Construct an explicit conformal map (biholomorphic equivalence) from the slit plane \(\Omega = \mathbb{C} \setminus (-\infty, 0]\) onto the open unit disk \(\mathbb{D} = \{w \in \mathbb{C} \mid |w| < 1\}\).
Solution
Use the principal square root on \[ \Omega=\mathbb C\setminus(-\infty,0]. \] If \(z=re^{i\theta}\) with \(-\pi<\theta<\pi\), define \[ \sqrt z=\sqrt r\,e^{i\theta/2}. \] Then \[ -\frac\pi2<\arg\sqrt z<\frac\pi2, \] so \(\Re\sqrt z>0\). Thus \[ z\longmapsto\sqrt z \] is a biholomorphism from \(\Omega\) onto the right half-plane \[ H=\{u:\Re u>0\}. \]
1. The Cayley transform\[ C(u)=\frac{u-1}{u+1} \]maps \(H\) biholomorphically onto \(\mathbb D\). Indeed, if \(u=x+iy\) with \(x>0\), then\[ |C(u)|^2 =\frac{(x-1)^2+y^2}{(x+1)^2+y^2}<1. \]
2. Therefore\[ F(z)=\frac{\sqrt z-1}{\sqrt z+1} \]is a conformal bijection \(\Omega\to\mathbb D\).
3. Solving for \(z\) gives\[ F^{-1}(w)=\left(\frac{1+w}{1-w}\right)^2, \qquad |w|<1. \]Since \((1+w)/(1-w)\) lies in the right half-plane, its square lies in \(\Omega\), so this is indeed the inverse.