The Exponential Form at Work
Chapter Twenty-Five
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 58 to 59 of 303
In one line
In exponential form every multiplication, division, power and root is one line of index arithmetic.
The five rules, gathered
(2 e^(i pi/6))(3 e^(i pi/3)) = 6 e^(i pi/2)
(12 e^(i pi/2))/(4 e^(i pi/6)) = 3 e^(i pi/3)
(2 e^(i pi/8))^4 = 16 e^(i pi/2)
conjugate(5 e^(i pi/7)) = 5 e^(-i pi/7)
1/(4 e^(i pi/3)) = (1/4) e^(-i pi/3)
Nothing there needs to be learned as a new rule. They are the index laws.
Roots, in exponential form
The root formula of the earlier chapter looks much less forbidding written this way. The n nth roots of r e to the it are given below, for k from 0 to n minus 1.
z = r^(1/n) e^(i(t + 2 pi k)/n)
The 2 pi k is there because e to the 2 pi i is 1, so multiplying the number by it changes nothing before the root is taken and changes everything after.
Worked: the cube roots of 27 e to the i pi.
The modulus of each root is the real cube root of 27, which is 3. The arguments are pi over three, pi over three plus 2 pi over three, and pi over three plus 4 pi over three, that is pi over three, pi, and five pi over three, the last being better written as minus pi over three.
3 e^(i pi/3) = 3/2 + 3i sqrt(3)/2
3 e^(i pi) = -3
3 e^(-i pi/3) = 3/2 - 3i sqrt(3)/2
(3 e^(i pi/3))^3 = -27
The last line is the check: the cube is 27 e to the i pi, which is minus 27, which is what we took the root of.
A worked question of the kind that is set
Express the quantity below in the form a + ib.
((1 + i)^6)/((1 - i sqrt(3))^4)
Convert both bases first. The number 1 + i is root two at 45 degrees; 1 minus i root three is 2 at minus 60 degrees.
1 + i = sqrt(2) e^(i pi/4)
1 - i sqrt(3) = 2 e^(-i pi/3)
Now the powers.
(sqrt(2) e^(i pi/4))^6 = 8 e^(3 i pi/2)
(2 e^(-i pi/3))^4 = 16 e^(-4 i pi/3)
Now divide: the moduli give 8 over 16, which is one half, and the arguments subtract.
3 pi/2 - (-4 pi/3) = 17 pi/6
Seventeen pi over six is more than a full turn, so subtract 2 pi to bring it into range, giving five pi over six.
(1/2) e^(5 i pi/6) = -sqrt(3)/4 + i/4
So the answer is minus root three over four plus i over four. Doing that same calculation by expanding the sixth power and the fourth power of Cartesian brackets is possible and would take the whole ten minutes you have.
The Exponential Form at Work
((1 + i)^6)/((1 - i sqrt(3))^4) = -sqrt(3)/4 + i/4
That last line is the same claim stated directly, and it was verified by machine, so the route above arrives where it should.
Where the form is genuinely used, not just convenient
A rotating quantity. An alternating voltage of amplitude V and angular frequency w is written V e to the i w t, and the whole of alternating-current circuit analysis is done that way: a resistor, a capacitor and an inductor all become a single complex number to divide by, and Ohm's law works again.
A signal's frequency content. e to the i w t is the one function that a linear system does not change the shape of, only the size and the timing. That property is why every transform in engineering is built out of complex exponentials, including the Laplace transform in the second half of this module.
A rotation in graphics. Multiplying by e to the it rotates by t. A rotation matrix is that statement written out in components.
The one trap
The index laws for a real exponential hold without qualification. For a complex one, the law that (e to the a) to the b equals e to the ab needs care when b is not an integer, because the left-hand side is multi-valued.
The safe rule, and it is enough for this paper: use the index laws freely for integer powers, and use the root formula, with its 2 pi k, whenever the index is fractional. The chapter on the logarithm of a complex number is where the multi-valuedness has to be faced properly.
Check yourself
(3 e^(i pi/4))(2 e^(i pi/4)) = 6i
(e^(i pi/3))^6 = 1
abs(7 e^(i pi/5)) = 7
(8 e^(i pi))^(1/3) = 2 e^(i pi/3)
(2 e^(i pi/6))/(e^(-i pi/6)) = 2 e^(i pi/3)
The fourth line gives only one of the three cube roots, the principal one. The other two are at pi over three plus 2 pi over three and pi over three minus 2 pi over three, and a question asking for "the cube roots" wants all three.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.