Find a conformal map from \(\mathbb { C } \backslash \{ x \in \mathbb { R } : | x | \geq 1 \}\) onto the open unit disk.
Solution
Let \[ D=\CC\sm\{x\in\RR:\abs x\geq1\} \] and define \[ M(z)=\frac{1+z}{1-z}. \]
1. The map\[ M:D\longrightarrow \CC\sm(-\infty,0] \]is a conformal bijection.
The only pole of \(M\) is \(z=1\), which is not in \(D\). Its inverse is \[ M^{-1}(w)=\frac{w-1}{w+1}. \]
On the two deleted rays away from the pole, \[ (-\infty,-1] \longmapsto (-1,0], \qquad (1,\infty) \longmapsto (-\infty,-1). \] The remaining deleted point \(z=1\) is the pole of \(M\). Conversely, if \(w\leq0\) is real and \(w\neq-1\), then \[ M^{-1}(w)\in\RR \] and \[ \abs{M^{-1}(w)}\geq1. \] Thus a finite point \(z\) lies in \(D\) exactly when \(M(z)\) does not lie on the nonpositive real axis. Since \(-1\) itself belongs to the deleted target slit, the inverse is defined at every point of \[ \CC\sm(-\infty,0]. \] Hence \(M\) is bijective between the displayed domains.
Finally, \[ M'(z)=\frac{2}{(1-z)^2}\neq0 \] on \(D\), so \(M\) is conformal.
2. On\[ \CC\sm(-\infty,0] \]take the principal logarithm\[ \Log w=\log\abs w+i\Arg w, \qquad -\pi<\Arg w<\pi, \]and put\[ s(w)=\exp\!\left(\frac12\Log w\right). \]Then\[ s:\CC\sm(-\infty,0]\longrightarrow R, \qquad R=\{\zeta\in\CC:\operatorname{Re}\zeta>0\}, \]is a conformal bijection.
The argument of \(s(w)\) is \[ \frac12\Arg w\in \left(-\frac{\pi}{2},\frac{\pi}{2}\right), \] so \(s(w)\in R\).
Conversely, if \(\zeta\in R\), then \[ -\frac{\pi}{2}<\arg\zeta<\frac{\pi}{2}. \] Thus \[ w=\zeta^2 \] has argument in \((-\pi,\pi)\) and lies in the slit plane, with \[ s(w)=\zeta. \] This proves surjectivity, and \(w=s(w)^2\) proves injectivity.
Also \[ s'(w)=\frac{s(w)}{2w}\neq0, \] so \(s\) is conformal.
3. The map\[ C:R\longrightarrow\DD, \qquad C(\zeta)=\frac{\zeta-1}{\zeta+1}, \]is a conformal bijection.
For \(\zeta=u+iv\) with \(u>0\), \[ \abs{\zeta+1}^2-\abs{\zeta-1}^2=4u>0, \] so \(\abs{C(\zeta)}<1\).
Its inverse is \[ C^{-1}(\eta)=\frac{1+\eta}{1-\eta}, \] and for \(\eta\in\DD\), \[ \operatorname{Re}C^{-1}(\eta) = \frac{1-\abs\eta^2}{\abs{1-\eta}^2} > 0. \] Hence \(C\) is bijective. Since \[ C'(\zeta)=\frac{2}{(\zeta+1)^2}\neq0 \] on \(R\), it is conformal.
4. Put\[ q(z)= s(M(z)) = \exp\!\left[ \frac12 \Log\!\left(\frac{1+z}{1-z}\right) \right], \]where the logarithm is the principal branch. A conformal bijection from \(D\)onto the unit disk is\[ \boxed{ F(z)=\frac{q(z)-1}{q(z)+1}. } \]
By steps <1>1–<1>3, \[ F=C\circ s\circ M \] is a composition of conformal bijections \[ D\xrightarrow{M}\CC\sm(-\infty,0] \xrightarrow{s}R\xrightarrow{C}\DD. \] Therefore \(F\) is a conformal bijection from \(D\) onto \(\DD\).
5. Q.E.D.
Step 4 gives the requested map.