Find a bijective conformal map from
\[ G : = \{ z \in \mathbb { C } : | z - 1 | < { \sqrt { 2 } } , ~ | z + 1 | < { \sqrt { 2 } } \} ~ \backslash ~ [ 0 , i ) \]
onto the open upper half plane.
Solution
Let \[ L= \{z\in\CC:\abs{z-1}<\sqrt2,\ \abs{z+1}<\sqrt2\} \] be the unslit lens and define \[ M(z)=\frac{z-i}{z+i}. \] Also put \[ W= \{w\in\CC:3\pi/4<\arg w<5\pi/4\}, \] where the argument on \(W\) is chosen in that interval.
1. The map\[ M:L\longrightarrow W \]is a conformal bijection.
Write \[ z=x+iy \] and \[ M(z)=u+iv. \] Since \[ \abs{z+i}^2=x^2+(y+1)^2, \] direct calculation gives \[ u = \frac{\abs z^2-1}{\abs{z+i}^2}, \qquad v = \frac{-2x}{\abs{z+i}^2}. \] Therefore \[ u+v = \frac{\abs z^2-1-2x}{\abs{z+i}^2} = \frac{\abs{z-1}^2-2}{\abs{z+i}^2} \] and \[ u-v = \frac{\abs z^2-1+2x}{\abs{z+i}^2} = \frac{\abs{z+1}^2-2}{\abs{z+i}^2}. \] Thus \[ z\in L \quad\Longleftrightarrow\quad u+v<0 \ \text{ and }\ u-v<0. \] These two inequalities describe exactly the wedge \[ 3\pi/4<\arg w<5\pi/4. \]
The inverse Möbius map is \[ M^{-1}(w)=i\frac{1+w}{1-w}. \] Since \(1\notin W\), it is defined on all of \(W\), and the displayed equivalences show that \(M^{-1}(W)\subseteq L\). Hence \(M\) is bijective.
Finally, \[ M'(z)=\frac{2i}{(z+i)^2}\neq0 \] on \(L\), because \(-i\) is a boundary point of the lens. Thus \(M\) is conformal.
2. The removed slit satisfies\[ M([0,i))=[-1,0). \]
For \[ z=iy, \qquad 0\leq y<1, \] one has \[ M(iy) = \frac{y-1}{y+1}. \] As \(y\) runs from \(0\) to \(1\), this runs monotonically from \(-1\) toward \(0\) without attaining \(0\). Hence the image is exactly \([-1,0)\).
3. The fourth-power map\[ P:W\longrightarrow \CC\sm(-\infty,0], \qquad P(w)=w^4, \]is a conformal bijection.
If \[ w=re^{i\theta}, \qquad 3\pi/4<\theta<5\pi/4, \] then \[ 3\pi<4\theta<5\pi. \] Subtracting \(4\pi\) gives an argument in \((-\pi,\pi)\), so \(w^4\) lies in the plane slit along the nonpositive real axis.
Conversely, for \[ \xi=Re^{i\phi}, \qquad R>0, \quad -\pi<\phi<\pi, \] there is a unique fourth root \[ w=R^{1/4}e^{i(\phi+4\pi)/4} \] whose argument lies in \((3\pi/4,5\pi/4)\). Thus \(P\) is bijective.
Since \[ P'(w)=4w^3\neq0 \] on \(W\), it is conformal.
4. The composition\[ Q(z)=M(z)^4-1 \]is a conformal bijection from the slit lens \(G\) onto\[ \CC\sm(-\infty,0]. \]
By steps <1>1–<1>3, \[ P\circ M \] maps the unslit lens \(L\) bijectively onto \[ \CC\sm(-\infty,0]. \] Step 2 gives \[ P(M([0,i))) = (0,1]. \] Therefore removing the slit from \(L\) removes exactly \((0,1]\) from the slit-plane image: \[ (P\circ M)(G) = \CC\sm(-\infty,1]. \] Subtracting \(1\) yields \[ Q(G)=\CC\sm(-\infty,0]. \] Translation preserves conformality and bijectivity.
5. On\[ \CC\sm(-\infty,0] \]take the principal logarithm and define\[ s(\xi) = \exp\!\left(\frac12\Log\xi\right). \]Then\[ s:\CC\sm(-\infty,0] \longrightarrow R, \qquad R=\{\eta\in\CC:\operatorname{Re}\eta>0\}, \]is a conformal bijection.
For the principal argument \[ -\pi<\Arg\xi<\pi, \] the argument of \(s(\xi)\) lies in \[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right), \] so \(s(\xi)\in R\).
Conversely, every \(\eta\in R\) has argument in that interval, so \[ \xi=\eta^2 \] has principal argument in \((-\pi,\pi)\) and satisfies \(s(\xi)=\eta\). Thus \(s\) is bijective.
Also \[ s'(\xi)=\frac{s(\xi)}{2\xi}\neq0, \] so it is conformal.
6. A bijective conformal map from \(G\) onto the upper half-plane is\[ \boxed{ F(z) = i\exp\!\left[ \frac12 \Log\!\left( \left(\frac{z-i}{z+i}\right)^4-1 \right) \right], } \]where \(\Log\) is the principal logarithm.
By steps <1>4–<1>5, \[ s\circ Q \] maps \(G\) conformally and bijectively onto the right half-plane. Multiplication by \(i\) sends the right half-plane onto \[ \mathcal H=\{w\in\CC:\operatorname{Im}w>0\}. \] Hence the displayed \(F\) is the required conformal bijection.
7. Q.E.D.
Step 6 gives the requested map.