Determine a one-to-one complex analytic mapping \(f\), other than \(f(z) = z\), that takes \(D = \{z \in \mathbb{C} : |z| < 1\}\) onto itself and satisfies \(f(1/3) = 1/3\).
Solution
One example is \[ \boxed{f(z)=\frac{3-5z}{5-3z}}. \]
1. Conjugating a nontrivial rotation gives a disk automorphism fixing \(1/3\).
Let \(a=1/3\) and define \[ \phi_a(z)=\frac{z-a}{1-az}. \] For \(|z|<1\), \[ 1-|\phi_a(z)|^2 =\frac{(1-a^2)(1-|z|^2)}{|1-az|^2}>0, \] so \(\phi_a\) maps the unit disk into itself. Direct substitution shows that \[ \phi_a^{-1}(w)=\frac{w+a}{1+aw}, \] which also maps the disk into itself. Hence \(\phi_a\) is a biholomorphic disk automorphism and \(\phi_a(a)=0\).
Now let \(R(w)=-w\), a nonidentity disk automorphism fixing zero, and set \[ f=\phi_a^{-1}\circ R\circ\phi_a. \] Then \(f\) is a one-to-one holomorphic map of \(D\) onto itself and \[ f(a)=\phi_a^{-1}(0)=a. \] Because conjugation preserves whether a map is the identity and \(R\) is not the identity, \(f\) is not the identity.
2. The conjugated map has the displayed explicit formula.
Using \(a=1/3\), \[ f(z)=\frac{a-\phi_a(z)}{1-a\phi_a(z)} =\frac{2a-(1+a^2)z}{1+a^2-2az} =\frac{3-5z}{5-3z}. \] Its denominator cannot vanish in the unit disk because its only zero is \(z=5/3\). The formula also gives directly \[ f(1/3)=1/3, \] and, for example, \(f(0)=3/5\ne0\), so the map is visibly nonidentity.