Let $f:D \longrightarrow{D}$ holomorphic, with $D$ is unitary disk.
Show that if $f$ has two fixed point, then $f$ is identity in $D$
I've done:
If $f(0)=0, f(a)=a, a\ne0$, as $|f(a)|=|a|$, per Schwarz Lemma $f(z)=cz$ with $c$ a complex number, then $a=f(a)=ca$, therefore $c=1$
I can't do the case, when $f(0)\ne0$
Sorry, my hint in the comments was wrong. Here's how to proceed: Call $a$, $b$ the fixed points and consider $\psi$ a Moebius transformation $\psi : D\to D$ such that $\psi(a)=0$ and consider $g= \psi \circ f \circ \psi^{-1}$. Then $g$ has fixed points $0$ and $\psi(b)$.