How to evaluate $\int (\frac{1}{z(\exp(z)-1)}dz$ with residue theorem. I am trying to evaluate the integral
 $\int (\frac{1}{z(\exp(z)-1)}dz$   $\mathbb{C}:|z|=1\}$
using Laurent theorem. $0$ is a singularity point and both denominators becomes $0$ when $z=0$. So it's hard to find a method to evaluate this using Residue theorem. Could someone please show me how to solve this.
Thank you.
 A: You have a double pole. The function is
$$\frac1z\left(z+\frac{z^2}2+\frac{z^3}{3!}+\cdots\right)^{-1}
=\frac1{z^2}\left(1+\frac{z}2+\frac{z^2}{6}+\cdots\right)^{-1}.$$
The residue is the $z^{-1}$-coefficient of this, that is the $z$-coefficient
of the power series for
$$\left(1+\frac{z}2+\frac{z^2}{6}+\cdots\right)^{-1}.$$
You need to extract that...
A: The easiest way to compute it is by using the formula for a residue of order 2:
$$
Res_{z=0}\left(\frac{1}{z(e^{z}-1)}\right)=\lim_{z\to 0}\frac{d}{dz}\left(z^2\frac{1}{z(e^{z}-1)}\right)=\lim_{z\to 0}\frac{d}{dz}\left(\frac{z}{e^{z}-1}\right)=\lim_{z\to 0}\frac{e^z-1-ze^z}{(e^z-1)^2}
$$
The last limit can be computed by expanding each term as a power series.
$$
\frac{e^z-1-ze^z}{(e^z-1)^2}=\frac{1+z+z^2/2-1-z(1+z)+o(z^3)}{(1+z-1)^2+o(z^4)}=-\frac{1}{2}+o(z^2)\to -\frac{1}{2}
$$
Hence the integral is just
$$
\int_{|z|=1}\frac{1}{z(e^z-1)}dz=2\pi iRes_{z=0}\left(\frac{1}{z(e^{z}-1)}\right)=-i\pi
$$
You could have also computed the firs couple of terms of the laurent series by expanding the exponential.
$$
\frac{1}{z(e^z-1)}=\frac{1}{z(z+z^2/2+o(z^3))}=\frac{1}{z^2}\frac{1}{1+z/2+o(z^2)}=\frac{1}{z^2}\left(1-z/2+o(z^2)\right)=\frac{1}{z^2}-\frac{1}{2z}+o(1)
$$
And you get the same residue.
