$z_0$ 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 $f$.

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.
