Evaluate the following integral:
$$\int_\gamma \frac{z}{e^z-1}dz$$
For:
Unfortunately, I'm really clueless... I've thought using Cauchy Theorem with Winding numbers, but there are 2 poles and one removable singularity inside the region, what can I do?
Thanks in advance.
You don't need to worry about the removable singularity. So, by the residue theorem, your integral is equal to$$2\pi i\left(1\times\operatorname{res}_{2\pi i}\left(\frac z{e^z-1}\right)+2\times\operatorname{res}_{-2\pi i}\left(\frac z{e^z-1}\right)\right),$$where the numbers $1$ and $2$ are the winding numbers of your loop with respect to $2\pi i$ and to $-2\pi i$ respectively. Now, you can compute the residues using the formula$$\operatorname{res}_a\left(\frac{f(z)}{g(z)}\right)=\frac{f(a)}{g'(a)}.$$