Let \(\Delta\subset\mathbb{C}\) denote the open unit disk.
Prove that for each \(\phi\in\operatorname{Aut}(\Delta)\) and each \(z\in\Delta\), \[ \frac{|\phi'(z)|}{1-|\phi(z)|^2}=\frac{1}{1-|z|^2}. \]
Let \(f:\Delta\to\Delta\) be an analytic map having two different fixed points \(z_1,z_2\in\Delta\). Prove that \(f(z)=z\) for all \(z\in\Delta\).
Solution
Fix \(z\in\Delta\). Let \[ A_z(\zeta)=\frac{\zeta-z}{1-\overline z\,\zeta} \] and similarly let \(A_{\phi(z)}\) send \(\phi(z)\) to \(0\). Then \[ \psi=A_{\phi(z)}\circ\phi\circ A_z^{-1} \] is an automorphism of \(\Delta\) fixing \(0\). By Schwarz’s lemma and its equality case, \[ \psi(\zeta)=e^{i\theta}\zeta, \qquad |\psi'(0)|=1. \] Using the chain rule and \[ |A_z'(z)|=\frac1{1-|z|^2}, \qquad |(A_z^{-1})'(0)|=1-|z|^2, \] we obtain \[ 1=|\psi'(0)| =\frac{|\phi'(z)|(1-|z|^2)}{1-|\phi(z)|^2}. \] Thus \[ \boxed{\frac{|\phi'(z)|}{1-|\phi(z)|^2} =\frac1{1-|z|^2}}. \]
Let \(z_1\ne z_2\) be fixed points of \(f\). Choose an automorphism \(A\) of \(\Delta\) with \(A(z_1)=0\), and set \[ g=A\circ f\circ A^{-1}. \] Then \(g:\Delta\to\Delta\) is holomorphic, fixes \(0\), and fixes the nonzero point \(A(z_2)\). Schwarz’s lemma gives \(|g(\zeta)|\le|\zeta|\), and equality at the nonzero fixed point forces \[ g(\zeta)=e^{i\theta}\zeta. \] Since \(A(z_2)\) is fixed and nonzero, \(e^{i\theta}=1\). Hence \(g\) is the identity, and therefore so is \(f\).