Let \(D\) be the region obtained by removing the interval \([0, 1)\) from the unit disk \(\abs{z} < 1\). Find a conformal map from \(D\) onto the open unit disk.
Solution
Let \[ D=\DD\sm[0,1). \] On \[ \CC\sm[0,\infty) \] take the logarithm branch \[ \Log z=\log\abs z+i\Arg z, \qquad 0<\Arg z<2\pi, \] and define \[ s(z)=\exp\!\left(\frac12\Log z\right). \]
1. The map\[ s:D\longrightarrow U, \qquad U=\{w\in\CC:\abs w<1,\ \operatorname{Im}w>0\}, \]is a conformal bijection.
For \(z\in D\), \[ \abs{s(z)}=\sqrt{\abs z}<1 \] and \[ 0<\arg s(z)=\frac12\Arg z<\pi. \] Thus \(s(z)\in U\).
Conversely, if \[ w\in U, \] then \[ 0<\arg w<\pi. \] Set \[ z=w^2. \] Then \(\abs z<1\) and \[ 0<\arg z<2\pi, \] so \(z\in D\). By the chosen branch, \[ s(z)=w. \] Thus \(s\) is onto, and the identity \[ z=s(z)^2 \] shows that it is injective.
Since \[ s'(z)=\frac{s(z)}{2z}\neq0 \] on \(D\), the map is conformal.
2. The map\[ M:U\longrightarrow Q_1, \qquad M(w)=\frac{1+w}{1-w}, \]where\[ Q_1=\{\zeta\in\CC:\operatorname{Re}\zeta>0,\ \operatorname{Im}\zeta>0\}, \]is a conformal bijection.
For \(w\in U\), \[ \operatorname{Re}M(w) = \frac{1-\abs w^2}{\abs{1-w}^2} > 0 \] and \[ \operatorname{Im}M(w) = \frac{2\operatorname{Im}w}{\abs{1-w}^2} > 0. \] Hence \(M(U)\subseteq Q_1\).
The inverse is \[ M^{-1}(\zeta)=\frac{\zeta-1}{\zeta+1}. \] For \(\zeta=u+iv\in Q_1\), \[ \abs{\zeta+1}^2-\abs{\zeta-1}^2=4u>0, \] so \(\abs{M^{-1}(\zeta)}<1\), and \[ \operatorname{Im}M^{-1}(\zeta) = \frac{2v}{\abs{\zeta+1}^2} > 0. \] Thus \(M^{-1}(Q_1)\subseteq U\), proving bijectivity.
Finally, \[ M'(w)=\frac{2}{(1-w)^2}\neq0 \] on \(U\), so \(M\) is conformal.
3. The squaring map\[ P:Q_1\longrightarrow\mathcal H, \qquad P(\zeta)=\zeta^2, \]where\[ \mathcal H=\{\xi\in\CC:\operatorname{Im}\xi>0\}, \]is a conformal bijection.
If \(\zeta=u+iv\in Q_1\), then \[ \operatorname{Im}(\zeta^2)=2uv>0. \] Conversely, each \(\xi\in\mathcal H\) has exactly one square root with argument in \[ \left(0,\frac{\pi}{2}\right), \] so exactly one square root lies in \(Q_1\). Hence \(P\) is bijective.
Since \[ P'(\zeta)=2\zeta\neq0 \] on \(Q_1\), it is conformal.
4. The Cayley map\[ C:\mathcal H\longrightarrow\DD, \qquad C(\xi)=\frac{\xi-i}{\xi+i}, \]is a conformal bijection.
For \(\xi=x+iy\) with \(y>0\), \[ \abs{\xi+i}^2-\abs{\xi-i}^2=4y>0, \] so \(\abs{C(\xi)}<1\). Its inverse is \[ C^{-1}(\eta)=i\frac{1+\eta}{1-\eta}, \] which maps \(\DD\) into \(\mathcal H\). Hence \(C\) is bijective. Also \[ C'(\xi)=\frac{2i}{(\xi+i)^2}\neq0 \] on \(\mathcal H\), so it is conformal.
5. Put\[ q(z)= \left( \frac{1+s(z)}{1-s(z)} \right)^2. \]A conformal bijection from the slit disk \(D\) onto the unit disk is\[ \boxed{ F(z)=\frac{q(z)-i}{q(z)+i}, \qquad s(z)=\exp\!\left(\frac12\Log z\right), \quad 0<\Arg z<2\pi. } \]
By steps <1>1–<1>4, \[ F=C\circ P\circ M\circ s \] is a composition of conformal bijections \[ D\xrightarrow{s}U\xrightarrow{M}Q_1 \xrightarrow{P}\mathcal H\xrightarrow{C}\DD. \] Therefore \(F\) is a conformal bijection from \(D\) onto \(\DD\).
6. Q.E.D.
Step 5 gives the requested map.