I wanted to prove the following.
Let $D\subseteq \mathbb{C}$ be a disk around some point $z_0\in\mathbb{C}$, $f,g:D\to \mathbb{C}$ holomorphic and injective such that $f(D)=g(D)$, $f(z_0)=g(z_0)$ and $f'(z_0)=g'(z_0)$. Then $f=g$.
I thought about finding some sequence in $D$ such that $f,g$ agree on the sequence (I don't need to worry about convergence since $D$ is bounded), but I don't really know where to start.
Also, although we haven't arrived to the Schwarz Lemma yet, I think it might be of some use.
So, if possible, how can the identity theorem and/or Schwarz lemma be used to solve this? And if not possible, which other ways are there?
Without loss of generality we may suppose that $D$ is the unit circle, and $z_0=0$, let $U=f(D)$. $g:D\to U$ is holomorphic and univalent so it is biholomorphic, i.e. it is invertible and its inverse $g^{-1}:U\to D$ is holomorphic. So, $h=g^{-1}\circ f:D\to D$ is holomorphic, and satisfies $h(0)=0$ and $h’(0)=1$. This, implies by Schwarz lemma that $h(z)=z$ for all $z$ in $D$. That is $g=f$. Done.