Entire function such that $\text{Im}f = \text{Re} ^4 f + 1$ is constant

Question

So I came across this problem:

Prove that if $f$ is entire, $u=\text{Re} f$, $v=\text{Im} f$ and $v(z)=u^4(z)+1$ $\forall z\in \mathbb{C}$, then $f$ is constant.

In this particular case, $v(z)\gt 0 \ \forall z\in \mathbb{C}$, so applying Liouville's theorem to $g=\exp(if)$ does the job, but I was wondering if there are any solutions that exploit the specific equality between $u$ and $v$ instead of just the positivity of $v$.

Answer 1

Note that Both Liouville and open mapping are overkills for this problem.

By Cauchy--Riemann $$u_x=v_y \\ u_y =-v_x$$ Using $v=u^4+1$ you get $$u_x=4u^3u_y \\ u_y=-4u^3u_x$$

Thus $$ u_x=-16u_x u^6 \\ u_y=-16u^6 u_y $$

This gives that for all $t \in C$ you either have $u(t)=0$ or $(u_x(t)=0, u_y(t)=0)$.

A simple continuity argument solves the problem:

The set $$M:=\{ t \in \mathbb C : u(t) =0 \}$$ is closed, and hence $N:= \mathbb C \backslash M$ is open. On $N$ you have $u_x=u_y=0$, and hence $u$ is constant on $N$.

Therefore, $u$ is constant on $M$ and constant on its complement, thus by continuity it is constant.

Answer 2

Every non-constant entire function is an open mapping. But, if $\operatorname{Im}(f)=\operatorname{Re}^4(f)+1$, $f(\mathbb C)$ is a subset of $\{x+yi\in\mathbb C\,|\,y=x^4+1\}$, whose interior is empty. Therefore, $f$ is constant.