How do these Möbius transforms arise?

Question

I am reading Henri Cartan's Elementary Theory of Analytic Functions of One or Several Complex Variables. Below is exercise 10 (p.111) in chapter 3:

Let $f$ be a holomorphic function in the disc $|z|<1$, such that $|f(z)|<1$ in this disc; if there exist two distinct points $a$ and $b$ in the disc such that $f(a)=a$ and $f(b)=b$, show that $f(z)=z$ in the disc. (Consider the function $g(z)=\frac{h(z)-a}{1-\bar{a}h(z)}$, with $h(z)=f\left(\frac{z+a}{1+\bar{a}z}\right)$, for which $g(0)=0$, $g\left(\frac{b-a}{1-\bar{a}b}\right)=\frac{b-a}{1-\bar{a}b}$, and $|g(z)|<1$ in the disc.)

The exercise itself is easy if we follow the hint. It's just a simple application of Schwarz's lemma. However, the constructions of $g$ and $h$ here seem too magical to me. How does one come up with the hint? Does it originate from some special trick about Möbius transform?

Answer 1

The idea is to consider $$ g = T \circ f \circ T^{-1} $$ where $T$ is an automorphism of the unit disk with $T(a) = 0$. Those automorphism are well-known and in fact Möbius transformations, see for example

• Proof that automorphisms of unit disk are of the form $\lambda \dfrac{z-a}{\overline{a}z-1}$
• Automorphisms of the Unit Disk

Then $g(0) = 0$ and $g(T(b)) = T(b)$, so that the Schwarz lemma can be applied to $g$.