Prove that \(\displaystyle{f(z)=-\frac{1}{2}\left(z+\frac{1}{z}\right)}\) is a conformal map from half disc \(\{z=x+iy:\ |z|<1,\ y>0\}\) to upper
half plane \({\mathbb H}=\{z=x+iy:\ y>0\}\).
Solution
Let \[ G=\{z:|z|<1,\ \operatorname{Im}z>0\}. \] For \(z=x+iy\), \[ \operatorname{Im}\left(z+\frac1z\right) =y-\frac{y}{|z|^2} =y\left(1-\frac1{|z|^2}\right)<0 \] on \(G\). Hence \[ \operatorname{Im}f(z)>0, \] so \(f(G)\subset\mathbb H\).
Also \[ f'(z)=-\frac12\left(1-\frac1{z^2}\right). \] Its zeros are \(z=\pm1\), which lie on the boundary, so \(f'\) never vanishes in \(G\).
To prove injectivity, suppose \(f(z_1)=f(z_2)\). Then \[ z_1+\frac1{z_1}=z_2+\frac1{z_2}, \] so \[ (z_1-z_2)\left(1-\frac1{z_1z_2}\right)=0. \] If \(z_1\ne z_2\), then \(z_1z_2=1\), impossible because \(|z_1z_2|<1\). Thus \(z_1=z_2\).
Finally, the boundary of \(G\) consists of \((-1,0)\cup(0,1)\) on the real axis and the upper semicircle \(|z|=1\). On the real intervals, \(f\) is real; on the upper semicircle, \(1/z=\overline z\), so again \(f(z)=-\operatorname{Re}z\) is real. As \(z\to0\) through \(G\), \(|f(z)|\to\infty\). Hence the boundary is mapped to the extended real line. Since \(f\) is injective and holomorphic with nonvanishing derivative, \(f(G)\) is a simply connected open subset of \(\mathbb H\) whose boundary in the sphere is the full extended real line. It must therefore be all of \(\mathbb H\).
Thus \(f:G\to\mathbb H\) is a conformal bijection.