TITLE: what are the residues at poles of $\frac{1}{1+\cosh{z}}$?
QUESTION [5 upvotes]: Consider the function $f(z)=\frac{1}{1+\cosh{z}}$. It has poles of order 2 at odd multiples of $\pi i$, but what are the residues at the poles? I've tried using $\frac{d}{dz} \Big((z-a)^2 f(z)\Big)$ for the residue at $a$, but get the answer to be 0, which I don't think is correct.

REPLY [4 votes]: Why do you think $0$ isn't correct? Since $\cosh (\pi\mathrm i+z)=-\cosh z$ is an even function of $z$, so is $f(\pi\mathrm i +z)$. Thus its Laurent series contains only even powers of $z$, and in particular doesn't contain a $z^{-1}$ term, so the residue is indeed $0$.