Find a conformal map from \(\Omega = \{z\in \mathbb{C} \mid |z-i| > 1,\, \operatorname{Re}(z) > 0\}\) to \(\mathbb{H} = \{w \in \mathbb{C} \mid \operatorname{Im}(w) > 0\}\).
Solution
Define \[ T(z)=\frac{z}{z-2i}. \] The boundary line \(\Re z=0\) and the circle \(|z-i|=1\) both pass through \(0\) and \(2i\), so \(T\) sends them to straight lines through \(0\) and \(\infty\).
Write \(z=x+iy\). A direct computation gives \[ T(z)=\frac{|z-i|^2-1+2ix}{|z-2i|^2}. \] Thus, for \(z\in\Omega\), \[ \Re T(z)>0 \quad\text{and}\quad \Im T(z)>0. \] Hence \(T(\Omega)\) lies in the first quadrant \[ Q=\{w:\Re w>0,\ \Im w>0\}. \] Conversely, \(T\) is a Möbius transformation and the two boundary arcs of \(\Omega\) map to the two boundary rays of \(Q\); equivalently, solving \[ z=\frac{2iw}{w-1} \] shows that every \(w\in Q\) has a preimage in \(\Omega\). Therefore \(T\) maps \(\Omega\) biholomorphically onto \(Q\).
The squaring map is biholomorphic from \(Q\) onto the upper half-plane. Consequently \[ F(z)=T(z)^2=\left(\frac{z}{z-2i}\right)^2 \] is a conformal map from \(\Omega\) onto \(\mathbb H\).