singularities and residues

**tenderline** · November 27th 2012, 11:35 AM

Consider the function $f(z)=\frac{z^3}{1-cosh(z)}$ . Find its singularities and compute residues.

I know the denominator vanishes for $z_k=2k\pi i, k$ integer. I first consider $k=0$ , so the function is analytic in $ $0<|z|<2\pi$ , and i can write in this punctured disc the following Laurent expansion:

starting from $cos(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n)!}$ i get $cosh(z)=cos(iz)=\sum_{n=0}^{\infty}\frac{z^{2n}}{( 2n)!}$ Hence $1-cosh(z)=\frac{z^2}{2!}-\frac{z^4}{4!}\ldots$
thus i can write $\frac{1}{1-\cosh(z)}=\frac{1}{-\frac{z^2}{2!}-\ldots}=-\frac{2}{z^2(1-h)}=-\frac{2}{z^2}(1+h+h^2\ldots)$ where $h=-\frac{2z^2}{2!}-\frac{2z^4}{4!}-\ldots$ .
So we have $\frac{1}{1-cosh(z)}=-\frac{2}{z^2}+\frac{4}{4!}+$ higther terms.
Finally, we get $\frac{z^3}{1-cosh(z)}=-2z+\frac{4z^3}{4!}$ + higther terms, fromw which i desume that $z_0=0$ is a removable singularity for f.

But now i don't knoe hoe to deal with $z_k$ with $k\neq 0$ . I imagine those to be all poles of order 2 for $f$ , but how to prove?

A lats question: is it correct to say: the poles $z_k$ accumulates to $\infty$ , hence $\infty$ is not an isolated singularity, thus i cannot compute $Res(f;\infty)$ ?

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