While reading a complex analysis textbook the following assertion came up
Since $f,g:D\equiv D(a,r) \to \mathbb{C}$ are analytic and injective functions such that $f(D)=g(D)$, $f(a)=g(a)$ and $f'(a)=g'(a)$ then $f=g$.
I do not think this follows directly! To prove it I think we need to use the Identity Principle. What I did is take a sequence $(z_n)_n \subset D $ such that $z_n \neq a$ and $z_n \to a$, then for each $n$ there exist $w_n \neq a$ such that $f(z_n)=g(w_n)$. By injectivity and since $f(a)=g(a)$, it follows that also $w_n \to a$. Now I am trying to use the condition $f'(a)=g'(a)$ to see that $z_n=w_n$ for infinitely many $n$'s and hence by the Identity Principle $f=g$. However this argumentation seems to lead nowhere.
Any help is very appreciated, perhaps there is a much more simple argument that do not uses anything that I have thought so far.
Without loss of generality we can assume $D$ to be the unit disk ($a=0$). Then $g^{-1}\circ f$ maps to unit disk to itself fixing zero. Now the Schwarz lemma says $$ |(g^{-1}\circ f)'(0)|=|(g^{-1})'\circ f(0)||f'(0)|\leq 1. $$ Differentiating $g^{-1}\circ g=z$ yields $(g^{-1})'\circ g\cdot g'=1$. Plug zero into this to show $$ (g^{-1})'\circ f(0)=(g^{-1})'\circ g(0)=\frac{1}{g'(0)}=\frac{1}{f'(0)}, $$ which means $(g^{-1}\circ f)'(0)=1$. Now the Schwarz lemma shows further that $g^{-1}\circ f(z)=az$ for some $|a|=1$, $f(z)=g(az)$, $f'(z)=ag'(az)$, and so $a=1$ by plugging in $z=0$. I believe this to be correct. I typed it up rather quickly.