I've been asked to compute the singularity type of $f(z) := \frac{1}{z}e^{-\frac{1}{z^2}} $.
Here's my reasoning:
$$ \frac{e^{-\frac{1}{z^2}}}{z} = z^{-1} \sum_{n=0}^\infty \big( -z^{-2} \big)^n \cdot \frac{1}{n!} = \sum_{n=0}^\infty (-1)^n \frac{z^{-2n-1}}{n!} = \sum_{n=0}^{-\infty} \frac{(-1)^{-n}}{(-n)!}z^{2n-1} = \sum_{n=-\infty}^\infty a_n z^n$$
with
$$ a_n := \begin{cases}
0 & \text{if } n\ge 0 \text{ or } (n<0 \text{ and } |n| \text{ is even})\\
\frac{-1}{(-n)!} & \text{if } n < 0 \text{ and } |n| \text{ is odd}\\
\end{cases}$$
so, since the Laurent series of the function has infinitely many $a_n$ with negative index different from $0$ I would claim that the function has an essential singularity in $0$. Still the singularity should be removable as
$$ \lim_{z \to 0} ~~z \cdot f(z) = 0$$
Where am I mistaken?
Best Answer
Since $f$ is analytic on $\mathbb{C}\setminus\{0\}$, the fact that $\lim_{z\to0}zf(z)=0$ would imply $g(z)=zf(z)=e^{-1/z^2}$ has a removable singularity at $0$.
But it is well known that the Taylor series at $0$ for the real function $$ h(x)=\begin{cases} 0 & \text{if $x=0$}\\ e^{-1/x^2} & \text{if $x\ne0$} \end{cases} $$ doesn't converge to $h$ except at $0$ (all the derivatives at zero are $0$).
In particular $0$ also cannot be a pole for $f$, so it's an essential singularity.
More simply (with real $t$) $$ \lim_{t\to0}(it)f(it)=\lim_{t\to0}e^{-1/(it)^2}=\lim_{t\to0}e^{1/t^2}=\infty $$