Show that an entire function $f$ is constant

Question

Let $f$ be a entire function such that $|\Re(f(z))| >M$ for some real number $M$. Show that $f$ is constant.

What I have to do is to show $f$ is bounded. hence it must be constant. But on which function should I apply Liouville theorem ?

Answer 1

If $\Re(z)>0$, then $$ \left\lvert \frac{1}{1+z} \right\rvert^2 = \frac{1}{(z+1)(\bar{z}+1)} = \frac{1}{z\bar{z}+2\Re(z)+1}<1, $$ since the other two terms in the denominator are positive, so $1/(1+(f(z)-M))$ will work.

Edit: Another possibility, depending on what you have to work with, is $e^{-f(z)}$, since $$ \lvert e^{-f(z)} \rvert^2 = e^{-(f(z)+\overline{f(z)})} = e^{-2\Re(f(z))}, $$ which is then bounded above by $e^{-2M}$.

Answer 2

Hint:)

$|f(z|\geq|{\bf Re}\, f(z)|>M\geq0$, define $g(z)=\dfrac{1}{f(z)}$ which is entire and bounded.