Consider the function $\displaystyle f(z)=\frac{z^3}{1-cosh(z)}$. Find its singularities and compute residues.
I know the denominator vanishes for $\displaystyle z_k=2k\pi i, k$ integer. I first consider $\displaystyle k=0$, so the function is analytic in \displaystyle 0<|z|<2\pi$, and i can write in this punctured disc the following Laurent expansion: starting from$\displaystyle cos(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n)!}$i get$\displaystyle cosh(z)=cos(iz)=\sum_{n=0}^{\infty}\frac{z^{2n}}{( 2n)!}$Hence$\displaystyle 1-cosh(z)=\frac{z^2}{2!}-\frac{z^4}{4!}\ldots$thus i can write$\displaystyle \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)$where$\displaystyle h=-\frac{2z^2}{2!}-\frac{2z^4}{4!}-\ldots$. So we have$\displaystyle \frac{1}{1-cosh(z)}=-\frac{2}{z^2}+\frac{4}{4!}+$higther terms. Finally, we get$\displaystyle \frac{z^3}{1-cosh(z)}=-2z+\frac{4z^3}{4!}$+ higther terms, fromw which i desume that$\displaystyle z_0=0$is a removable singularity for f. But now i don't knoe hoe to deal with$\displaystyle z_k$with$\displaystyle k\neq 0$. I imagine those to be all poles of order 2 for$\displaystyle f$, but how to prove? A lats question: is it correct to say: the poles$\displaystyle z_k$accumulates to$\displaystyle \infty$, hence$\displaystyle \infty$is not an isolated singularity, thus i cannot compute$\displaystyle Res(f;\infty)\$?