If $e^f$ is bounded. Prove $f$ is constant.

Question

$f$ is an entire function.

If $ e^f $ is bounded, prove $f$ is constant.

I am stuck as I am not sure how to approach this, although I know Liouville's Theorem must be used and I must show f is bounded.

Answer 1

Since $e^f$ is bounded and entire, it is constant. So, for some $w\in\mathbb C\setminus\{0\}$, you have $(\forall z\in\mathbb C):e^{f(z)}=w$. Now, let $\omega$ be a logarithm of $w$; then, for each $z\in\mathbb C$, $f(z)=\omega+2\pi in$, for some integer $n$. But $f$ is continuous and $\mathbb C$ is connected. So, $f(\mathbb C)$ is connected and therefore you must have the same $n$ for every $z$. In particular, $f$ is constant.

Answer 2

Since $f$ is entire, as a composition map $e^f$ is entire. Since $e^f$ is bounded, by Liouville's Theorem $e^f$ is constant. But this is possible only if $f$ is constant.