I am studying complex analysis, and have ran into some trouble. I saw this sample problem:
Let $f(z)=u(x,y)+iv(x,y)$ an entire function such that $u(x,y)\leq K$ for all points in the complex plane. Show that $u(x,y)$ where $a$ is a constant function.
As I saw, the solution goes as follows:
We take the function $h(z)=e^{f(z)}$. That function is entire, as it is made of two entire functions. Also, $|h(z)|=|e^{f(z)}|=|e^{u(x,y+iv(x,y)}|=e^{u(x,y)}\leq e^K$
So far I understand what is going on, since all I did was exponentiate the function and use some properties. However, now comes the tricky part for me:
We have to apply Liouville's Theorem, obtaining that $h(z)=e^{f(z)}$ is a constant function. Then, $|h(z)|=e^{u(x,y)}=K $, and thus $u(x,y)=constant$
But I don't understand that last part. I know that Liouvilles theorem is: $$|f'(z)|\leq \frac{M}{r}$$
But I don't know how it can be applied here to show that $u(x,y)$ is constant.