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 \(\abs{F(z)} = 1\).
\(F: \DD \to \DD\) is bijective.
Hint: Calculate \(F \circ F\).
Solution
Goal: (a) Show \(\abs{\frac{w - z}{1 - \bar w z}} < 1\) for \(\abs z, \abs w < 1\) (with equality when \(\abs z = 1\) or \(\abs w = 1\)), for \(\bar z w \neq 1\); (b) show \(F(z) = \frac{w - z}{1 - \bar w z}\) maps \(\DD\) to itself, is holomorphic, sends \(0 \mapsto w\), \(w \mapsto 0\), the circle to the circle, and is bijective.
1. Compute \(\abs{w - z}^2 - \abs{1 - \bar w z}^2 = (\abs w^2 - 1)(1 - \abs z^2)\).
\(\abs{w - z}^2 = \abs w^2 + \abs z^2 - w\bar z - \bar w z\) and \(\abs{1 - \bar w z}^2 = 1 + \abs w^2 \abs z^2 - \bar w z - w \bar z\); subtracting gives \(\abs w^2 + \abs z^2 - 1 - \abs w^2\abs z^2 = (\abs w^2 - 1)(1 - \abs z^2)\).
2. (a): For \(\abs z, \abs w < 1\), \(\abs{\frac{w - z}{1 - \bar w z}} < 1\).
By 1, \(\abs{w - z}^2 - \abs{1 - \bar w z}^2 = (\abs w^2 - 1)(1 - \abs z^2) < 0\) (both factors negative), so \(\abs{w - z} < \abs{1 - \bar w z}\); dividing by the positive \(\abs{1 - \bar w z}\) (nonzero because \(\abs{\bar w z} = \abs w \abs z < 1\)) gives the claim.
3. (a): If \(\abs z = 1\) or \(\abs w = 1\), then \(\abs{\frac{w - z}{1 - \bar w z}} = 1\).
If \(\abs z = 1\), then \(\abs{1 - \bar w z} = \abs z \abs{1 - \bar w z} = \abs{z - \bar w z^2}\)… more directly: \(\abs{1 - \bar w z} = \abs{\bar z - \bar w} = \abs{z - w}\) since \(\bar z = 1/z\); indeed \(\abs{1 - \bar w z} = \abs{\overline{1 - \bar w z}} = \abs{1 - w\bar z}\), and \(\abs{1 - w\bar z} = \abs{z - w}\) because \(\abs z = 1\) (multiply by \(\abs z\): \(\abs{z - w\abs z^2} = \abs{z - w}\)). So numerator and denominator have equal modulus.
The case \(\abs w = 1\) is symmetric: \(\abs{1 - \bar w z} = \abs{w - z\abs w^2} = \abs{w - z}\).
4. (b): \(F\) is holomorphic on \(\DD\).
\(F\) is a rational function whose denominator \(1 - \bar w z\) does not vanish on \(\DD\) (as \(\abs{\bar w z} \leq \abs w \abs z < 1\) for \(\abs w < 1\), \(\abs z < 1\)); hence it is holomorphic there.
5. (b): \(F(0) = w\) and \(F(w) = 0\).
\(F(0) = \frac{w - 0}{1 - 0} = w\), and \(F(w) = \frac{w - w}{1 - \bar w w} = \frac{0}{1 - \abs w^2} = 0\) (valid since \(\abs w < 1\)).
6. (b): \(F\) maps \(\DD\) into itself and \(\abs z = 1 \implies \abs{F(z)} = 1\).
By 2, \(\abs{F(z)} < 1\) for \(\abs z < 1\); by 3 with the roles as given, \(\abs{F(z)} = 1\) for \(\abs z = 1\) (equality case \(\abs z = 1\)).
7. (b): \(F: \DD \to \DD\) is bijective.
\(F(F(z)) = \frac{w - F(z)}{1 - \bar w F(z)}\).
Clearing the inner denominator \(1 - \bar w z\) gives \[ F(F(z)) = \frac{w(1 - \bar w z) - (w - z)}{(1 - \bar w z) - \bar w(w - z)} = \frac{z(1 - \abs w^2)}{1 - \abs w^2} = z \] for \(\abs w < 1\). Thus \(F \circ F = \mathrm{id}_{\DD}\), so \(F\) is a bijection.
8. Q.E.D.
<1>2–<1>3 prove (a) and <1>4–<1>7 prove (b).