- Let \(D = \{ z \in \mathbf { C } : | z | < 1 \}\) and P and Q be distinct points in D. Prove the following statement: If f and g are conformal (or equivalently biholomorphic) self-maps of D, \(f ( P ) = g ( P )\) and \(f ( Q ) = g ( Q )\) , then \(f \equiv g\)
Solution
1. The comparison map is conjugate to a disk map fixing zero and another point.
Since \(g\) is biholomorphic, \(h=g^{-1}\circ f\) is a holomorphic disk self-map. The two assumed equalities give \(h(P)=P\) and \(h(Q)=Q\). Set \[ \phi(z)=\frac{z-P}{1-\overline Pz},\qquad \phi^{-1}(w)=\frac{w+P}{1+\overline Pw}. \] Direct substitution gives the inverse identities, and \[ 1-|\phi(z)|^2= \frac{(1-|P|^2)(1-|z|^2)}{|1-\overline Pz|^2}>0 \quad(z\in D). \] The corresponding identity for \(\phi^{-1}\) shows that both maps preserve \(D\). Thus \(H=\phi\circ h\circ\phi^{-1}\) is a holomorphic disk self-map fixing zero and \(q=\phi(Q)\ne0\).
2. The normalized map is the identity.
Schwarz’s lemma gives \(|H(w)|\leq|w|\) [@SS03]. The quotient \(H(w)/w\) extends holomorphically at zero by the Taylor expansion and is bounded by one. Its value at \(q\) is \(H(q)/q=1\). The maximum modulus principle therefore makes the quotient identically one [@SS03]. Hence \(H(w)=w\) on \(D\), and undoing the conjugation gives \(h=\operatorname{id}_D\). Since \(h=g^{-1}\circ f\), this is exactly \(f=g\) on the whole disk.