Real and imaginary parts of a complex-valued function

Question

How do you get a complex-valued function $ f(z) = f(x+iy) = \frac{z^{s-1}}{e^{-z}-1}, $

where $s$ is a constant complex number and $z$ is a complex variable, into the form: $ f(x+iy) = a(x,y) + ib(x,y)$?

Thank You,

C.A

Answer 1

I'd ask for upvotes to offset my mental health bills after typing this, but I'm sure you'll all need it to pay your eye doctors.

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Allow me to give you a taste of how hopelessly tedious, cumbursome, and unfruitful it is to expand expressions like these into real and imaginary parts. If it helps you solve an integral, I will eat my shoe.

Let $z=x+iy$ and $s=\rho+i\sigma$.

$$\frac{z^{s-1}}{e^{-z}-1}=\frac{(x+iy)^{(\rho-1)+i\sigma}}{e^{-x-iy}-1}=\frac{\exp\big[((\rho-1)+i\sigma)\log(x+iy)\big]}{e^{-x-iy}-1}$$

$$=\frac{\exp\big[((\rho-1)+i\sigma)\log\left(\sqrt{x^2+y^2}e^{i\arctan(y/x)}\right)\big]}{e^{-x}(\cos(-y)+i\sin(-y))-1} $$

$$=\frac{\exp\big[((\rho-1)+i\sigma)\left(\frac{1}{2}\log(x^2+y^2)+i\arctan\frac{y}{x}\right)\big]}{(e^{-x}\cos y-1)-ie^{-x}\sin y}$$

$$\frac{\exp\left[\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)+i\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)\right]}{(e^{-x}\cos y-1)-ie^{-x}\sin y}$$

$$\scriptsize =\exp\left[\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right]\frac{\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)+i\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)-ie^{-x}\sin y} $$

$$\tiny =\frac{\exp\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)^2+(e^{-x}\sin y)^2}\left[\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)+i\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)\right]\left((e^{-x}\cos y-1)+ie^{-x}\sin y\right)$$

$$\tiny = \frac{\exp\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)^2+(e^{-x}\sin y)^2}\left[\begin{matrix}\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)(e^{-x}\cos y-1)-\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)e^{-x}\sin y \\ +i\left(\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)e^{-x}\sin y+\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)(e^{-x}\cos y-1)\right)\end{matrix}\right].$$

Therefore,

$$\small a=\frac{\exp\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)^2+(e^{-x}\sin y)^2}\left[\begin{matrix}\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)(e^{-x}\cos y-1) \\ -\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)e^{-x}\sin y\end{matrix} \right]$$

and

$$\small b=\frac{\exp\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)^2+(e^{-x}\sin y)^2}\left[\begin{matrix}\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)e^{-x}\sin y \\ +\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)(e^{-x}\cos y-1)\end{matrix}\right] $$

Answer 2

I think we can calculate a(x,y) and b(x,y) in the way,

a(x,y)=1/2[f(z)+f*(z)],
b(x,y)=1/2i[f(z)-f*(z)].