == = ⇒ =−+
()()().az iaz aifzcz ic c z i−−∴ = =++ () ()112233 3afi fic = ⇒ ==∵ ac⇒ =.
()()(),az izifzcz i z i−−∴ = =++ here 0a≠
8) Let 1234,,,zz zz be four distinct points in ∞. Then show
that ()1234,,,zz zzis a real number iff all four points lie
on a circle.
(Hint: Define :s∞∞→ by ()()234,,,sz z z z z= Show that ()s∞
is a circle. Here ()1s−= the set of all z such that ()234,,,zz z z is real.
9) Prove that all the points Cz∈satisfying 124z
z+=+ lie in a circle.
Find its radius and centre (2009) munotes.in
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10) Find the image of the circle x2+ y2+2x=0 in the complex plane
under the transformation w=z1
(2008)
11) Find the Mobius transform which maps the point s z=1, I ,-1
onto the points Mobius transformation (2008)
12) Let H={z 0) Im(/>∈z C }and let D={z 1||/<∈z C }. Find the
Mobius transformation g s.t .g(H)=D and g(i)=0 . Justify your claims
(2007)
13) Show that Mobius transformation has 0 and ∞as its only fixed
points if and only if it is dilation (magnification ) (2007)
14) Show that Mobius transformation has ∞ as its only fixed points
if and only if it is a translation (2007)
15) Find the Mobius transform which maps the real axis ∪∞R onto
the circle 1=z. (2006)
16) Fix a,b,c,d C∈with c 0≠.show that ∞→ →++zascadczbaz
17) Verify that the Mobius transformation 1izwiz+=+ m a p s t h e
exterior of the circle 1z= in the z-plane into the upper half plane ()Im 0w> in the w-plane.
18) Find the image of the circle 33zi − = i n t h e c o m p l e x p l a n e
under the transformation1wz=. Illustrate the results graphically.
19) Find the image of an infinite strip 1142y ≤ ≤ in the complex plane
under the transformation1wz =
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