Separating an Inverse Function Into Real and Imaginary Parts
Chapter Thirty-Six
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 80 to 81 of 303
In one line
Set the inverse function equal to u + iv, take the function of both sides, and compare parts.
The question this answers
A standard examination question reads: if tanh inverse of (x + iy) equals u + iv, find u and v. Or: separate sin inverse of (a + ib) into real and imaginary parts.
The logarithmic forms of the previous two chapters will do it, but they lead into surds of complex numbers and are hard to control. There is a better method, and it is the one to use.
The method
Do not try to work forwards from the logarithm. Work backwards.
- Let the answer be u + iv.
- Apply the forward function to both sides, so the inverse disappears.
- Separate the forward function of u + iv into real and imaginary parts, using the results already proved.
- Compare parts with the given number. You now have two real equations in u and v.
- Solve them, usually by taking a ratio or by using an identity.
The whole trick is step two: an inverse function is hard to manipulate and its forward partner is easy, so get rid of the inverse first.
Worked: tanh inverse of x + iy
Let u + iv be the answer, so that x + iy is the hyperbolic tangent of u + iv.
x + i y = tanh(u + i v)
Now use the separation of tanh from the chapter on the tangent of a complex angle.
tanh(u + i v) = (sinh(2u) + i sin(2v))/(cosh(2u) + cos(2v))
Comparing parts gives the two equations below.
x = sinh(2u)/(cosh(2u) + cos(2v))
y = sin(2v)/(cosh(2u) + cos(2v))
Now the standard manoeuvre. Form x squared plus y squared, and separately form 2x over (1 minus x squared minus y squared). The algebra is routine and the results are these.
tan(2v) = 2y/(1 - x^2 - y^2)
tanh(2u) = 2x/(1 + x^2 + y^2)
So:
u = atanh(2x/(1 + x^2 + y^2))/2
v = atan(2y/(1 - x^2 - y^2))/2
Those two are the answer, and they are the pair MU's own reading list prints. Notice the symmetry: the same expression appears in both with x and y exchanged and one sign changed, which is a good way to recall them.
Worked: sin inverse of a complex number
Let u + iv be sin inverse of (a + ib), so a + ib is the sine of u + iv.
a + i b = sin(u + i v)
Use the separation of the sine.
sin(u + i v) = sin(u) cosh(v) + i cos(u) sinh(v)
Comparing parts:
a = sin(u) cosh(v)
b = cos(u) sinh(v)
Two equations, two unknowns. Eliminate v by using cosh squared minus sinh squared equals one: divide the first by sin u and the second by cos u, then square and subtract.
Separating an Inverse Function Into Real and Imaginary Parts
a^2/sin(u)^2 - b^2/cos(u)^2 = 1
That is an equation in u alone. It is a quadratic in sin squared u once you clear the denominators, and solving it gives u; putting u back into either original equation gives v.
The same elimination the other way, using sin squared plus cos squared equals one, gives an equation in v alone.
a^2/cosh(v)^2 + b^2/sinh(v)^2 = 1
Both forms are examinable, and which you use depends on which unknown the question asks for first.
Worked, with numbers
Find the real and imaginary parts of tanh inverse of i.
Here x = 0 and y = 1, so the formulae give the following.
2(0)/(1 + 0 + 1) = 0
So the real part u is half of tanh inverse of zero, which is zero. And for the imaginary part, 1 minus 0 minus 1 is zero, so the argument of the arctangent is 2 divided by 0, which is unbounded, and the arctangent of that is pi over two. So v is half of pi over two, which is pi over four.
atanh(i) = i pi/4
Check it directly: the hyperbolic tangent of i pi over four is i times the tangent of pi over four, which is i times one, which is i.
tanh(i pi/4) = i
The answer is right.
The two checks to run on any answer
Put the imaginary part to zero. If y = 0 the answer must reduce to the ordinary real inverse function of x, with no imaginary part. Any formula that fails that test has been copied down wrongly.
Apply the forward function. One substitution, as in the worked example above, and you know.
Check yourself
tanh(i pi/4) = i
sin(i) = i sinh(1)
asin(i) = i asinh(1)
atanh(0) = 0
tanh(atanh(1/2)) = 1/2
The third line is the answer to "separate sin inverse of i", and it says the answer is purely imaginary: u is zero and v is sinh inverse of one, which is log(1 + root 2). That is the sort of answer to expect when the number you start from is purely imaginary.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.