Find a bijective conformal map from \(G\) to \(\mathbb{H} = \{w \in \mathbb{C} \mid \operatorname{Im}(w) > 0\}\), where \[G \coloneqq \{z \in \mathbb{C} \mid |z-1| < \sqrt{2}, \, |z+1| < \sqrt{2}\} \setminus [0, i).\]
Solution
The two circles \(|z-1|=\sqrt2\) and \(|z+1|=\sqrt2\) meet orthogonally at \(\pm i\). Set \[ T(z)=\frac{z-i}{z+i}. \] Then \(T(i)=0\), \(T(-i)=\infty\), so the two circular boundary arcs become rays. Since \(T(0)=-1\), the unslit lens maps to the sector \[ \frac{3\pi}{4}<\arg w<\frac{5\pi}{4}. \] Moreover, for \(0\le t<1\), \[ T(it)=\frac{t-1}{t+1}\in[-1,0), \] so the slit maps to that radial segment.
1. Rotate by \(-1\):\[ \zeta=-T(z). \]Then\[ -\frac\pi4<\arg\zeta<\frac\pi4, \qquad \zeta\notin(0,1]. \]Squaring maps this slit sector biholomorphically onto\[ \{u:\Re u>0\}\setminus(0,1]. \]
2. The Cayley transform\[ v=\frac{u-1}{u+1} \]maps the right half-plane to \(\mathbb D\) and \((0,1]\) to \((-1,0]\). Thus\[ v\in\mathbb D\setminus(-1,0]. \]The principal square root maps this slit disk biholomorphically onto the right half-disk\[ H_+=\{s:|s|<1,\ \Re s>0\}. \]
3. Finally,\[ q=\frac{s-i}{s+i} \]maps the imaginary diameter of \(H_+\) to the negative real axis and the semicircular boundary to the negative imaginary axis; hence it maps \(H_+\) biholomorphically onto the third quadrant. Squaring then maps that quadrant biholomorphically onto the upper half-plane.
Therefore the composition \[ z\mapsto T(z)\mapsto -T(z)\mapsto(-T(z))^2 \mapsto \frac{(-T(z))^2-1}{(-T(z))^2+1} \mapsto \sqrt{\frac{(-T(z))^2-1}{(-T(z))^2+1}} \mapsto q\mapsto q^2 \] with the principal square root is a bijective conformal map \(G\to\mathbb H\).