Proving $f$ is holomorphic or antiholomorphic if $e^f$ and $f$ are harmonic Suppose we have a harmonic function $f$ such that $e^f$ is also harmonic. I want to prove $f$ is either holomorphic or antiholomorphic. 
I tried using the fact that $\Delta u= 4\partial_z(\partial_{\overline{z}} u)=0$ but I didn't get anywhere with that and I assumed that does not imply $\partial_{\overline{z}}=0$ or $\partial_z =0$. 
I actually have the same problem but with $f^2$ instead of $e^f$ and I suppose it is similar, so I hope I can do one knowing how to do the other one.
 A: Let $D \subset \Bbb C$ be a domain and $f: D \to \Bbb C$ a function
with continuous derivatives up to order $2$.
$\exp$ is holomorphic, i.e. $\partial_{\overline{z}} e^z = 0$,
therefore the chain rule for Wirtinger derivatives
gives
$$
 \partial_z (e^f) = e^f \partial_z f \, ,\\
 \partial_{\overline{z}} (e^f) = e^f \partial_{\overline{z}}f \, .
$$
It follows that
$$
\Delta e^f= 4\partial_z(\partial_{\overline{z}} e^f)
= 4\partial_z( e^f \partial_{\overline{z}} f)
= 4 \left(e^f (\partial_z f)(\partial_{\overline{z}} f) + e^f \partial_z(\partial_{\overline{z}} f \right)
= 4 e^f (\partial_z f)(\partial_{\overline{z}} f) + e^f \Delta f \, .
$$
In particular, if $f$ and $e^f$ are harmonic then
$$
(\partial_z f(z))(\partial_{\overline{z}} f(z)) = 0
$$
for all $z \in D$. Both $\partial_z f(z)$ and $\partial_{\overline{z}} f$
are harmonic in $D$, therefore one of them must be
identically zero in $D$ (see for example product of harmonic functions), which means that $f$ is anti-holomorphic or holomorphic.
The same approach works if $f$ and $f^2$ are harmonic, or even
for $f$ and $\phi \circ f$, if $\phi$ is holomorphic in $\Bbb C$ and not linear.
