If $f$ has a pole of order $N$ at $z=a$ then $\exp{f}$ has an essential singularity at $z=a$.

Question

I want to show that if $f$ has a pole of order $N$ at $z=a$ then $\exp{f}$ has an essential singularity at $z=a$.

My idea is to write $f(x)=\sum_{k=-N}^{\infty} a_k (z-a)^k$ so that $$\exp f(x) = \sum_{n\mathop=0}^{\infty} \frac 1 {n!}\left(\sum_{k =-N}^{\infty}a_k (z-a)^k\right)^n$$ where $N \ge 1$.

If I can show that there are infinitely many negative degree terms in the expansion then I am done. So I just need to show that there won't be some weird cancellation that could occur. I am not sure what the best way of going about this is though.

I realise there may be better ways of proving the statement but I want to see if this can be made to work.

Answer 1

It's better to observe this: since $f$ has a pole at $a$, then $\lim_{z\to a}f(z)=\infty$ and therefore, for each $w\in\mathbb C$, you can find a sequence $(z_n)_{n\in\mathbb N}$ of complex numbers such that $\lim_{n\to\infty}z_n=\infty$ and that $\lim_{n\in\mathbb N}e^{f(z_n)}=w$. Therefore, $e^f$ has an essential singularity at $1$.

Answer 2

If $f$ has a pole at $a$, then for each $\newcommand{\ep}{\varepsilon}\ep>0$, the image $f(B'_\ep(a))$ of the deleted $\ep$-neighbourhood of $a$ contains a set of the form $\{z:|z|> R\}$. So the image $\exp(f(B_{\ep(a)}))$ contains the set $\{\exp(z):|z|>R\}$. This contains sequences tending to zero and to $\infty$, so $f(z)$ cannot tend to a finite limit, nor to $\infty$, as $z\to a$, so has neither a removable singularity nor a pole.