- (10 points) Let \(D \subset \mathbb { C }\) be the unit disk and \(\Omega \subset \mathbb { C }\) a bounded, simply connected domain. If \(f _ { 1 } : D \to \Omega\) and \(f _ { 2 } : D \to \Omega\) are holomorphic bijections so that \(f _ { 1 } ( 0 ) = f _ { 2 } ( 0 )\) ， then how are \(f _ { 1 }\) and \(f _ { 2 }\) related to each other?
Solution
They differ by precomposition with a unique rotation: \[ \boxed{f_2(z)=f_1(\lambda z)\quad(z\in D),\qquad |\lambda|=1.} \] Moreover \(\lambda=f_2'(0)/f_1'(0)\).
1. Their composition is an automorphism of the disk fixing zero.
An injective holomorphic function has nonzero derivative; the holomorphic inverse function theorem consequently makes the inverse of each \(f_j\) holomorphic on \(\Omega\) [@SS03]. Thus \(H=f_1^{-1}\circ f_2\) is a holomorphic bijection \(D\to D\), its inverse is \(f_2^{-1}\circ f_1\), and \(H(0)=0=H^{-1}(0)\) by the common-value assumption. Schwarz’s lemma applied to both maps gives \[ |H(z)|\leq|z|=|H^{-1}(H(z))|\leq|H(z)|. \] Hence \(|H(z)|=|z|\) throughout \(D\) [@SS03].
2. The equality forces precisely the stated rotation freedom.
The quotient \(Q(z)=H(z)/z\) extends holomorphically to zero with value \(H'(0)\), by the Taylor expansion of \(H\). It has modulus one at each nonzero point and modulus at most one everywhere by continuity. The maximum modulus principle therefore makes \(Q\) a constant \(\lambda\) with \(|\lambda|=1\) [@SS03]. This gives \(H(z)=\lambda z\) and \(f_2(z)=f_1(\lambda z)\). Differentiating at zero gives \(f_2'(0)=\lambda f_1'(0)\). Since \(f_1'(0)\ne0\), this also proves uniqueness of \(\lambda\).
Conversely, for every \(|\lambda|=1\), precomposition of \(f_1\) with \(z\mapsto\lambda z\) is a holomorphic bijection onto the same domain and has the same value at zero. Thus every stated rotation is possible, and there is no further freedom.