TITLE: If $f$ is entire and $e^f$ is constant $f$ is constant
QUESTION [6 upvotes]: What I am trying to prove is that if $f$ is entire and $e^f$ is constant $f$ is constant. This is my attempt but I just can't go forward to get anything meaningful. $$e^{f(z)}=c$$ where $c\in \mathbb{C}$ is a constant. Then the set of solutions for $f(z)$ is {$\log|c|+i(Arg(c)+2k\pi)|k\in \mathbb{Z}$}. But then how do I show that $f$ is constant here after. Any help will be appreciated. Thanks

REPLY [3 votes]: By the chain rule, we can take the derivative of both sides to get
$$e^{f(z)} = c$$
$$f'(z)e^{f(z)}=0$$
$$f'(z)\cdot c = 0$$
$$f'(z)=0$$
This implies $f(z)$ is constant.