Application of Liouville's theorem exercise 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.
 A: 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 $f$.
A: If $\Im(f(z)) \ge 0$, then 
$$
     1 = \Im(i) \le \Im(i)+\Im(f(z))=\Im(i+f(z))\le |\Im(i+f(z))|\le |i+f(z)|.
$$
So $g(z)=1/(f(z)+i)$ is holomorphic because $i+f(z)$ is never $0$, and
$$
       1\le |i+f(z)| \implies \frac{1}{|i+f(z)|} \le 1\implies |g(z)| \le 1.
$$
That makes $g$ constant. So $g(z)=C$ for all $z$ and some $C\ne 0$. Solving for $f$ gives
$$
         f(z) = \frac{1}{C}-i.
$$
A: Pick some $w_0$ so that $e^{i w_0} =z_0$.
Then for all $z \in \mathbb C$ you have
$$e^{i f(z)}=e^{i w_0} \Rightarrow f(z)=w_0+2k_z \pi  \,;\, k_z \in \mathbb Z\,.$$
Now use continuity of $f$ to deduce that $k_z$ is constant. 
Here is a big hint for this: for each $k \in \mathbb Z$ you can easily prove that the set 
$$f^{-1}(w_0+2k \pi)$$
is open and closed.
