$f \in Aut (\mathbb{D})$ with two fixed points is the identity I have been working on this homework problem for my complex analysis class for some time to no avail.  The questions asks for me to show that for $f \in Aut (\mathbb{D})$ such that $f(z_1) = z_1$ and $f(z_2) = z_2$ for $z_1 \ne z_2 \in \mathbb{D}$, $f$ is the identity on $\mathbb{D}$: $f(z) = z$ for all $z \in \mathbb{D}$.
If we assume that one of the fixed points, say $z_1$ is zero so that $f(0) = 0$, then the result follows nicely from Schwarz' Lemma.  However, in the case when both $z_1$ and $z_2$ are non-zero I am having difficulty.  It seems like this case should somehow be able to be deduced from the $z_1 = 0$ case, but I have not found out how to do this yet.
As always, any guidance would be much appreciated!   

Can you compose your function with another one such that the composition satisfies Schwarz?
