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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?

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

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  • $\begingroup$ I haven't learned this before. It makes a lot of sense. Thanks very much! $\endgroup$ – William McGonagall Oct 13 '19 at 10:58

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