Find residues of $f(z)=\frac{1}{(e^{z}-1)^{2}}$

Question

How to find the residues of $f(z)=\dfrac{1}{(e^{z}-1)^{2}}$

I have found that the poles $z=2\pi i n$.

But when I apply the formula $\dfrac{1}{(m-1)!}\lim\limits_{z\rightarrow z_{0}}\dfrac{d^{m-1}}{dz^{m-1}}\left((z-z_{0})^{m}f(z)\right)$,

I got $\lim\limits_{z\rightarrow z_{0}}\dfrac{d}{dz}(z-z_{0})^{2}\dfrac{1}{(e^{z}-1)^{2}}$,

which is so complicated that I cannot get the limit value.

Answer 1

Hint: This function is periodic with period $2\pi i$, therefore all residues on $z=2\pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:

Edit: $$\lim_{z\to0}\dfrac{d}{dz}z^2\dfrac{1}{(e^{z}-1)^2}=\lim_{z\to0}\frac{2z\left(e^z-1-ze^z\right)}{(e^z-1)^3}=\color{blue}{-1}$$

Answer 2

Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.

$\frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$\frac{1}{z^2(1+z/2)^2}$=$\frac{1-z+o(z^2)}{z^2}$

By defination residue is coefficent of 1/z implies -1.