The standard way to transform elliptic coordinates $(\mu, \nu)$ $\ to$ Cartesian coordinates $(x,y)$:
$x = a \cosh(\mu) \cos(\nu)$
$y = a \sinh(\mu) \sin(\nu)$
Is there any way to get the transformation $(x,y)$ to $(\mu,\nu)$? Meaning is there a way to find:
$\mu = f(x,y)$
$\nu = g(x,y)$
I'm guessing that it would involve $\sinh^{-1}$'s and $\cosh^{-1}$'s, if it was possible to do this at all.
1 Answer
According to this the complex form of the system is $x+iy=\cosh(\mu+i\nu).$ To me that was a surprise but it checks easily using the definitions of sinh, cosh, and even/odd properties of sine and cosine. In a formula reference book I found that $$\cosh^{-1}(z)=\ln(z+\sqrt{z^2-1})=i \arccos z.$$ I'm not expert enough to say anything about choosing the log branch for the log or the arccosine here.
Note: One has to use the fact that $\cosh z=\cosh -z$ when inverting via $\cosh^{-1}$ in order to assure the relation $\mu \ge 0$ in the transformation. Given $z$ if left as $x+iy,$ extracting the squareroot inside the log is messy to say the least.
