How can i compute the residue of:
$$f(z) = \frac{z-1}{1+\cos(\pi z)}$$
at $z_k=2k+1, k\neq 0$.
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@Did Good. So, I will delete the comments. Why not edit? – Pedro Feb 07 '13 at 19:42
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Set $w=z-(2k+1)$, then at $z=2k+1$, $w=0$. $$ \begin{align} \frac{w+(2k+1)-1}{1-\cos(\pi w)} &=(w+(2k+1)-1)\frac1{\frac{\pi^2}{2}w^2+O\left(w^4\right)}\\ &=(w+2k)\frac1{w^2}\left(\frac2{\pi^2}+O\left(w^2\right)\right)\\ &=\frac{4k}{\pi^2}\frac1{w^2}+\frac2{\pi^2}\frac1w+O(1) \end{align} $$ Thus, the residue at $w=0$, aka $z=2k+1$, is $\frac2{\pi^2}$.
robjohn
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Hint:
Points $z_k=2k+1$ are poles of order $2$ for $f(z) = \dfrac{z-1}{1+\cos(\pi z)}$ with residues
$$\operatorname{res}\limits_{z=z_k}{f(z)}=\lim\limits_{z\to{z_k}}\dfrac{d}{dz}{\left(\dfrac{(z-z_k)^2 (z-1)}{1+\cos{\pi z}}\right)} = \dfrac{2}{\pi^{2}}.$$
M. Strochyk
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yes i know this, and i tried to use a formula like $\lim_{z\rightarrow z_k}\frac{d}{dz}[(z-z_k)^2f(z)]$, but i think i may use series expansion: i get $f(z)=\frac{z-z_k}{(z-z_k)^2[\frac{\pi^2}{2}-\ldots]}+\frac{2k}{(z-z_k)^2[\frac{\pi^2}{2}-\ldots]}$. Can i conclude now that the residue is $\frac{2}{\pi^2}$? – Federica Maggioni Feb 07 '13 at 19:56
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@FedericaMaggioni: substitution makes the notation cleaner. I use the series expansion and big-O notation in my answer. – robjohn Feb 08 '13 at 19:03