# Show that $0$ is an essential singularity of $e^{\frac{1}{z}}$

I am working on the following exercise:

Show that $$0$$ is an essential singularity of $$e^{\frac{1}{z}}$$.

I am new to Laurent Series, so I would be grateful if you could have a look at my attempt:

I would say that it is enough to note that (by definition) $$e^{\frac{1}{z}} = \sum_{n=0}^\infty \frac{1}{z^n n!},$$ which is already the Laurent Series of $$e^{\frac{1}{z}}$$. It is obvious that the main part of this Laurent Series does not vanish, so the singularity is essential. Is this enough or did I forget to check any preconditions?

• You showed that the Laurent series contains infinite many terms with $z^m$ with negative $m$ , which is enough to show that a singularity is essential. – Peter Mar 4 at 19:05
• Did you mean $e^{1/z}$? – José Carlos Santos Mar 4 at 19:06
• @JoséCarlosSantos and Ted Shifrin: Sorry, I made a typo, I meant $e^{1/z}$, sorry for the inconvenience – 3nondatur Mar 4 at 19:11
• @3nondatur There is still one $e^{-z}$ left. But, yes, what you did is enough. – José Carlos Santos Mar 4 at 19:12