# How do these Möbius transforms arise?

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?

• Unless I am mistaken, it should be $h(z)=f\left(\frac{z+a}{1+\bar{a}z}\right)$. – Martin R Oct 13 '19 at 10:52
• @MartinR Thanks for catching the typo. It's fixed now. – William McGonagall Oct 13 '19 at 10:59

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