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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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    Well, you can always modify this map by multiplying by some constant and substituting $\alpha z$ for $z$. And yes, it should be R > 1.2012-11-14

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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 $z\mapsto z/C$.