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Which is the concrete formula for the conformal mapping (normalized at infinity), acting from $\mathbb C \backslash D^*$ onto

$\mathbb C\backslash[-1, 1]$?

Here $\mathbb C$ denotes the set of all complex numbers and $D^*$ denotes the closed unit disk of the complex plane.

Also, I would be interested in references containing many examples of such of conformal mappings, by replacing the interval $[-1, 1]$ with other various subsets of the complex plane.

Thanks a lot.

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closed as too localized by Andreas Thom, Pete L. Clark, Yemon Choi, Willie Wong, S. Carnahan Nov 13 '10 at 12:34

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You can find this information in almost any complex analysis book. –  S. Carnahan Nov 10 '10 at 16:54
    
In more detail: compose the Koebe function planetmath.org/encyclopedia/KoebeFunction.html with an appropriate Moebius transformation –  Yemon Choi Nov 10 '10 at 18:15
    
See also en.wikipedia.org/wiki/Joukowsky_transform –  S. Carnahan Nov 13 '10 at 12:36
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1 Answer 1

Finally I found that the conformal mapping is given by the formula f(z)=(1/2)(z + 1/z). Thanks any whay.

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