Let \(z, w \in \CC\) with \(\bar z w \neq 1\). Prove that \[ \abs{w-z \over 1 - \bar w z} < 1 \quad\text{ if } \abs{z}<1,~ \abs{w} < 1 \] with equality when \(\abs{z} = 1\) or \(\abs{w} = 1\).
Prove that for a fixed \(w\in \DD\), the mapping \(F: z\mapsto {w-z \over 1 - \bar w z}\) satisfies
- \(F\) maps \(\DD\) to itself and is holomorphic.
- \(F(0) = w\) and \(F(w) = 0\).
- \(\abs{z} = 1\) implies \(\abs{F(z)} = 1\).
Solution: part 1
\[ 0 &\leq (1 - \abs{w}^2)(1-\abs{z}^2) \\ \implies \abs{w}^2 + \abs{z}^2 &\leq 1 + \abs{w}^2 \abs{z}^2 \\ \implies \abs{w}^2 + \abs{z}^2 - 2\Re(\bar{w} z) &\leq 1 + \abs{w}^2 \abs{z}^2 - 2\Re(\bar{w} z) \\ \implies \abs{w-z}^2 &\leq \abs{1-\bar{w}z}^2 .\] Note that if either \(\abs{w}^2 = 1\) or \(\abs{z}^2 = 1\) then the first line is an equality, yielding equality in the final line.
Solution: part 2
That \(F: \DD\to \DD\): follows from the inequality, since \(\abs{z}, \abs{w} < 1\) for \(z,w\in \DD\). Holomorphicity: follows from the fact that rational expressions of holomorphic functions are holomorphic away from where the denominators vanish.
Then just note that \(\abs{\bar{w} z} \leq \abs{w}\abs{z} < 1\), so \(\abs{1 - \bar{w} z} > 0\).
\(F(0) = {w-0 \over 1-0} = w\)
\(F(w) = {w-w\over 1 - \bar w w} = 0\)
\(\abs{z} = 1\) yields equality in part 1.
Other notes: \(F\) is bijective on \(\DD\): \[ F(F(z)) &= {w - \qty{w-z\over 1-\bar w z} \over 1 - \bar{w}\qty{w-z\over 1 - \bar w z} } \\ &= {w(1-\bar w z) - (w-z) \over (1-\bar w z) - \bar w (w-z)} \\ &= {z-\abs{w}^2 z \over 1 - \abs{w}^2 }\\ &= z .\]