Is $f$ constant if $e^f$ is constant?

Question

How would one go about showing that $f$ is constant if $e^f$ is constant by Liouville's theorem? The original problem asked to prove the entire function $f = u + iv$ was constant if the real part of the function was bounded, $|u|\leq M$ for all $z$.

I got as far as proving $e^f$ was a constant through the following \begin{equation} |e^f| = |e^{u+iv}| = |e^u| = e^{|u|} \leq e^M \end{equation} Now how would I continue to prove $f$ is constant? Can I just directly assume that $f$ is a constant? Something tells me the next step is so simple but I do not know what it could be.

Thank you for your time and Thanks in advanced for any feedback.

Answer 1

Hint: Since $\exp$ is a non-constant entire function, $\{z: \exp(z) = w\}$ is a discrete set for any $w$.

Answer 2

Its very simple: When $g(z):=e^{f(z)}$ is constant then $g'(z)=e^{f(z)} f'(z)\equiv0$, which implies $f'(z)\equiv0$; whence $f$ is constant.