Prove that \(\displaystyle{f(z)=-\frac{1}{2}\left(z+\frac{1}{z}\right)}\) is a conformal map from the half disc \[ \{z=x+iy:\ |z|<1,\ y>0\} \] to \(\HH \da \{z=x+iy:\ y>0\}\).
Solution
The map \(f\) is holomorphic on the half disc, and \(f'(z)=-{1\over2}\qty{1-z^{-2}}\neq0\) there since \(z\neq\pm1\), so it suffices to show that \(f\) maps the half disc bijectively onto \(\HH\). Consider the images of arcs \(\gamma_R(t) \da Re^{it}\) for \(t\in (0, \pi)\) and \(0<R<1\), which partition the upper half disc: \[ f(Re^{it}) = - {1\over 2}\qty{ \qty{R+\inverseof{R}}\cos(t) + i(R-\inverseof{R})\sin(t) } = H_R\cos(t) + iV_R\sin(t) ,\] where \(H_R \da -{1\over 2}\qty{R+\inverseof{R}}\) and \(V_R \da {1\over 2}(\inverseof{R}-R)\). Here \(H_R^2-V_R^2=1\), \(\abs{H_R}>1\) and \(V_R>0\). So as \(t\) ranges through \((0, \pi)\), \(f\circ\gamma_R\) traces the upper half of the ellipse \(E_R\) with semi-axes \(\abs{H_R}\) and \(V_R\) and foci \(\pm1\), once and injectively.
As \(R\) decreases from \(1\) to \(0\), \(V_R\) increases from \(0\) to \(\infty\). The confocal ellipses \(\theset{E_R}_{0<R<1}\) are pairwise disjoint and their union is \(\CC\setminus[-1,1]\): a point \(w\notin[-1,1]\) lies on the unique such ellipse with \(\abs{w-1}+\abs{w+1}=2\abs{H_R}\). So each \(w\in\HH\) lies on exactly one upper half-ellipse, and has exactly one preimage in the half disc.