From an old complex analysis prelim:
Suppose that $\Omega\not= \mathbb{C}$ is a simply connected region, $a$ and $b$ are distinct points on $\Omega$, $f$ is a conformal map of $\Omega$ on $\mathbb{D}$ with $f(a)=0$, and $g$ is holomorphic in $\Omega$ with $g(\Omega)\subseteq \mathbb{D}$ and $g(b)=0$.
- Prove that $|g(a)|\leq |f(b)|$.
- If equality holds, prove that $g$ is a conformal map of $\Omega$ onto $\mathbb{D}$.
I think this has to do with the Riemann mapping theorem, but I don't know how to apply it. This is a tough problem.
$f:\Omega\rightarrow \mathbb D$ is conformal , therefore it is univalent and it has an inverse: $f^{-1}:\mathbb D\rightarrow \Omega.$
We now need an analytic function $\mathbb D\rightarrow\mathbb D$ which maps $0$ to $0$ in order to use Schwarz's lemma.
First, consider the Möbius Transform of the Disk, $M:\mathbb D\rightarrow\mathbb D,$ which maps $g(a)$ to $0$ and $0$ to $g(a).$ (More precisely, $M(z)= (g(a)-z)/ (1-\overline{g(a)}z).$)
Then the analytic function $M\circ g\circ f^{-1}:\mathbb D\rightarrow \mathbb D$ maps $0$ to $0$, thus , using Schwarz's lemma we obtain : $$|(M\circ g\circ f^{-1})(z)|\leq|z|\Rightarrow$$ $$|M(g(z))|\leq|f(z)|$$ For $z=b$: $$|M(g(b))|\leq|f(b)|\Rightarrow |M(0)|\leq |f(b)|\Rightarrow $$ $$|g(a)|\leq|f(b)|$$
Now, if equality holds, then again from Schwarz's lemma : $$(M\circ g\circ f^{-1})(z)=e^{i\theta}z\Rightarrow$$ $$g(z)=M^{-1}\left(e^{i\theta}f(z)\right)$$ which is conformal, since it is a composition of conformal maps.