# Singularity Point Of $f(z)=e^{\frac{1}{z}}$ at $z_0=0$

When looking at the laurent series of $f(z)=e^{\frac{1}{z}}$ we get:

$$\sum_{n=0}^{\infty}\frac{1}{n!z^n}$$ or $$\sum_{n=0}^{\infty}\frac{z^{-n}}{n!}$$

The principal part is infinite, so it is essential singularity.

Is there a way to characterize the essential singularity with a limit?

the $\lim_{z\to z_0}|f(z)|=L$ is removable singularity and $\lim_{z\to z_0}|f(z)|=+ \infty$ is a pole, what about the essential singularity?

• $$\not\exists n\in\mathbb N\text{s.t.}\lim_{z\to z_0}(z-z_0)^nf(z)=0$$ – Simply Beautiful Art Jun 30 '17 at 13:12
• According to Picard, the limit does not exist at all (even allowing $\infty$) – Hagen von Eitzen Jun 30 '17 at 13:14
• @HagenvonEitzen What is that supposed to mean? – Simply Beautiful Art Jun 30 '17 at 13:15

A characterization of essential singularities through limits is this: an analytic function as an essential singularity at $z_0$ if and only if, for every $w\in\mathbb C$, there is a sequence $(z_n)_{n\in\mathbb N}$ such that $\lim_{n\in\mathbb N}z_n=z_0$ and that $\lim_{n\in\mathbb N}f(z_n)=w$. This is basically the Casorati-Weierstrass theorem.