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 \(F(w) = 0\).
\(|F(z)| = 1\) if \(|z| = 1\).
\(F: {\mathbb D} \mapsto {\mathbb D}\) is bijective.
Hint: Calculate \(F \circ F\).
Tie’s Extra Questions: Fall 2015### Tie’s Extra Questions: Fall 2015
Solution
For (a), compute \[ |1-\bar wz|^2-|w-z|^2 =(1-|w|^2)(1-|z|^2). \] Hence if \(|w|<1\) and \(|z|<1\), the right-hand side is positive and therefore \[ \left|\frac{w-z}{1-\bar wz}\right|<1. \] If either \(|w|=1\) or \(|z|=1\), the right-hand side is zero, giving equality of the two moduli and hence \[ \left|\frac{w-z}{1-\bar wz}\right|=1. \]
Now fix \(w\in\mathbb D\) and write \[ F(z)=\frac{w-z}{1-\bar wz}. \] The denominator does not vanish on \(\mathbb D\), so \(F\) is holomorphic there. Part (a) gives \(F(\mathbb D)\subseteq\mathbb D\). Direct substitution gives \[ F(0)=w,\qquad F(w)=0, \] and part (a) also gives \(|F(z)|=1\) for \(|z|=1\).
Finally, \[ F(F(z))=z. \] Indeed, substituting the formula for \(F(z)\) and simplifying gives the identity. Thus \(F\) is its own inverse, so it is bijective from \(\mathbb D\) onto itself. Therefore \(F\in\operatorname{Aut}(\mathbb D)\).