This is an exericse about the Schwarz Lemma:
Let $\Omega \subseteq \mathbb{C}$ be a simply connected domain and
$z_0 \in \Omega \neq \mathbb{C}$. Furthermore, let $f, g: \Omega \to \Omega$ biholomorphic such that\begin{align} f(z_0) = g(z_0), \qquad f'(z_0) = g'(z_0). \end{align}
Proof that $f = g$.
The hint states to use Schwarz's Lemma as well as Riemann's mapping theorem. In fact later only allows me to use Schwarz's Lemma (in this case). Could someone give me a proper (second) hint how to use Schwarz's Lemma to show that $f = g$? Should I consider using $h:= f-g$?
Best Answer
Since $\Omega\subsetneq\Bbb C$ is simply connected, then by the Riemann mapping theorem we know that it exists $h:\Omega\to\Delta$ biholomorphic such that $h(z_0)=0,\; h'(z_0)\neq0$, where $\Delta$ is the open unit disk.
Consider then $$ F:=h\circ g^{-1}\circ f\circ h^{-1}\colon\Delta\to\Delta $$ which is a biholomorphism such that $F(0)=0$ and $|F(z)|<1$ on the whole $\Delta$; thus by Schwartz Lemma we get \begin{align*} &|F(z)|\le|z|\;\;\forall z\in\Delta. \end{align*} Schwartz applies to $F^{-1}$ as well letting us to achieve $|F^{-1}(w)|\le|w|\;\;\forall w \in\Delta$.
Since every $w\in\Delta$ can be written as $F(z)$ for a unique $z\in\Delta$, we get
$$ |z|\le|F(z)|\;\;\; \forall w\in\Delta $$ and thus $$ |F(z)|=|z|\;\;\forall z\in\Delta. $$ So Schwarz implies $F(z)=az$ for some $a\in\Bbb C,\;|a|=1$.
Taking the derivative of this last one at $0$ one and exploiting your hypotesis on $f$ and $g$, you get that $a=1$ and thus $f=g$.