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 \(\abs{F(z)} = 1\).
\(F\) is a bijection.
Solution
Part 1: See Spring 2021.1 above.
Part 2, holomorphicity: This is meromorphic: it is a rational function of \(z\) (a quotient of polynomials in \(z\)), hence meromorphic, and it has a singularity only at \(z\) such that \(\bar w z = 1\). This can only happen if \(z, w \in S^1\): taking the modulus yields \[ \bar w z = 1 \implies \abs{w}^2\abs{z}^2 = 1 ,\] and moreover since \(\abs{w}^2 \leq 1\) and \(\abs{z}^2\leq 1\), the only way this product can be one is when \(\abs{w}^2 = \abs{z}^2 = 1\). This also forces \(z=1/\bar w\).
The claim is that the singularity \(1/\bar w\) is removable. Note that \(1\over w = \bar w\) on the circle, so \(1/\bar{w} = \bar{\bar w} = 2\), so \[ \qty{ z- \inverseof{\bar{w}} } \qty{w-z \over 1-\bar w z} &= \qty{\bar w z - 1 \over \bar w} \qty{w-z \over 1-\bar w z} \\ &= \inverseof{\bar{w}}(w-z) \\ &= w(w-z) \\ &\converges{z\to \inverseof{\bar w}=w }\to 0 .\]
Part 2, being a bijection: This follows from finding an explicit inverse, using that \(F^2 = \id\): \[ F(F(z)) &= \frac{w- \qty{ \frac{w-z}{1-\bar{w} z} } }{1-\bar{w} \qty{ \frac{w-z}{1-\bar{w} z} } } \\ &= \frac{w(1-\bar{w} z)-(w-z)}{q-\bar{w} z-\bar{w}(w-z)} \\ &= \frac{w-|w|^{2} z-w+z}{1-\bar{w} z-|w|^{2}+\bar{w} z} \\ &= \frac{z\left(1-|w|^{2}\right)}{1-|w|^{2}} \\ &= z .\]
Part 2, being an involution: A direct check shows that \(F(w) = 0\), since the numerator vanishes, and \(F(0) = {w - 0 \over 1 - 0} = w\).
Part 3, preserving the circle: Follows from the estimate in part 1.