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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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up vote 4 down vote accepted

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 $0$.

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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
Thanks for this! – Harry Macpherson Mar 27 '12 at 9:38
Lesson not to doubt oneself without good reason, and not to scratch out your answer because of it. (What kind of idiot fails an exam more than once because he did this and was too slow to write something else? I don't know >_> ) – Vandermonde Aug 17 '15 at 20:36

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