$z_0$ non-removable singularity of $f\Rightarrow z_0$ essential singularity of $\exp(f)$ 

Let $z_0$ be a non-removable isolated singularity of $f$. Show that $z_0$ is then an essential singularity of $\exp(f)$.


Hello, unfortunately I do not know how to proof that. To my opinion one has to consider two cases:


*

*$z_0$ is a pole of order $k$ of $f$.

*$z_0$ is an essential singularity of $f$.
 A: We can also look at it from the other side.
If $z_0$ is a removable singularity of $e^f$, then $\lvert e^{f(z)}\rvert < K$ in some punctured neighbourhood of $z_0$. Since $\lvert e^w\rvert = e^{\operatorname{Re} w}$, that means $\operatorname{Re} f(z) < K'\; (= \log K)$ in a punctured neighbourhood $\dot{D}_\varepsilon(z_0)$ of $z_0$, and that implies that $z_0$ is a removable singularity of $f$. (Were it a pole, $f(\dot{D}_\varepsilon(z_0))$ would contain the complement of some disk $D_r(0)$; were it an essential singularity, each $f(\dot{D}_\varepsilon(z_0))$ would be dense in $\mathbb{C}$ by Casorati-Weierstraß; in both cases $\operatorname{Re} f(z)$ is unbounded on $\dot{D}_\varepsilon(z_0)$.)
If $z_0$ were a pole of $e^f$, it would be a removable singularity of $e^{-f}$, hence $z_0$ would be a removable singularity of $-f$ by the above, hence $z_0$ would be a removable singularity of $f$, and therefore a removable singularity of $e^f$ - contradiction.
A: The above proof is very nice, here I provide another idea. The real question is that if $f$ has a pole, then $e^f$ has an essential singularity.
Assume that the conclusion is false, then there exist an integer $k$ and an analytic function $g$ in some disc $|z|<\delta$, for example, $g(0)\neq0$, such that $e^f=z^kg$,
$$f'=\frac{(e^f)'}{e^f}=\frac{k}{z}+\frac{g'}{g}$$
integrating it on the contour $|z|=\frac{\delta}{2}$ we find $k=0$, since $g$ is analytic. Using the fact that $g(0)\neq0$ we can find another analytic function $h$ in the disc $|z|<\delta$(actually we should shrink this disc), such that $e^f=g=e^h$, so $f=h+2n\pi i$, $f$ has a removable singularity, contradiction!
A: Under aforementioned conditions the function $f(z)$ may be expanded about $z = z_0$ 
up to a certain radius (radius of convergence) namely
\begin{equation*}
f(z) = \sum_{k= -\infty}^{+ \infty} a_k (z-z_0)^k = 
\sum_{k= 1}^{+\infty} \frac{a_{-k}}{ (z-z_0)^k} + \sum_{k= 0}^{+ \infty} a_k (z-z_0)^k = f_0(z) + f_1(z);
\end{equation*}
here we note the principal part $f_0(z)$ and the non-principal holomorphic part $f_1(z)$.
Because the isolated singularity of $f(z)$ at $z=z_0$ is not removable, it means that we have either a pole of order $n \in \mathbb{N}$ or an essential singularity.
In case of a pole at least $a_{-n} \ne 0$, in case of an essential singularity infinitely many coefficients $a_{-k},\ k \in \mathbb{N}$ are non-zero - 
in both cases the principal part $f_0(z)$ is not trivially equivalent to zero.
We may inspect $g(z) = e^{f(z)}$ and its expression about $z=z_0$ namely
\begin{equation*}
g(z) = e^{f(z)} %= e^{ f_0(z) + f_1(z)} = e^{f_0(z)} \cdot e^{f_1(z)} = 
= \sum_{j=0}^{+\infty} \frac{(f_0(z))^j}{j!}  \cdot e^{f_1(z)} = 
\sum_{j=0}^{+\infty} \frac{1}{j!} \left( \sum_{k= 1}^{+\infty} \frac{a_{-k}}{ (z-z_0)^k} \right)^j  \cdot e^{f_1(z)} = g_0(z) \cdot g_1(z);
\end{equation*}
here we note the non-trivial multiplicand $g_0(z)$ and the leftover $g_1(z) = e^{f_1(z)}$ 
is a holomorphic non-vanishing function in a neighborhood of $z=z_0$. 
However it is clear that the principal part of $g_0(z)$ exhibits an infinitude of summands in the obtained expression - 
it is equivalent then to $g(z)$ having an essential singularity at $z=z_0$.
