- (20 points) Let D denote the unit disc \(\{ z : | z | < 1 \}\) . Determine a holomorphic mapping f of D onto itself for which \(\begin{array} { r } { f ( \frac { 1 } { 2 } ) = - \frac { 1 } { \pi } } \end{array}\)
Solution
Let \[ \phi(z)=\frac{z-1/2}{1-z/2} \] and put \(b=-1/\pi\). Define \[ \boxed{f(z)=\frac{\phi(z)+b}{1+b\phi(z)} =\frac{\displaystyle\frac{z-1/2}{1-z/2}-1/\pi} {\displaystyle1-\frac1\pi\frac{z-1/2}{1-z/2}}.} \]
1. The two factors used in the formula are disk automorphisms.
For real \(a\in(-1,1)\), the Möbius map \[ \phi_a(z)=\frac{z-a}{1-az} \] satisfies \[ 1-|\phi_a(z)|^2 =\frac{(1-a^2)(1-|z|^2)}{|1-az|^2}>0 \qquad(|z|<1), \] and has inverse \[ \phi_a^{-1}(w)=\frac{w+a}{1+aw}. \] Thus \(\phi=\phi_{1/2}\) is a biholomorphic map of the disk onto itself and \(\phi(1/2)=0\).
Since \(|b|=1/\pi<1\), the map \[ \psi_b^{-1}(w)=\frac{w+b}{1+bw} \] is also a disk automorphism. Its denominator cannot vanish for \(|w|<1\) because \(|bw|<1\).
2. Their composition has the required value and is onto.
The displayed function is precisely \[ f=\psi_b^{-1}\circ\phi. \] It is therefore holomorphic and bijective from \(D\) onto \(D\). Moreover, \[ f(1/2)=\psi_b^{-1}(\phi(1/2))=\psi_b^{-1}(0)=b=-\frac1\pi. \] Thus it satisfies every condition in the problem.