An entire function whose imaginary part is bounded is constant

Question

I need to prove the following question

Question

Let $g(z)$ be an entire function such that there exists an $\alpha > 0$ such that $\left|\operatorname{Im}(g(z))\right| \le \alpha$. Prove that $g(z)$ is a constant function.

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My first thought was to use Liouville's theorem. As that states, for an entire function $g$, if $g$ is bounded, then $g$ is constant.

So if we could prove that $g$ is bounded, then by Liouville's theorem, $g$ is constant.

I have attempted to prove that $g$ is bounded, but I am struggling. Some help would be much appreciated.

Answer 1

Hint: compose it with the function $z\mapsto e^{iz}$

Answer 2

This is little picard theorem: https://en.wikipedia.org/wiki/Picard_theorem

Answer 3

The Little Picard Theorem states that if $f$ is an entire, non-constant function from $\mathbb C$ to $\mathbb C$, then $f$ takes on either every value in $\mathbb C$ or every value but one.

There is more than one value that your function doesn’t take on, so cannot be both entire and non-constant.

The “or every value but one” is necessary here because $e^z$ is an entire, non-constant function that is never $0$.