TITLE: If $f=u+iv$ is an entire function such that $u^2\geq v^2,$ then $f$ is constant
QUESTION [6 upvotes]: Let $f=u+iv$ be an entire function such that $u^2(z) \geq v^2(z), \forall z \in  \mathbb{C}.$ Could anyone advise me how to prove $f \equiv$ constant $?$ Hints will suffice. Thank you.

REPLY [3 votes]: $$
u^2-v^2=\mathrm{Re}\,f^2\ge 0,
$$
but the function $g=f^2+1$ also entire, and 
$$
\lvert g\rvert=\lvert\, f^2+1\rvert\ge\mathrm{Re}\,(\,f^2+1) \ge 1.
$$
Hence 
$$
h=\frac{1}{g},
$$
entire and bounded.