evaluate the complex integration

Question

How to evaluate the below integral

$$\oint_{c} \frac{dz}{e^{z}-1}$$ where $C$ is the circle $|z|=1$

Can this be done by Cauchy's formula? If yes how? Or do I need to do something else in order to solve this

Answer 1

$$e^z-1=z+\frac{z^2}{2!}+\ldots\implies\frac1{e^z-1}=\frac1{z\left(1+\frac z2+\ldots\right)}=\frac1z\left(1-\frac z2+\ldots\right)$$

So what's the residue of $\;\frac1{e^z-1}\;$ at $\;z=0\;$, the only singularity inside $\;|z|=1\;$ ? Now use the residue theorem.