Find a conformal map from \(D : = \left\{ z \in \mathbb { C } : | z - i | > 1 , \mathrm { R e } ( z ) > 0 \right\}\) onto the open upper half plane.
Solution
Let \[ D=\{z\in\CC:\abs{z-i}>1,\ \operatorname{Re}z>0\} \] and let \[ Q_3=\{w\in\CC:\operatorname{Re}w<0,\ \operatorname{Im}w<0\}. \] Define \[ M(z)=\frac{z}{2i-z}. \]
1. The map \(M\) sends \(D\) into \(Q_3\).
Write \[ z=x+iy\in D \] and put \[ \Delta=\abs{2i-z}^2=x^2+(2-y)^2>0. \] Multiplying numerator and denominator by the conjugate of \(2i-z\) gives \[ \operatorname{Re}M(z) = \frac{1-\abs{z-i}^2}{\Delta} < 0 \] because \(\abs{z-i}>1\), and \[ \operatorname{Im}M(z) = \frac{-2x}{\Delta} < 0 \] because \(x>0\). Hence \(M(z)\in Q_3\).
2. The map\[ M:D\longrightarrow Q_3 \]is a conformal bijection with inverse\[ M^{-1}(w)=\frac{2iw}{1+w}. \]
Solving \[ w=\frac{z}{2i-z} \] for \(z\) gives the displayed inverse.
Conversely, let \[ w=u+iv\in Q_3 \] and set \[ z=\frac{2iw}{1+w}. \] Since \(u<0\) and \(v<0\), one has \(w\neq-1\), so the formula is defined. Put \[ \Gamma=\abs{1+w}^2=(1+u)^2+v^2. \] Direct calculation gives \[ \operatorname{Re}z = \frac{-2v}{\Gamma} > 0. \] Also \[ z-i=i\frac{w-1}{w+1}, \] so \[ \abs{z-i}^2 = \frac{\abs{w-1}^2}{\abs{w+1}^2}. \] Because \[ \abs{w-1}^2-\abs{w+1}^2=-4u>0, \] we obtain \[ \abs{z-i}>1. \] Thus \(z\in D\), proving that the displayed inverse maps \(Q_3\) into \(D\).
Finally, \[ M'(z)=\frac{2i}{(2i-z)^2}\neq0 \] on \(D\), so \(M\) is conformal.
3. The squaring map\[ S:Q_3\longrightarrow\mathcal H, \qquad S(w)=w^2, \]where\[ \mathcal H=\{\zeta\in\CC:\operatorname{Im}\zeta>0\}, \]is a conformal bijection.
If \[ w=u+iv\in Q_3, \] then \[ \operatorname{Im}(w^2)=2uv>0, \] so \(S(Q_3)\subseteq\mathcal H\).
Every \(\zeta\in\mathcal H\) has argument \[ 0<\theta<\pi. \] Its two square roots have arguments \[ \frac{\theta}{2} \qquad\text{and}\qquad \frac{\theta}{2}+\pi. \] Exactly the second lies in \[ \pi<\arg w<\frac{3\pi}{2}, \] which is \(Q_3\). Hence every \(\zeta\in\mathcal H\) has exactly one square root in \(Q_3\), proving bijectivity.
Since \[ S'(w)=2w\neq0 \] on \(Q_3\), the map is conformal.
4. A conformal bijection from \(D\) onto the upper half-plane is\[ \boxed{ F(z) = \left(\frac{z}{2i-z}\right)^2 }. \]
By steps <1>2–<1>3, \[ F=S\circ M \] is a composition of conformal bijections \[ D\xrightarrow{M}Q_3\xrightarrow{S}\mathcal H. \] Therefore \(F\) is a conformal bijection from \(D\) onto \(\mathcal H\).
5. Q.E.D.
Step 4 gives the requested map.