Let \(U\) be the upper half plane \(\{z\in\mathbb{C}\mid\operatorname{Im}(z)>0\}\) with the slit \(\{iy\mid 0<y\le 1\}\) removed. Find a conformal equivalence from \(U\) to \(\mathbb{D}\).
Solution
Let \[ U=\{\operatorname{Im}z>0\}\setminus\{iy:0<y\le1\}. \] The squaring map is injective on the upper half-plane and maps it onto \(\mathbb C\setminus[0,\infty)\). The removed slit is mapped to \([-1,0)\). Therefore \[ z\longmapsto z^2 \] maps \(U\) biholomorphically onto \(\mathbb C\setminus[-1,\infty)\), and hence \[ z\longmapsto z^2+1 \] maps \(U\) biholomorphically onto \(\mathbb C\setminus[0,\infty)\).
On \(\mathbb C\setminus[0,\infty)\) choose the square-root branch with \(0<\arg w<2\pi\). Then \[ S(w)=\sqrt w \] has argument in \((0,\pi)\), so it maps this slit plane biholomorphically onto the upper half-plane \(\mathbb H\).
Finally the Cayley map \[ C(\zeta)=\frac{\zeta-i}{\zeta+i} \] maps \(\mathbb H\) biholomorphically onto \(\mathbb D\). Thus one explicit conformal equivalence is \[ \boxed{\displaystyle \Phi(z)=\frac{\sqrt{z^2+1}-i}{\sqrt{z^2+1}+i},} \] where the square root is the branch just specified.