Let \(z, w\) be complex numbers, such that \(\bar{z} w \neq 1\). Prove that \[\abs{\frac{w - z}{1 - \bar{w} z}} < 1 \; \; \; \mbox{if} \; |z| < 1 \; \mbox{and}\; |w| < 1,\] and also that \[\abs{\frac{w - z}{1 - \bar{w} z}} = 1 \; \; \; \mbox{if} \; |z| = 1 \; \mbox{or}\; |w| = 1.\]
Prove that for fixed \(w\) in the unit disk \(\mathbb D\), the mapping \[F: z \mapsto \frac{w - z}{1 - \bar{w} z}\] satisfies the following conditions:
\(F\) maps \(\mathbb D\) to itself and is holomorphic.
\(F\) interchanges \(0\) and \(w\), namely, \(F(0) = w\) and \(F(w) = 0\).
\(\abs{F(z)} = 1\) if \(|z| = 1\).
\(F: {\mathbb D} \mapsto {\mathbb D}\) is bijective.
Hint: Calculate \(F \circ F\).
Solution
The key identity is \[ |1-\overline wz|^2-|w-z|^2 =(1-|w|^2)(1-|z|^2). \] Indeed, expanding both squared moduli gives \[ 1-\overline wz-w\overline z+|w|^2|z|^2 -|w|^2+w\overline z+\overline wz-|z|^2, \] which factors as claimed.
If \(|z|<1\) and \(|w|<1\), the right-hand side is positive, so \[ |w-z|<|1-\overline wz|, \] and therefore \[ \left|\frac{w-z}{1-\overline wz}\right|<1. \] If \(|z|=1\) or \(|w|=1\), the right-hand side is \(0\), so the two moduli are equal and the quotient has modulus \(1\) whenever it is defined.
Now fix \(w\in\mathbb D\) and put \[ F(z)=\frac{w-z}{1-\overline wz}. \] Because \(|\overline wz|<1\) for \(z\in\mathbb D\), the denominator never vanishes there, so \(F\) is holomorphic. The first part shows that \(F(\mathbb D)\subset\mathbb D\), and directly \[ F(0)=w, \qquad F(w)=0. \] It also shows \(|F(z)|=1\) for \(|z|=1\).
Finally, a direct substitution gives \[ F(F(z))=z. \] Thus \(F\) is its own inverse. Consequently \(F:\mathbb D\to\mathbb D\) is bijective, hence an automorphism of the unit disk.