I was originally looking for a conformal map that maps a punctured unit disc to unit disc. The only answer I can find lead to this resource.
The final step of the answer given rely on a conformal map that maps an ellipse to a unit disc. Although we know such a map exist by Riemann Mapping Theorem, is there any way to write down the map explicitly (let's say, length of axes are given)?
The only related formula I can find is Joukowski transformation which does the other way around.
The Joukowski map produces a slit ellipse, and is of no help here.
For a full ellipse there is no simple formula, but an interesting procedure, using the so-called Bergman kernel. See
Peter Henrici, Applied and computational complex analysis, Volume 3, Wiley 1986, pp. 529–552. The case of the ellipse is treated on p. 546 and p. 550.
The shape of the ellipse is specified by a parameter $k$ in the range $0 < k < 1$, called the elliptic modulus. The incomplete elliptic integral of the first kind (or inverse Jacobi AM function) is defined as $$ F(\phi,k) := \int_0^\phi \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}} \, . $$ Let $c>0$ be another parameter. Then, the following function $$ f(z):=c \sin \left( \frac{\pi}{2} \, \frac{F\left(\arcsin\left(k^{-\frac{1}{2}}z\right), k\right)}{F\left(\frac{\pi}{2},k\right)}\right) $$ maps the unit disk $|z|<1$ to an elliptic region $\frac{x^2}{a^2}+\frac{y^2}{b^2}<1$.
For example, if $k=\frac{1}{\sqrt{2}}$, using Maple I generated the plot:
The parameters $k$ and $c$ can be computed from the semiaxes $a > b$ using the following formulas: $c=\sqrt{a^2-b^2}$ and $$ k = 4 \sqrt{q} \left( \frac{(1+q^2)(1+q^4)(1+q^6)\cdots}{(1+q)(1+q^3)(1+q^5)\cdots}\right)^4 \quad \text{where } q = \left( \frac{a-b}{a+b} \right)^2 \, . $$
Reference: H. A. Schwarz. Gesammelte mathematische Abhandlungen, Band 2. Springer, Berlin (1890), pp. 102-107 (in Italian).
