Find and classify isolated singularities of $f(z) = \frac{z}{1-e^{z^2}}$ and calculate residues on them

Question

Obviously, all isolated singularities will be of the form $z=\sqrt{2 \pi i k}$ for $k \in \mathbb{Z}$ but I don't know how to classify. I tried expanding $\frac{1}{f} = \frac{1-e^{z^2}}{z}$ to $- \sum_{n=1}^{\infty} \frac{z^{2n -1}}{n!}$ which is of the form $zg(z)$ where $g$ is analytic near $0$ and $g(0) \neq 0$ so I can say $0$ is a simple zero of $\frac{1-e^{z^2}}{z}$ hence a simple pole of $\frac{z}{1-e^{z^2}}$. But how do I generalize?

Answer 1

For zero:

$$f(z)=\frac z{-z^2-\frac{z^4}{2!}+-\ldots}=-\frac1{z+\frac{z^3}2+\ldots}=-\frac1{z\left(1+\frac{z^2}2+\ldots\right)}=$$

$$=-\frac1z\left(1-\left(1+\frac{z^2}2+\ldots\right)+\left(1+\frac{z^2}2+\ldots\right)^2\right)-\ldots\implies z=0$$

is at most a simple pole, but since $\;\lim\limits_{z\to0}f(z)\;$ isn't finite then it is a simple pole, and it's residue is

$$\lim_{z\to0}z(f(z))\stackrel{\text{l'Hospital}}=\lim_{z\to0}\frac{2z}{-2ze^{z^2}}=-1$$

at $\;z=z_k:=\sqrt{2k\pi i}\;$: choosing the branch of the square root for which $\;\sqrt1=1\;$ , we get:

$$\lim_{z\to z_k}(z-z_k)f(z)\stackrel{\text{l'H}}=\lim_{z\to z_k}\frac{z+z-z_k}{-2ze^{z^2}}=\frac{z_k}{-2z_ke^{z_k^2}}=-\frac12$$