I need to find the singularity points of the following function: $$f(z)=\frac{z}{1-e^{z^2}}$$
and determine whether it is a removeable singularity/a pole/an essential singularity, and find the residue.
I know the point $0$ is a singularity, but I'm not sure how to find the function's Laurent series, or how to determine whether $\;\lim\limits_{z \to 0} \frac{z}{1-e^{z^2}}\;$ exists. I'm pretty new at the subject of singularity points and only did basic examples.
On
There is a lemma which says the following:
An analytic function, $f(z)$ has zeros of order $n$ at $z_{0}$ if $f(z_{0})=f'(z_{0})...=f^{n-1}(z_{0})=0$ and $f^{n}(z_{0}) \neq0$.
Simply put, $f(z_{0})=0$ and $n$ is the first positive integer such that $f^{n}(z_{0}) \neq0$.
In your question, we first let $p(z)=z$ and $q(z)=1-e^{z^{2}}$. The zeros are obtained when we set $$1-e^{z^{2}}=0.$$ Clearly, $z=0$ is a zero of order 1.
Now, we consider $p(z_{0}) =p(0)=z|_{z_{0}=0}=0$.
Finally, $p'(z_{0})=1 \neq 0$. By the lemma, $p(z)$ has a zero, $z=0$ of order 1.
Thus, we conclude that $f(z)$ has a zero, $z=0$ of order $1-1=0$.
That is, $z=0$ is a removable singularity.
The zeros of $e^z-1$ are of order $1$ at $z \in 2i \pi \mathbb{Z}$,
thus the poles of $\frac{1}{e^z-1}$ are of order $1$ at $z \in 2i \pi \mathbb{Z}$
the poles of $\frac{z}{e^z-1}$ are of order $1$ at $z \in 2i \pi n, n \ne 0$
the poles of $\frac{z^2}{1-e^{z^2}}$ are of order $1$ at $z^2 = 2i \pi n,n \ne 0$ so $z = i^m (1+i) \sqrt{ \pi n},n > 0$
and the poles of $\frac{z}{1-e^{z^2}}$ are of order $1$ at $z = i^m (1+i) \sqrt{ \pi n}, n \ge 0$
$$ $$
therefore the Cauchy integral formula shows $\frac{z}{1-e^{z^2}}$ has a Laurent series converging for $0 < |z| < \sqrt{2\pi}$, and another one different for each $n$ converging for $ \sqrt{2 n\pi} < |z| < \sqrt{2(n+1)\pi}$
the Taylor series of $\frac{z}{e^z-1}$ is given by the Bernoulli numbers and $\frac{z}{e^z-1}+\frac{z}{2}$ is even means $B_{2k+3} = 0$