The Joukowsky transformation with $z = x+i y$ $$ f(z) = \frac{1}{2}\left(z + \frac{1}{z}\right) $$ maps circles to ellipses. I wonder what is the connection of this transform to the Elliptic coordinate system ? $$ x = a \ \cosh \mu \ \cos \nu \\ y = a \ \sinh \mu \ \sin \nu $$
Connection between Joukowsky transform and elliptic coordinate system
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transformation
coordinate-systems
1 Answers
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We may parametrise the $z$-plane without zero by the "exponential polar coordinates" $z=e^{u+iv}$, where $u \in \mathbb{R}$ and $0 \leqslant v < 2\pi$. Examining the effect that $z \mapsto w = f(z)$ has on this, we find $$ f(e^{u+iv}) = \frac{1}{2}(e^{u+iv}+e^{-u-iv}) = \cosh{(u+iv)} = \cosh{u}\cos{v}+i\sinh{u}\sin{v}, $$ which are elliptic coordinates with $a=1$. Replacing the Joukowsky transform with $(z+a^2/z)/2$ and using polar coordinates $ae^{u+iv}$ instead gives the more general elliptic coordinates with $a \neq 1$.