Exercise: If $f$ has a pole on $z_0$ then $e^f$ has an infinity order pole on $z_0$
(when I mean an infinity order pole I mean that the Laurent series of $e^f$ has infinite many terms $a_k$ where $k<0$ equivalence the limit of $e^f$ at $z_0$ doesn't exist, I am sorry I don't know the definition in English)
My attempt: I tried to show that indeed the $\underset{z\rightarrow z_0}{\lim}e^{f(z)} $ does not exist.
Since $f$ has a pole on $z_0$ I assume its order is $k \in \mathbb{N}$ and i get $$f(z)=\sum_{n=0}^{+\infty}a_n(z-z_0)^n + \sum_{n=0}^{k}a_{-n}\frac{1 }{(z-z_0)^{n} }$$
then
$$e^{f(z)}=e^{\sum_{n=0}^{+\infty}a_n(z-z_0)^n + \sum_{n=0}^{k}a_{-n}\frac{1 }{(z-z_0)^{n} }}$$
$$\underset{z\rightarrow z_0}{\lim}e^{f(z)}=\underset{z\rightarrow z_0}{\lim}e^{\sum_{n=0}^{+\infty}a_n(z-z_0)^n + \sum_{n=0}^{k}a_{-n}\frac{1 }{(z-z_0)^{n} }}$$
since $\underset{z\rightarrow z_0}{\lim}\sum_{n=0}^{+\infty}a_n(z-z_0)^n\rightarrow 0$ and $\underset{z\rightarrow z_0}{\lim}\sum_{n=0}^{k}a_{-n}\frac{1 }{(z-z_0)^{n} }\rightarrow \infty $
$$\underset{z\rightarrow z_0}{\lim}e^{f(z)}=e^{\infty}$$ and we know $\underset{z\rightarrow \infty}{\lim}e^{z}$ doesn't exist.
Is my attempt correct ?
Best Answer
I think you mean the essential singularity. To prove that $z_0$ is the essential singularity of $f(z)$, you should prove that neither $\lim_{z\rightarrow z_0}f(z)$ nor $\lim_{z\rightarrow z_0}\frac{1}{f(z)}$ exits. In this problem, $\lim_{z\rightarrow z_0^{+}}e^{f(z)}=\infty$ and $\lim_{z\rightarrow z_0^{-}}e^{f(z)}$ doesn't exit, so both $\lim_{z\rightarrow z_0}e^{f(z)}$ and $\lim_{z\rightarrow z_0}\frac{1}{e^{f(z)}}$ don't exit, so $z_0$ is the essential singularity of $e^{f(z)}$