Simple proof for the result:$a$ is a removable singularity of $e^f$ iff $a$ is a removable singularity of $f$.

Solution 1:

If $e^f$ has a removable singularity at $a$ then $e^f = g$ where $g$ is holomorphic in $|z-a| < R$. In particular, $g$ is locally bounded, so that $$ |g(z)| < K \quad \text{for } |z - a| < R/2 \, . $$ for some real constant $K > 0$. It follows that for $ 0 < |z-a| < R/2$ $$ \operatorname{Re}f(z) = \log |g(z)| < \log K \implies \operatorname{Re} (\log K - f(z)) > 0 \, . $$ Now consider the function $h(z) = T(\log K - f(z))$ where $T(z) = \frac{z-1}{z+1}$ is the Möbius transformation which maps the right halfplane onto the unit disk.

• Riemann's theorem on removable singularities shows that $h$ has a removable singularity at $z=a$.
• The maximum modulus principle shows that the continuation function $\hat h$ satisfies $\hat h(a) \ne 1$.

Then $$ \hat f(z) = \log K - T^{-1}(\hat h(z)) = \log K - \frac{1+\hat h(z)}{1-\hat h(z)} $$ is an analytic continuation of $f$ at $z=a$, i.e. $f$ has a removable singularity at that point.