what are the residues at poles of $\frac{1}{1+\cosh{z}}$?

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.

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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 $0$.
Thanks! This gives a quicker way of seeing that the residue is 0. But the next part of the question is to take the rectangular contour with corners at $\pm n\pi$ and $\pm n\pi + 2n\pi i$ and integrate $\frac{e^{ivz}}{1+\cosh{z}}$ around this contour. What is the limit of this integral as $n\rightarrow\infty$? Since all the residues are 0, the integral seems to be 0 regardless of $n$, so the question is trivial. Or am I missing something? – Harry Macpherson Mar 26 '12 at 16:10
@Harry: What's $v$? Assuming that it's a non-zero constant, no, the residues of that function aren't zero; $\mathrm e^{\mathrm iv(\mathrm i\pi + z)}$ has a linear term, and that times the $z^{-2}$ term of $f(\pi\mathrm i+z)$ gives a $z^{-1}$ term. – joriki Mar 26 '12 at 16:20