If $f=u+iv$ is an entire function such that $u^2\geq v^2,$ then $f$ is constant

Question

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.

Answer 1

let $g(z)=f(z)-2i$, we have $|g(z)|\geq 1$ for all $z\in \Bbb C$, now let $h(z)=\dfrac{1}{g(z)}$, $h$ is entire and bounded hence it is constant.

(the inequality): $|g(z)|^2=u^2(z)+v^2(z)-4v(z)+4\geq 2v^2(z)-4v(z)+4=2(v(z)-1)^2+2\geq 1$

Answer 2

There did not exists $z_0 ,z_1\in\mathbb{C}$ such that $f(z_0) =i$ and $f(z_1 )=2i$ and hence by Little Picard theorem $f$ must be constant.

Answer 3

Observe that $f^2 = (u+iv)(u + iv) = (u^2 - v^2) + i(2uv)$. Since $f^2$ is entire, $u^2 - v^2$ is harmonic. Thus $u^2 - v^2$ is a nonnegative harmonic function.

Answer 4

Express $f$ in polar coordinates, $f(z)=re^{iϕ}$ Then the inequality implies that ϕ is restricted to the interval $[0,π/4]$.

Answer 5

$$ u^2-v^2=\mathrm{Re}\,f^2\ge 0, $$ 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.