Let \(a>b>0\), and let \(E\subset\mathbb C\) be the ellipse with major axis \([-a,a]\) and minor axis \([-ib,ib]\). Let \(I\subset[-a,a]\) be the line segment joining the foci of \(E\), and let \(B\) be the annular region between \(E\) and \(I\).
Find \(R>1\) and an explicit conformal map \[ f:A(R)=\{z:1<|z|<R\}\longrightarrow B. \]
Hint: use a map of the form \(z+c/z\).
Solution
Let \[ c=\sqrt{a^2-b^2}, \qquad R=\frac{a+b}{c} =\sqrt{\frac{a+b}{a-b}}. \] Then \(c\) is the focal distance of the ellipse, and an explicit conformal map is \[ \boxed{ F(z)=\frac c2\left(z+\frac1z\right), \qquad 1<|z|<R.} \]
1. The number \(R\) satisfies \(R>1\) and\[ \frac c2\left(R+R^{-1}\right)=a, \qquad \frac c2\left(R-R^{-1}\right)=b. \]
Since \(a>b>0\), \[ c=\sqrt{(a-b)(a+b)}>0 \] and \[ R=\frac{a+b}{\sqrt{(a-b)(a+b)}} =\sqrt{\frac{a+b}{a-b}}>1. \] Also \[ R^{-1}=\sqrt{\frac{a-b}{a+b}}=\frac{a-b}{c}. \] Therefore \[ \frac c2(R+R^{-1}) =\frac12((a+b)+(a-b))=a, \] and similarly \[ \frac c2(R-R^{-1}) =\frac12((a+b)-(a-b))=b. \]
2. The inner boundary circle \(|z|=1\) is mapped onto the focal segment\[ I=[-c,c]. \]
For \(z=e^{i\theta}\), \[ F(e^{i\theta}) =\frac c2(e^{i\theta}+e^{-i\theta}) =c\cos\theta. \] As \(\theta\) runs from \(0\) to \(2\pi\), \(c\cos\theta\) runs over the segment \([-c,c]\).
The foci of the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] are at \(\pm\sqrt{a^2-b^2}=\pm c\), so this is exactly the segment \(I\) from the statement.
3. The outer boundary circle \(|z|=R\) is mapped onto the ellipse \(E\).
For \(z=Re^{i\theta}\), \[ \begin{aligned} F(z) &=\frac c2\left(Re^{i\theta}+R^{-1}e^{-i\theta}\right)\\ &=\frac c2(R+R^{-1})\cos\theta +i\frac c2(R-R^{-1})\sin\theta. \end{aligned} \] By 1 this becomes \[ F(Re^{i\theta}) =a\cos\theta+ib\sin\theta, \] which parametrizes the ellipse with major axis \([-a,a]\) and minor axis \([-ib,ib]\).
4. The derivative of \(F\) does not vanish on \(A(R)\).
We have \[ F'(z)=\frac c2\left(1-\frac1{z^2}\right). \] Thus \(F'(z)=0\) only at \(z=\pm1\), both of which lie on the inner boundary and not in the open annulus \(1<|z|<R\).
5. The map \(F\) is injective on \(A(R)\).
Suppose \(F(z_1)=F(z_2)\). Then \[ z_1+z_1^{-1}=z_2+z_2^{-1}. \] Multiplying by \(z_1z_2\) and factoring gives \[ (z_1-z_2)(z_1z_2-1)=0. \] Hence either \(z_1=z_2\) or \(z_2=1/z_1\).
If \(z_1\in A(R)\), then \(|z_1|>1\), so \[ |1/z_1|<1. \] Thus the second possibility cannot also lie in \(A(R)\). Therefore \(z_1=z_2\), proving injectivity.
6. The image \(F(A(R))\) is exactly the annular region \(B\) between \(I\)and \(E\).
For every radius \(1<r<R\), the same calculation as in 3 shows that \(F(|z|=r)\) is the ellipse \[ x=\frac c2(r+r^{-1})\cos\theta, \qquad y=\frac c2(r-r^{-1})\sin\theta, \] with foci \(\pm c\).
As \(r\) increases from \(1\) to \(R\), the semimajor and semiminor axes \[ \frac c2(r+r^{-1}), \qquad \frac c2(r-r^{-1}) \] increase continuously from \((c,0)\) to \((a,b)\). Hence these nested confocal ellipses foliate exactly the region between the focal segment \(I\) and the outer ellipse \(E\).
Thus \[ F(A(R))=B. \]
7. Therefore \(F:A(R)\to B\) is conformal.
By 4, \(F\) is locally conformal; by 5 it is injective; and by 6 its image is \(B\). Hence \(F\) is an analytic homeomorphism from \(A(R)\) onto \(B\).