Let \(D=\{z\in\mathbb C:|z|<1\}\). Prove that every conformal bijection of \(D\) onto \(D\) is a linear fractional transformation.
An injective holomorphic map merely into \(D\) need not be linear fractional. For example, \(q(z)=z/2+z^2/8\) satisfies \(|q(z)|<5/8<1\) on \(D\), and \[ q(z)-q(w)=(z-w)\left(\frac12+\frac{z+w}{8}\right). \] The second factor has modulus greater than \(1/4\) for \(z,w\in D\), proving injectivity. If \(q(z)=(Az+B)/(Cz+E)\) with \(AE-BC\ne0\), multiplication gives a polynomial identity. Its cubic coefficient forces \(C=0\), and its quadratic coefficient then forces \(E=0\), contradicting the determinant condition. Thus it is not linear fractional; surjectivity is essential to the classification.
Solution
The maps are exactly \[ f(z)=\lambda\frac{z-a}{1-\overline a z}, \qquad a\in D,\quad |\lambda|=1. \]
1. A disk automorphism reduces the problem to a bijection fixing zero.
Surjectivity gives \(a\in D\) with \(f(a)=0\). Define \[ \phi_a(z)=\frac{z-a}{1-\overline a z}, \qquad \phi_a^{-1}(w)=\frac{w+a}{1+\overline a w}. \] Their denominators are nonzero on \(D\), and direct substitution shows that they are inverse. Moreover \[ 1-|\phi_a(z)|^2 =\frac{(1-|a|^2)(1-|z|^2)}{|1-\overline a z|^2}>0 \qquad(z\in D). \] The same identity with \(-a\) proves that the inverse maps \(D\) into \(D\). Thus \(H=f\circ\phi_a^{-1}\) is a conformal bijection of \(D\) onto \(D\) and \(H(0)=0\). Its inverse is holomorphic by the holomorphic inverse function theorem [@SS03], and also fixes zero.
2. The normalized map is a rotation, giving the full classification.
Schwarz’s lemma applied to \(H\) and \(H^{-1}\) gives \[ |H(z)|\leq|z|=|H^{-1}(H(z))|\leq|H(z)|. \] Hence \(|H(z)|=|z|\). The quotient \(H(z)/z\) extends holomorphically across zero and has modulus at most one. It has modulus one at every nonzero interior point, so the maximum modulus principle makes it a constant \(\lambda\) of modulus one [@SS03]. Therefore \(H(z)=\lambda z\) and \(f=H\circ\phi_a\) has the displayed form. Conversely, every displayed map is the composition of \(\phi_a\) with a rotation, so is a conformal bijection of \(D\) onto \(D\).