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.
what are the residues at poles of $\frac{1}{1+\cosh{z}}$?
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residue-calculus
1 Answers
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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 $0$.
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0Lesson 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 >_> ) – 2015-08-17