If $f$ is entire and $|Re(f)||Im(f)|$ is bounded then $f$ is constant

Question

I want to prove the statement "If $f$ is entire and $|Re(f)||Im(f)|$ is bounded then $f$ is constant". I could not solve this but here are some of my ideas. If we write $f=u+iv$, the $f^2=(u^2-v^2)+i(2uv)$ which is also entire. Therefore, $f^2$ is an entire function with a bounded imaginary part, which I assume means that $f^2$ is constant. Hence, $f$ is constant. I know this is not a proof but this is what I have. Can anyone provide a proper proof? Thank you

Answer 1

If $f^2$ has bounded imaginary part, then $f^2$ maps $\mathbb{C}$ into a strip $S = \{x + i y : |y| \le M\}$ for some $M$. This strip is conformally equivalent to the disk by a map $g$, so the function

$$g\circ f^2 : \mathbb{C} \to \mathbb{D}$$

is bounded. By Liouville, it is constant. Since $g$ is a bijection, this implies $f^2$ is constant. It's a good exercise in Cauchy-Riemann equations to prove that $f^2$ constant implies $f$ is constant.