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Find a conformal bijection $f(z):\mathbb{C}\setminus D\rightarrow \mathbb{C}\setminus E(a, b)$ where $E(a, b)$ is the ellipse $\{x + iy : \frac{x^2}{a}+\frac{y^2}{b}\leq1\}$

Here $D$ denotes the closed unit disk.

I hate to ask having not given the question a significant amount of thought, but due to illness I missed several classes, really need to catch up, and the text book we're using (Ahlfors) doesn't seem to have anything on the mapping of circles to ellipses except a discussion of level curves on pages 94-95. and I can't figure out how to get there through composition of the normal elementary maps (powers, exponential and logarithmic), and fractional linear transformations take circles into circles and are therefore useless for figuring this out.

I prefer hints, thanks in advance.

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$z \mapsto \tfrac{1}{2} \left( z + \tfrac{1}{z} \right)$ maps $\{ z \mid |z| > R \}$ for each $R > 0$ biholomorphically to the exterior of some ellipse (depending on $R$) with foci $z = \pm 1$. Modify this map based on the foci of the given ellipse. – WimC Nov 13 '12 at 5:45
This is the map discussed, but I don't understand how this could work, because this map takes a given circle to an ellipse with a fixed major and minor axis based only upon the radius of the original circle. I need for a fixed radius to go to an arbitrary ellipse, with axes a and b. This means I need to find a way to make $a=r+\frac{1}{r} and b=r-\frac{1}{r} satisfy any value even they are both functions of r - how would one show this is generally possible? – Quantumpencil Nov 14 '12 at 1:47
Also this map is not conformal at 1 and -1, how can I map the boundary of the unit disk to anything? – Quantumpencil Nov 14 '12 at 1:50
Well, you can always modify this map by multiplying by some constant and substituting $\alpha z$ for $z$. And yes, it should be $R > 1$. – WimC Nov 14 '12 at 6:04

The conformal map $z\mapsto z+z^{-1}$ sends $\{|z|>R\}$ onto the exterior of ellipse with semi-axes $A=R+R^{-1}$ and $B=R-R^{-1}$. Note that $A^2-B^2=4$. Thus, you should multiply the given $a,b$ by a constant $C$ such that $(Ca)^2-(Cb)^2=4$, then solve $Ac=R+R^{-1}$ for $R$. After applying the map given above, the final step is $z\mapsto z/C$.

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