Show that if \(f: \Delta\to\Delta\) is a biholomorphism with \(f(0) = 0\) then \(f\) is a rotation.
Solution
By Schwarz, \(\abs{f(z)} \leq \abs{z}\), and if we write \(h \da \inverseof{f}\) then \(h(0) = 0\) and \(\abs{h(z)} \leq \abs{z}\) as well. Schwarz says \(f\) will be a rotation if there is any \(z_0\) such that \(\abs{f(z_0)} = \abs{z_0}\). Write \(f(z) = w\), we’ll then show that in fact \(\abs{f(z)} = \abs{z}\) for all \(z\in \DD\). \[ \abs{z} = \abs{(h\circ f)(z)} = \abs{h(w)} \leq \abs{w} = \abs{f(z)}\leq \abs{z} .\]
Solution: Shorter version
By Schwarz, \(\abs{f'(0)} \leq 1\) – the claim is that we have equality, so that by Schwarz \(f(z) = \lambda z\) for some \(\abs{\lambda} = 1\). Use that \(f'(0) \neq 0\) since \(f\) is a bijection near zero and if \(g\da \inverseof{f}\) then \(g'(0) = 1/f'(0)\). Moreover Schwarz applies to \(g\), so \(1 \geq \abs{g'(0)} \geq {1\over \abs{f'(0)} }\), forcing \(\abs{ f'(0) } = 1\). By the equality clause in the Schwarz lemma for \(f\), \(f\) is rotation.