Holomorphic function $f:B(0,1) \to B(1,4)$ with $f(0)=3$ and $f(\frac{1}{2})=1$ Problem
Find all holomorphic functions $f:B(0,1) \to B(1,4)$ that verify $f(0)=3$ and $f(\frac{1}{2})=1$.
I have no idea how to attack this problem,since it comes after an exercise related to Schwarz lemma, I thought that maybe the idea is to somehow apply the lemma in this problem as well, but if that is the case, I don't know how could I use it. I would appreciate suggestions.
 A: I think the idea is to transform the problem into something susceptible to the Schwarz lemma.
Consider $g(z) = \frac{1}{4}(f(z) - 1)$.
This function maps the unit disk to the unit disk, with $g(0) = 1/2$, and $g(1/2) = 0$.
Composing with the Möbius transformation
$$T(z) = \frac{z- \frac{1}{2}}{1 - \frac{1}{2}z}$$
gives a map of the unit disk to itself which fixes the origin and takes the point $z = 1/2$ to $z = -1/2$.
So the by the Schwarz lemma the composite $T \circ g$ is $(T \circ g)(z) = -z$.
Working backwards yields
$$
f(z) = 4 \frac{2z-1}{z-2} + 1
$$
A: Let's consider $g(z)=(1/4)(f(z)-1)$; this maps the unit disk $D=B(0,1)$ back to itself. Note that $g(0)=1/2$, $g(1/2)=0$, so if we swap $0$, $1/2$ with the Mobius transformation
$$
\varphi(z) = \frac{2z-1}{z-2} ,
$$
and introduce $h(\varphi(z))=g(z)$, then $h:D\to D$, $h(0)=0$, $h(1/2)=1/2$. By the Schwarz Lemma, $h(w)=w$. Unwrapping the definitions, we see that
$$
f(z) = 4g(z)+1 = 4\varphi(z) + 1 = \frac{9z-6}{z-2}
$$
is the only solution.
