I think I found a solution, but it seems a little bit long and complicated. Tell me what you think of it.
For $a \in D$, we define on $\mathbb{C}\setminus\left\lbrace\dfrac{1}{\bar{a}}\right\rbrace$ the function $\phi_a : z \mapsto \dfrac{z-a}{1-\bar{a}z}$.
It is well known that :
- $\phi_a$ is an automorphism of $D$
- $\phi_a(\partial D)=\partial D$
- $\phi_a^{-1}=\phi_{-a}$
Then, let us consider $g=\phi_{w_0}\circ f \circ \phi_{z_0}^{-1}=\phi_{w_0}\circ f \circ \phi_{-z_0}$. It is clear that $g(D)=D$ and $g(\partial D)=\partial D$.
We have $g(0)=\phi_{w_0}\circ f \circ \phi_{-z_0}(0)=\phi_{w_0} \circ f (z_0) = \phi_{w_0} (w_0)=0$.
Thus, we can apply Schwarz lemma to $g$. We have $|g(z)|\leq |z|$ for all $z\in D$ and $g'(0)\leq 1$.
Moreover, if $g'(0)=1$ or if there exist $z\in D^*$ such that $|g(z)|=|z|$, then $g=R_\theta$ for a certain $\theta\in\mathbb{R}$, where $R_\theta:z\mapsto e^{i\theta}z$. Then, we would have $f=\phi_{-w_0} \circ R_\theta \circ \phi_{z_0}$, which would end the demonstration (developping this expression, we find that $f$ would be of the wanted form for some $a$ function of $z_0,w_0$ and $\theta$).
We will then procedeed by contradiction : let us suppose that $g'(0)<1$ and $|g(z)|<|z|$ for every $z\in D^*$.
We define $h(z)=\dfrac{g(z)}{z}$ for $z\in \bar{D}^*$ and $h(0)=g'(0)$. Then :
- $h$ is holomorphic on $D$, continuous on $\bar{D}$
- $h(D)=D$ (because $g(z)<z$ for $z \in D^*$ and $g'(0)<1$)
- $h(\partial D)=\partial D$
$|h|$ is continous on the compact $\bar{D}$ so it possesses a minimum. It is clear that this minimum is in $D$. So by the maximum modulus principle, $h$ must cancel in $D$.
Yet, $h(0)=g'(0)\neq 0$, because $f'(0)\neq 0$ (composition of derivation gives $g'(0)$ directly proportionnal to $f'(0)$). Thus, there exists $v \in D^*$ such that $h(v)=0$ and then $g(v)=0$, or again $f(\phi_{-z_0}(v))=w_0$. This is in contradiction with the fact that $f^{-1}(w_0)=\lbrace z_0 \rbrace$, which ends the demonstration.