Exhibit a conformal map from \(\{z\in\CC\mid\abs{z}<1,\ \Re z>0\}\) onto \(\DD=\{z\in\CC\mid\abs{z}<1\}\).
Solution
Let \[ H=\{z\in\CC:\abs{z}<1,\ \Re z>0\}, \qquad Q=\{w:\Re w<0,\ \Im w<0\}, \] and define \[ M(z)=\frac{z-i}{z+i}, \qquad S(w)=w^2, \qquad C(\zeta)=\frac{\zeta-i}{\zeta+i}. \]
1. \(M\) maps \(H\) conformally onto the open third quadrant \(Q\).
The boundary of \(H\) consists of the diameter \[ \{iy:-1<y<1\} \] and the right semicircle \(\{\abs{z}=1,\ \Re z>0\}\). For \(z=iy\) on the diameter, \[ M(iy)=\frac{y-1}{y+1}<0, \] so this boundary arc maps to the negative real ray.
A Möbius transformation maps generalized circles to generalized circles. Since the unit circle passes through \(i\) and \(-i\), and \[ M(i)=0,\qquad M(-i)=\infty, \] its image is a line through \(0\). At the point \(z=1\) of the right semicircle, \[ M(1)=-i, \] so the right semicircle maps to the negative imaginary ray. Finally, \[ M\left(\frac12\right)=-\frac35-\frac45 i, \] which lies in \(Q\). The Möbius transformation \(M\) is a homeomorphism of the Riemann sphere carrying \(\partial H\) onto \(\partial Q\), and it carries the point \(\tfrac12\in H\) into \(Q\), so it maps \(H\) bijectively and conformally onto \(Q\).
2. \(S\) maps \(Q\) conformally onto the upper half-plane.
Every \(w\in Q\) has a unique argument \[ \pi<\arg w<\frac{3\pi}{2}. \] Thus \[ 2\pi<\arg(w^2)<3\pi, \] which modulo \(2\pi\) is exactly the range \((0,\pi)\). Hence \(w^2\) lies in the upper half-plane. Conversely, every point of the upper half-plane has exactly one square root whose argument lies in \((\pi,3\pi/2)\), so \(S\) is bijective from \(Q\) onto the upper half-plane. Its derivative \(2w\) never vanishes on \(Q\), so it is conformal there.
3. \(C\) maps the upper half-plane conformally onto \(\DD\).
For \(\Im\zeta>0\), \[ \abs{\zeta-i}<\abs{\zeta+i}, \] so \(\abs{C(\zeta)}<1\). The inverse Möbius transformation \[ C^{-1}(u)=i\,\frac{1+u}{1-u} \] satisfies \[ \Im C^{-1}(u)=\frac{1-\abs{u}^2}{\abs{1-u}^2}>0 \] for \(\abs{u}<1\), so \(C\) is a bijection from the upper half-plane onto \(\DD\). Möbius transformations are conformal off their pole.
4. The map\[ \boxed{ F(z)=C(S(M(z))) =\frac{\left(\frac{z-i}{z+i}\right)^2-i} {\left(\frac{z-i}{z+i}\right)^2+i}} \]is a conformal bijection from \(H\) onto \(\DD\).
Steps <1>1–<1>3 show that \(M\), \(S\), and \(C\) are conformal bijections \(H\to Q\), \(Q\to\{\Im\zeta>0\}\), and \(\{\Im\zeta>0\}\to\DD\). Their composite \(F\) is therefore a conformal bijection \(H\to\DD\).
5. Q.E.D.
Step 4 exhibits the required conformal map.