If $f$ is entire and $\exp(f(z))$ is a polynomial, then $f$ is constant.

Question

In a recent question that was just deleted, @danielfischer gave at the end of his answer the following exercise: for entire $f$, $$e^{f(z)} \text{ is a polynomial} \iff f \text{ is constant}$$

I was thinking about how to prove this...my first idea was to use that $\text{Re}(f)$ is majored by some $\log$ function as $z\to \infty$. Any ideas?

Answer 1

Hint: Consider the type of the singularity of $e^{f(z)}$ in $\infty$.

Different hint: Does $e^{f(z)}$ have any zeros?

The first hint generalises to show that $e^{h(z)}$ for analytic $h$ never can have a pole in an isolated singularity of $h$.

Answer 2

It can be solved using more elementary tools.

If $\mathrm{e}^{f(z)}$ is a non-constant polynomial, then it has a root (due to the Fundamental Theorem of Algebra), which contradicts the fact that the exponential does not vanish.