- Prove that \(f ( z ) = - { \textstyle \frac { 1 } { 2 } } ( z + 1 / z )\) is a conformal map from the half-disc \(\left\{ z = x + i y : | z | < 1 , \ y > 0 \right\}\) to the upper half-plane.
[Hint: The equation \(f ( z ) = w\) reduces to the quadratic equation \(z ^ { 2 } + 2 w z + 1 = 0\) 2 which has two distinct roots in C whenever w \(\neq \pm 1\) . This is certainly the case if \(w \in \mathbb { H } . ]\)
Solution
Let \[ D_+=\{z:|z|<1,\ \Im z>0\}, \qquad f(z)=-\frac12\left(z+\frac1z\right). \] For \(z=x+iy\in D_+\), \[ \Im\left(z+\frac1z\right) =y-\frac{y}{|z|^2} =y\left(1-|z|^{-2}\right)<0, \] so \(\Im f(z)>0\). Hence \(f(D_+)\subseteq\mathbb H\).
If \(f(z)=f(w)\), then \[ z+z^{-1}=w+w^{-1}, \] so \[ (z-w)\left(1-\frac1{zw}\right)=0. \] Since \(|zw|<1\), the second factor cannot vanish; therefore \(z=w\). Thus \(f\) is injective.
Now fix \(\eta\in\mathbb H\). The equation \(f(z)=\eta\) is \[ z^2+2\eta z+1=0. \] Its two roots are reciprocal. They are nonreal because \(\eta\notin\mathbb R\). Exactly one root lies in the upper half-plane, since if \(z\) is nonreal then \(1/z\) has imaginary part of the opposite sign. Let \(z\) be the upper-half-plane root. If \(|z|>1\), then the formula above gives \[ \Im f(z)<0, \] contrary to \(f(z)=\eta\in\mathbb H\). Thus \(|z|<1\), so \(z\in D_+\). Therefore \(f\) is surjective onto \(\mathbb H\).
Finally \[ f'(z)=-\frac12(1-z^{-2}), \] and \(f'(z)=0\) only at \(z=\pm1\), neither of which lies in \(D_+\). Hence \(f\) is conformal from \(D_+\) onto \(\mathbb H\).