Find a conformal map from \(D = \{ z : \ | z | < 1 , \ | z - 1 / 2 | > 1 / 2 \}\) to the unit disk \(\ \Delta = \{ z :\) \(| z | < 1 \}\)
Solution
Let \[ D=\left\{z\in\CC:\abs z<1,\ \abs{z-\frac12}>\frac12\right\} \] and define \[ M(z)=\frac{1+z}{1-z}. \] Also put \[ S=\{w\in\CC:0<\operatorname{Re}w<1\}. \]
1. The map\[ M:D\longrightarrow S \]is a conformal bijection.
For \(z\neq1\), \[ \operatorname{Re}M(z) = \frac{1-\abs z^2}{\abs{1-z}^2}. \] Hence \[ \abs z<1 \quad\Longleftrightarrow\quad \operatorname{Re}M(z)>0. \] Moreover, \[ \begin{aligned} 1-\operatorname{Re}M(z) &= \frac{\abs{1-z}^2-(1-\abs z^2)}{\abs{1-z}^2}\\ &= \frac{2(\abs z^2-\operatorname{Re}z)}{\abs{1-z}^2}\\ &= \frac{2\left(\abs{z-\frac12}^2-\frac14\right)} {\abs{1-z}^2}. \end{aligned} \] Thus \[ \abs{z-\frac12}>\frac12 \quad\Longleftrightarrow\quad \operatorname{Re}M(z)<1. \] These two equivalences show that \(M(D)\subseteq S\).
Solving for \(z\) gives \[ M^{-1}(w)=\frac{w-1}{w+1}. \] For \(w\in S\) one has \(w\neq-1\), and applying the two displayed equivalences to \(z=M^{-1}(w)\) gives \[ \abs z<1, \qquad \abs{z-\frac12}>\frac12. \] Hence \(M^{-1}(S)\subseteq D\), proving bijectivity.
Finally, \[ M'(z)=\frac{2}{(1-z)^2}\neq0 \] on \(D\), so \(M\) is conformal.
2. The map\[ E:S\longrightarrow\mathcal H, \qquad E(w)=e^{\pi i w}, \]where\[ \mathcal H=\{\zeta\in\CC:\operatorname{Im}\zeta>0\}, \]is a conformal bijection.
Write \[ w=u+iv, \qquad 0<u<1. \] Then \[ E(w)=e^{-\pi v}e^{i\pi u}, \] whose argument lies in \((0,\pi)\). Therefore \(E(w)\in\mathcal H\).
Conversely, every \[ \zeta=re^{i\theta}\in\mathcal H, \qquad r>0, \quad 0<\theta<\pi, \] has the preimage \[ w=\frac{\theta}{\pi}-i\frac{\log r}{\pi}\in S. \] Thus \(E\) is onto.
If \(E(w_1)=E(w_2)\), then \[ w_1-w_2\in2\ZZ. \] The real parts of \(w_1,w_2\) both lie in \((0,1)\), so their difference has real part in \((-1,1)\). Hence the only possible element of \(2\ZZ\) is \(0\), and \(w_1=w_2\). Thus \(E\) is injective.
Since \[ E'(w)=\pi i e^{\pi i w}\neq0, \] the map is conformal.
3. The map\[ C:\mathcal H\longrightarrow\DD, \qquad C(\zeta)=\frac{\zeta-i}{\zeta+i}, \]is a conformal bijection.
For \(\zeta=x+iy\) with \(y>0\), \[ \abs{\zeta+i}^2-\abs{\zeta-i}^2=4y>0, \] so \[ \abs{C(\zeta)}<1. \] Writing \(\eta=C(\zeta)\) and solving for \(\zeta\) gives \[ C^{-1}(\eta)=i\frac{1+\eta}{1-\eta}, \] and the inverse Cayley map sends \(\DD\) into \(\mathcal H\). Hence \(C\) is bijective. Its derivative \[ C'(\zeta)=\frac{2i}{(\zeta+i)^2} \] never vanishes on \(\mathcal H\), so \(C\) is conformal.
4. A conformal bijection from the given domain onto the unit disk is\[ \boxed{ F(z) = \frac{ \exp\!\left(\pi i\frac{1+z}{1-z}\right)-i }{ \exp\!\left(\pi i\frac{1+z}{1-z}\right)+i }. } \]
By steps <1>1–<1>3, \[ F=C\circ E\circ M \] is a composition of conformal bijections \[ D\xrightarrow{M}S\xrightarrow{E}\mathcal H\xrightarrow{C}\DD. \] Therefore \(F\) is a conformal bijection from \(D\) onto the unit disk.
5. Q.E.D.
Step 4 gives the requested map.