Analytic complex function in a half plane

Question

I have a holomorphic function $f$ in the whole $\mathbb{C}$ and $f(\mathbb{C})$ in a half plane. I have to prove that $f$ is constant. I read about a conformal map, but I am at the beginning of my studies, so my "instruments" are very simple.

Answer 1

Let's say it's the upper half-plane. Notice that $\frac1{2i+f}$ is holomorphic with image contained inside the unit circle, allowing you to use Liouville.

Answer 2

I will assume without loss of generality that $f(\mathbb{C})\subset \mathbb{H}$ (the upper-half plane). The Cayley transform, $g(z)=\frac{z-i}{z+i}$ is a very useful map that bijectively maps $\mathbb{H}$ to $\mathbb{D}$ the unit disc in the complex plane (you should verify this if you haven't seen this map before. Observe that $g\circ f$ is a holomorphic map that maps $\mathbb{C}$ into $\mathbb{D}$ and is therefore bounded. It follows from Liouville's theorem that $g\circ f(z)=c$ for all $z$ and some constant $c$. Therefore, because $g$ is bijective, we see that $f(z)$ is constant.