Prove that real f(3)>3

Question

Let $f$ be a conformal mapping of the domain $$ \Omega = \{z:\Re(z)>0\}-(0,1] $$ onto the domain $\{z:\Re(z)>0\}$ so that $f(2)=2$, $f^\prime (2)>0$. Prove that we have real $f(3)>3$.

I know one special case $f= \frac{2}{\sqrt3} \sqrt{z^2-1} $. But what's the general proof? (Maybe using schwarz lemma?)

Answer 1

There is a unique conformal mapping between the two simply connected region satisfying $f(2)=2$ and $f'(2)>0$. Thus, the one you found is the only one, with these properties - see the formulation of Riemann Mapping Theorem.