# If $z=a$ is not the removable singularity of $f$, show that $e^{f(z)}$ has essential singularity at $z=a$.

Let $f$ be analytic on $0<|z-a|<R$ for some $R>0$. If $z=a$ is not removable singularity of $f$, show that $e^{f(z)}$ has an essential singularity at $z=a$.

Since it's not removable singularity, it is either a pole of $f$ or essential singularity of $f$. The case of $z=a$ is a pole I can prove it. For the case of essential singularity, I write $$f(z)=\sum\limits_{k=0}^\infty c_k(z-a)^k+\sum\limits_{j=1}^{\infty}d_j\frac{1}{(z-a)^j}.$$
Call first series $g$ and second series $h$, then $e^{f(z)}=e^{g(z)}e^{h(z)}$. It is obvious that $e^{g(z)}$ is analytic at $z=a$ and $e^{g(a)}\neq 0$.Thus, it can't write as the form $(z-a)g_1(z)$ where $g_1$ is analytic at $z=a$. Now, we focus on $e^{h(z)}$. $$e^{h(z)}=\sum\limits_{k=0}^\infty \frac{\sum\limits_{j=1}^\infty [d_j\frac{1}{(z-a)^{j}}]^k}{k!}$$ Then, in order to claim $z=a$ has essential singularity of $f$, I tried to write $e^{h(z)}$ as $$e^{h(z)}=\sum\limits_{s=0}^\infty e_s\frac{1}{(z-a)^s} (\star)$$ However, it seems that it shall be able to write as the form as $(\star)$, but do we need to worry about whether the series converges and whether we can change the double series? Or, is it possible to prove $z=a$ is essential singularity of $f$ avoiding to face this problem? Thanks.

If $\exp(f)$ has a pole at $a$ of order $m$, then by the argument principle: $$-m = \int_{|z-a|=\varepsilon}\frac{e^{f(z)}f'(z)}{e^{f(z)}}\,dz = \int_{|z-a|=\varepsilon}f'(z)\,dz = 0.$$ Contradiction.
• It is so cool! It seems that this proof also works for the case: $z=a$ has a pole of $f$. Am I right? Apr 5, 2017 at 15:58
Suppose $e^{f(z)}$ had a pole at $a$. This would mean that $|e^{f(z)}| = e^{\text{Re}\, f(z)}\ge M$ for all $z$ with $0<|z-a|<\delta$. Can this happen if $f(z)$ has an essential singularity at $a$? [Hint: Casorati-Weierstrass.]