Let $f$ be a holomorphic function such that $Im(f(z))$ is positive for all $z$. Prove that $f$ is constant.
Liouville's theorem states that if an entire function is bounded, then it must be constant. Let $f(z)=a(z)+ib(z)$ So I consider the function $e^{if(z)}=e^{ia-b}=e^{ia}e^{-b}$. Taking the absolute value $$|e^{if(z)}|=$$ $$|e^{ia}e^{-b}|=$$ $$|e^{ia}||e^{-b}|=$$ $$|e^{-b}|$$ $$\leq 1$$
From here one deduces $e^{if(z)}=z_0$, for $z_0 \in \mathbb C$.
I don't know how can I conclude from here that $f(z)$ must be constant. If all of these was taking place in $\mathbb R$, then $e^{f(x)}=k \implies f(x)=log(k)$. But in the complex case I have branches of logarithms and that confuses me a little bit.
Best Answer
The Moebius transformation $$T:\quad w\mapsto{w-i\over w+i}$$ maps the upper half plane onto the unit disk $D$. Therefore the function $$g(z):=T\bigl(f(z)\bigr)$$ is entire and bounded. By Liouville's theorem $g$ has to be constant, and so is $f$.