Let \[ E(a,b)=\left\{x+iy:\frac{x^2}{a^2}+\frac{y^2}{b^2}\le1\right\}, \qquad a>b>0. \] Find an explicit conformal homeomorphism \[ \mathbb C\setminus\Delta\longrightarrow\mathbb C\setminus E(a,b). \]
Solution
Set \[ \alpha=\frac{a+b}{2}, \qquad \beta=\frac{a-b}{2}. \] Then \[ \boxed{ F(z)=\alpha z+\frac{\beta}{z} =\frac{a+b}{2}z+\frac{a-b}{2z} } \] is a conformal homeomorphism from the exterior disk onto the exterior of \(E(a,b)\).
1. The unit circle is mapped homeomorphically onto the boundary ellipse.
For \(z=e^{i\theta}\), \[ \begin{aligned} F(e^{i\theta}) &=\alpha(\cos\theta+i\sin\theta) +\beta(\cos\theta-i\sin\theta)\\ &=(\alpha+\beta)\cos\theta +i(\alpha-\beta)\sin\theta\\ &=a\cos\theta+ib\sin\theta. \end{aligned} \] This is the standard one-to-one parametrization of \[ \partial E(a,b) =\left\{x+iy:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\right\}. \]
2. The map \(F\) is conformal and injective on \(\{z:|z|>1\}\).
Since \[ \frac{\beta}{\alpha}=\frac{a-b}{a+b}<1, \] the equation \[ F'(z)=\alpha-\frac{\beta}{z^2}=0 \] would force \[ |z|^2=\frac{\beta}{\alpha}<1. \] Thus \(F'\) has no zero for \(|z|>1\).
For injectivity, suppose \(F(z_1)=F(z_2)\). Then \[ 0 =\alpha(z_1-z_2)+\beta\left(\frac1{z_1}-\frac1{z_2}\right) =(z_1-z_2)\left(\alpha-\frac{\beta}{z_1z_2}\right). \] If \(z_1\ne z_2\), then \[ z_1z_2=\frac{\beta}{\alpha}, \] which is impossible when \(|z_1|,|z_2|>1\). Hence \(z_1=z_2\).
3. The image is exactly \(\mathbb C\setminus E(a,b)\).
The rational map \(F\) extends continuously to the exterior closed disk in the Riemann sphere, \[ \{z:|z|\ge1\}\cup\{\infty\}, \] with \(F(\infty)=\infty\). The injectivity argument in 2 remains valid when \(|z_1|,|z_2|\ge1\), and 1 gives injectivity on the boundary circle. Thus this extension is a homeomorphism onto its image.
By 1 its boundary is \(\partial E(a,b)\). The Jordan curve theorem gives exactly two complementary components of this ellipse. The interior of the exterior closed disk on the sphere, \[ \{z:|z|>1\}\cup\{\infty\}, \] contains \(\infty\), so it maps onto the complementary component containing \(\infty\), namely \[ \widehat{\mathbb C}\setminus E(a,b). \] Restricting to finite points gives \[ F:\mathbb C\setminus\Delta \longrightarrow \mathbb C\setminus E(a,b) \] as required.