$\oint_C \frac{z^2}{\exp z + 1} dz$ using residue theorem 
Evaluate this integral using residues,
  $$\oint_C \frac{z^2}{\exp z + 1} dz,$$ where $C$ is the contour bounded by $|z| = 4$.

My attempt: 
pole of $\exp z + 1 = +\pi i, -\pi i$. Now what is the order of the pole? 
 A: You have correctly stated that the singularities of your integrating function come as :
$$e^z + 1 = 0$$
but you've calculated the outcome wrong. Specifically, it is :
$$e^z + 1 = 0 \Leftrightarrow z= 2n\pi i + \pi i, \space n\in \mathbb Z$$
Over the contour in which you're integrating, you'll need to check how many of these poles will fall in. Specifically, you have to check the "length" of each number (as the absolute value over reals).
Truly then, the only poles residing in your integrating contour $C : |z| < 4$ are the poles $\pm \space \pi i$, since for any other value of $n$ the length of the imaginary number is larger than 4.
The poles $\pm \pi i$ are simple poles since $(e^z)' \neq 0$ and thus you can calculate the needed residues for the calculation of the integral as : 
$$\text{Res} \bigg(\frac{f(z)}{g(z)}, z= \pm \space \pi i\bigg) = \frac{f(\pm \space \pi i)}{g'(\pm\space\pi i)} $$
where $f(z) = z^2$ and $g(z) = 1/(e^z + 1)$.
A: Notice that 
$$e^z+1=e^{\pi i+ (z-\pi i)}+1=-e^{(z-\pi i)}+1=-\sum_{k=0}^{\infty}\frac{(z-\pi i)^k}{k!}+1=-(z-\pi i)+O((z-\pi i)^2).$$
Hence, in a neighbourhood of $z=\pi i$,
$$\frac{z^2}{\exp z + 1}=\frac{((z-i\pi)+i\pi)^2}{-(z-\pi i)+o(z-\pi i)}
=\frac{\pi^2}{z-\pi i}+O(1).$$
So what is the order of the pole $z=\pi i$? What is the residue at $\pi i$?
A very similar approach works around the other pole inside $|z|<4$, that is $z=-\pi i$.
A: Since $\exp'(z)\neq0$, they're simple poles and theefore you can compute the residues there using the formula$$\operatorname{res}_a\left(\frac{f(z)}{g(z)}\right)=\frac{f(a)}{g'(a)}$$and then you can apply the residue theorem.
