Singularity type of $\frac{1}{z} e^{-\frac{1}{z^2}}$

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?

• It's essential, $z\cdot f(z)$ has no limt as $z \to 0$. [Take $z$ purely imaginary to see it blow up.] – Daniel Fischer Jul 31 '15 at 14:34
• Consider the function $f(x)=e^{-1/x^2}$ for $x>0$ and $f(x)=0$ for $x\leq 0$. This function is infinitely, continuously differentiable on $\mathbb R$ but $f^{(k)}(0)=0$ for all integers $k\geq 0$. The function $f$ does not have a power series representation at $0$. – SamM Jul 31 '15 at 14:36

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 $t$) $$\lim_{t\to0}(it)f(it)=\lim_{t\to0}e^{-1/(it)^2}=\lim_{t\to0}e^{1/t^2}=\infty$$