Let \[ A=\{z\in S^1:\operatorname{Im}z\ge0\}. \] Find a conformal map \[ f:\mathbb C\setminus\Delta\longrightarrow\mathbb C\setminus A. \]
Solution
One such map is \[ \boxed{ f(z)=\frac{z(z+1+i)}{1+(1-i)z} }. \]
1. The Möbius map\[ Q(z)=\frac{z-1}{z+1} \]maps the exterior domain \(E=\{z:|z|>1\}\) conformally onto the punctured right half-plane\[ H_R\setminus\{1\}, \qquad H_R=\{\zeta:\operatorname{Re}\zeta>0\}. \]
For \(|z|>1\), \[ \operatorname{Re}Q(z) =\frac{|z|^2-1}{|z+1|^2}>0. \] Conversely, if \(\operatorname{Re}\zeta>0\) and \(\zeta\ne1\), then \[ Q^{-1}(\zeta)=\frac{1+\zeta}{1-\zeta} \] is finite and satisfies \[ \left|\frac{1+\zeta}{1-\zeta}\right|^2-1 =\frac{4\operatorname{Re}\zeta}{|1-\zeta|^2}>0. \] Thus \(Q\) is a conformal bijection from \(E\) onto \(H_R\setminus\{1\}\); the omitted point \(1\) is the image of \(\infty\) on the Riemann sphere.
2. The Möbius map\[ M(w)=i\frac{1-w}{1+w} \]maps \(\mathbb C\setminus A\) conformally onto\[ \bigl(\mathbb C\setminus[0,\infty)\bigr)\setminus\{-i\}. \]
For \(w=e^{i\theta}\) with \(0\le\theta\le\pi\), \[ M(w) =i\frac{1-e^{i\theta}}{1+e^{i\theta}} =\tan\frac\theta2. \] Hence, on the Riemann sphere, \[ M(A)=[0,\infty], \] where \(w=-1\) maps to \(\infty\). Also \[ M(\infty)=-i. \] Since \(M\) is a Möbius transformation, deleting \(A\) and then deleting the source point \(\infty\) gives exactly \[ M(\mathbb C\setminus A) =\bigl(\mathbb C\setminus[0,\infty)\bigr)\setminus\{-i\}. \]
3. The map\[ T(\zeta) =-\left(\frac{\zeta+i}{\sqrt2}\right)^2 \]is a conformal bijection\[ H_R\setminus\{1\} \longrightarrow \bigl(\mathbb C\setminus[0,\infty)\bigr)\setminus\{-i\}. \]
The affine map \[ P(\zeta)=\frac{\zeta+i}{\sqrt2} \] is an automorphism of \(H_R\) and sends \[ 1\longmapsto p:=\frac{1+i}{\sqrt2}. \] The map \[ S(u)=-u^2 \] is a conformal bijection from \(H_R\) onto \(\mathbb C\setminus[0,\infty)\): its inverse is the branch \[ u=\sqrt{-t} \] with positive real part. Finally, \[ S(p)=-p^2=-i. \] Therefore \(T=S\circ P\) has the asserted restriction to the punctured domains.
4. The composition\[ f=M^{-1}\circ T\circ Q \]is therefore a conformal bijection from \(\mathbb C\setminus\Delta\) onto \(\mathbb C\setminus A\), and it simplifies to the displayed rational function.
Since \[ M^{-1}(t)=\frac{i-t}{i+t}, \] we have \[ f(z) =\frac{i+\frac12(Q(z)+i)^2}{i-\frac12(Q(z)+i)^2}. \] Now \[ Q(z)+i =\frac{(1+i)(z+i)}{z+1}, \] so \[ \frac12(Q(z)+i)^2 =i\frac{(z+i)^2}{(z+1)^2}. \] Hence \[ \begin{aligned} f(z) &=\frac{(z+1)^2+(z+i)^2}{(z+1)^2-(z+i)^2}\\ &=\frac{z(z+1+i)}{1+(1-i)z}. \end{aligned} \] Steps <1>1–<1>3 show that every factor in the composition is conformal and bijective on the indicated domain, so this \(f\) has precisely the required source and target.