Mobius Transformation and Schwarz's Lemma

Question

My goal is to find a Mobius transformation $g$ that sends $K : |z| < R$ bijectively to itself, and also sends $0$ to $a \in R$. To my knowledge, there is a theorem that says the following:

For a disc $K: |z| \leq 1$ and with $f:K \rightarrow K$ being analytic on $|z| <1$, $f$ is of the following form:

$$f(z) = e^{i\alpha} \frac {z-a}{1-\overline a z}$$

My idea was that to accomplish what I need to, I have to generalize the proof of this theorem to a disc of radius $R$, rather than the unit disc. How would I be able to do this?

Answer 1

Consider function $g: D_R \to D_R, $ $g(z) =R\cdot f\left(\frac{z}{R}\right)$

Answer 2

There is a theorem which states that every Mobius map from the unit disk to itself looks like

$$e^{i\phi}\frac{z-a}{1-\overline{a}z}$$

for some angle $\phi$ and complex number $a$ with $|a|\le 1$.

In particular, if $|a|\le1$, there is a Mobius map of the unit disk sending $0\mapsto a$, it is

$$f_a(z)=\frac{a-z}{1-\overline{a}z}.$$

If you define $g_a(z)=R\cdot f_{a/R}(z/R)$, then $f_a$ is the Mobius map of the disk of radius $R$ which sends $0\mapsto a$, where now $|a|\le R$. We may write it out as

$$g_a(z)=R^2\frac{a-z}{R^2-\overline{a}z}. $$