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 $h=-\frac{2z^2}{4!}-\frac{2z^4}{6!}-\ldots$.
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, from which i desume that $z_0=0$ is a removable singularity for f.
But now i don't know how 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 last question: is it correct to say: the poles $z_k$ accumulates to $\infty$, hence $\infty$ is not an isolated singularity, thus i cannot compute $Res(f;\infty)$?
Best Answer
You can perform the same procedure. The Taylor series for cosine near $z = 2k\pi$ is the same as its series at $z = 0$ because of periodicity. So
$$ \cos z = \sum_{n=0}^{\infty} (-1)^n \frac{(z-2k\pi)^{2n}}{(2n)!}, $$
hence
$$ \begin{align*} \cosh z &= \sum_{n=0}^{\infty} (-1)^n \frac{(iz-2k\pi)^{2n}}{(2n)!} \\ &= \sum_{n=0}^{\infty} \frac{(z+i2k\pi)^{2n}}{(2n)!}. \end{align*} $$
Then rewrite the numerator of $f(z)$ as
$$ \begin{align*} z^3 &= (z+i2k\pi-i2k\pi)^3 \\ &= (z+i2k\pi)^3 - i6k\pi(z+i2k\pi)^2 - 12 k^2\pi^2 (z + i2k\pi) + i2k^3\pi^3. \end{align*} $$