Find and classify the singularities of $f(z) = \frac{1}{e^{z^2}-1}$

I am trying to find and classify the singularities of $f(z) = \frac{1}{e^{z^2}-1}$. I'd also like to find the residue for any poles. So far, I have found that the singularities are of the form $z=\sqrt{2\pi k i}$ for $k \in \mathbb{Z}$.

I believe that when $z = 0$, I can write $$\frac{1}{e^{z^2}-1} = \left(z^2\left(1 + \frac{z^2}{2!} + \frac{z^4}{3!} + ...\right)\right)^{-1}$$ so then there is a double pole at $z=0$ and the residue there is $$\frac{d}{dz}\left(\left(1 + \frac{z^2}{2!} + \frac{z^4}{3!} + ...\right)^{-1}\right)$$ evaluated at 0, which is 0. Does that seem correct?

But, when $z=\sqrt{2 \pi k i}$ for $k \neq 0$, I think I have a removable singularity, but I'm not sure how to show this.

• As $\frac1{f(z)}\to 0$ at these points, the singularity is certainly not removable Commented Jul 12, 2018 at 2:00

Here we have three cases: $z=0$, $e^{z^2}=1$ and singularity at infinity as $z=\infty$. The series \begin{align} f(z) &=\dfrac{1}{e^{z^2}-1}\\ &=\dfrac{1}{z^2+\frac{1}{2!}z^4+\frac{1}{3!}z^6+\cdots}\\ &=\dfrac{1}{z^2}\cdot\dfrac{1}{1+\frac{1}{2!}z^2+\frac{1}{3!}z^4+\cdots}\\ &=\dfrac{1}{z^2}\left(1-\frac{1}{2!}z^2+\frac{z^4}{12}+\cdots\right)\\ &=\dfrac{1}{z^2}-\frac{1}{2}+\frac{z^2}{12}+\cdots \end{align} shows that $z=0$ is a pole of order $2$ with residue $0$, as you found. Also the series $$f(\dfrac1z)=z^2-\dfrac12+\dfrac{1}{12z^2}+\cdots$$ shows that the infinity is an essential singularity, and in fact there is no $k$ such that $\displaystyle\lim_{z\to0}z^kf(\dfrac1z)<\infty$.
As Hagen von Eitzen points out in the comments, the singularity at $z=\sqrt{2\pi i k}$ for $k\neq 0$ cannot be removable. Inspired by this answer, I've done the following:
Let $g(z) = e^{z^2}$, and $z_0 = \sqrt{2\pi k i}$, $k \neq 0$. Then note $$2(2 \pi k i)^{1/2} = g'(z_0) = \lim_{z \to z_0} \frac{e^{z^2}-e^{z_0^2}}{z-z_0} = \lim_{z \to z_0}\frac{e^{z^2} - 1}{z-\sqrt{2 \pi k i}}$$ Hence $$\lim_{z \to \sqrt{2\pi k i}}(z-\sqrt{2 \pi k i}) \frac{1}{e^{z^2}-1} = \frac{1}{2(2\pi k i)^{1/2}}$$
Thus at $z=\sqrt{2\pi k i}$, $k \neq 0$, we have a simple pole with the residue given in the last line.
• The power in your answer is $\dfrac12$ rather than $\dfrac32$, I think. Commented Jul 12, 2018 at 10:04