Complex - Entire functions

How can I prove this. I could not use $\Im(w)<0$ condition in Liouville's theorem.

Let $f(z)$ be an entire function and assuming that $f(z)$ does not take values in $\Im(w)<0$ show that $f$ is identically zero.

Thanks.

• You have to assume non-constant. – Davide Giraudo Jun 4 '12 at 16:42
• Consider $g\circ f$ where $g$ is a suitable fractional linear transformation. – user31373 Jun 4 '12 at 16:45
• What can you say about $g(z) = \frac{f(z)}{f(z)+i}$. – copper.hat Jun 4 '12 at 16:51
• How g can be used in thm ? – Brhn Jun 4 '12 at 17:00
• $g$ is bounded and entire. What can you say about such functions? And what does that say about $f$? – copper.hat Jun 4 '12 at 17:06

Let $g(z)=e^{if(z)}$. Then $g$ is entire and $$|g(z)|=|\exp((\Re f(z)+i\Im f(z))i)| =|\exp\left(i\Re f(z)-\Im f(z)\right)|=e^{-\Im f(z)}\leq 1.$$ By Liouville theorem, $g$ is constant hence $e^{if(z)}=C$ and $f'(z)e^{if(z)}=0$ so $f$ is constant (but not necessarily $0$).
Suppose $f$ is non constant. Then $\mathcal{Im}(f)$ is open so we can assume $\mathcal{Im}(f) \subset I(w) > 0$
$\varphi :z \mapsto \frac{z - i }{z + i}$ is a bijection from $\mathbb{D}$ to $I(w) > 0$
So $\varphi^{-1} \circ f$ is an entire function which is bounded so by Liouville it is constant and then $f$ is constant.