Find a conformal map from \(\mathbb { C } \backslash \{ x \in \mathbb { R } : x \leq 0 \}\) onto the open unit disk.
Solution
Let \[ D=\CC\sm(-\infty,0]. \] On \(D\) take the principal logarithm \[ \Log z=\log\abs z+i\Arg z, \qquad -\pi<\Arg z<\pi, \] and define \[ s(z)=\exp\!\left(\frac12\Log z\right). \]
1. The map\[ s:D\longrightarrow R, \qquad R=\{w\in\CC:\operatorname{Re}w>0\}, \]is a conformal bijection.
For \(z\in D\), \[ -\frac{\pi}{2} < \arg s(z) = \frac12\Arg z < \frac{\pi}{2}. \] Hence \[ \operatorname{Re}s(z)>0, \] so \(s(D)\subseteq R\).
Conversely, let \(w\in R\). Then \[ -\frac{\pi}{2}<\arg w<\frac{\pi}{2}. \] Set \[ z=w^2. \] Then \[ -\pi<\arg z<\pi, \] so \(z\in D\), and the chosen branch satisfies \[ s(z)=w. \] Thus \(s\) is onto. Since \[ z=s(z)^2, \] it is also injective.
Finally, \[ s'(z)=\frac{s(z)}{2z}\neq0 \] on \(D\), so \(s\) is conformal.
2. The map\[ C:R\longrightarrow\DD, \qquad C(w)=\frac{w-1}{w+1}, \]is a conformal bijection.
For \(w=u+iv\) with \(u>0\), \[ \abs{w+1}^2-\abs{w-1}^2=4u>0, \] so \[ \abs{C(w)}<1. \]
The inverse is \[ C^{-1}(\eta)=\frac{1+\eta}{1-\eta}. \] For \(\eta\in\DD\), \[ \operatorname{Re}C^{-1}(\eta) = \frac{1-\abs\eta^2}{\abs{1-\eta}^2} > 0. \] Thus \(C^{-1}(\DD)\subseteq R\), proving bijectivity.
Also \[ C'(w)=\frac{2}{(w+1)^2}\neq0 \] on \(R\), so \(C\) is conformal.
3. A conformal bijection from the slit plane \(D\) onto the unit disk is\[ \boxed{ F(z) = \frac{s(z)-1}{s(z)+1}, \qquad s(z)=\exp\!\left(\frac12\Log z\right), \quad -\pi<\Arg z<\pi. } \]
By steps <1>1–<1>2, \[ F=C\circ s \] is a composition of conformal bijections \[ D\xrightarrow{s}R\xrightarrow{C}\DD. \] Therefore \(F\) is a conformal bijection from \(D\) onto \(\DD\).
4. Q.E.D.
Step 3 gives the requested map.