singularities and residues

• Nov 27th 2012, 11:35 AM
tenderline
singularities and residues
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 $Res(f;\infty)$?