Conformal map of ellipse interior to a disk

This post will present the conformal map between the interior of an ellipse and the unit disk.

Given an ellipse centered at the origin with semi-major axis a and semi-minor axis b. We will assume without loss of generality that a² − b² = 1 and so the foci are at ±1.

Hermann Schwarz published the conformal map from the ellipse to the unit disk in 1869 [1, 2].

The map is given by

$f(z) = \sqrt{k}\,\, \text{sn} \left( \frac{2K}{\pi} \sin^{-1}(z) \right)$

where sn is the Jacobi elliptic function with parameter k². The constants k and K are given by

$\begin{align*} q &= (a + b)^{-4} \\ k &= \left(\frac{\theta_2(q)}{\theta_3(q)}\right)^2 \\ K &= \frac{\pi}{2} \theta_3(q)^2 \end{align*}$

where θ₂ and θ₃ are theta constants, the value so the theta functions θ₂(z, q) and θ₃(z, q) at z = 1.

Conformal maps to the unit disk are unique up to rotation. The map above is the unique conformal map preserving orientation:

$\begin{align*} f(0) &= 0 \\ f(\pm a) &= \pm 1 \\ f(\pm bi) &= \pm i \\ \end{align*}$

Inverse map

The inverse of this map is given by

$g(w) = \sin\left(\frac{\pi}{2K} \,\, \text{sn}^{-1} \left(\frac{w}{\sqrt{k}}\right) \right)$

The inverse of the sn function with parameter m can be written in terms of elliptic integrals.

$\text{sn}^{-1}(z; m) = F\left(\sin^{-1}(u; m)\right)$

where F is the incomplete elliptic integral of the first kind and m is the parameter of sn and the parameter of F.

Plot

I wanted to illustrate the conformal map using an ellipse with aspect ratio 1/2. To satisfy a² − b² = 1, I set a = 2/√3 and b = 1/√3. The plot at the top of the post was made using Mathematica.

[1] H. A. Schwarz, Über eigige Abbildungsaufgaben, Journal für di reine und angew. Matheamatik, vol 70 (1869), pp 105–120

[2] Gabor Szegö. Conformal Mapping of the Interior of an Ellipse onto a Circle. The American Mathematical Monthly, 1950, Vol. 57, No. 7, pp. 474–478

Inverse map

Plot

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