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.

  • $\begingroup$ EDIT: I made a very stupid typo in the above. It's fixed now. $\endgroup$ Commented Nov 15, 2015 at 4:50
  • 1
    $\begingroup$ These (edited) equations don't give as simple an elimination of $\mu$ as in my previous comment, which I'll now delete. $\endgroup$
    – coffeemath
    Commented Nov 15, 2015 at 7:11
  • $\begingroup$ Related: math.stackexchange.com/questions/277140/… $\endgroup$ Commented Jul 28, 2017 at 9:24

2 Answers 2


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.

  • $\begingroup$ I don't really understand what you mean by the note. How do I use it to invert the relation involving the $\cosh$? $\endgroup$ Commented Oct 12, 2021 at 8:55

The inversion is $$\mu = \text{arccosh}\frac{\sqrt{(x+a)^2+y^2}+\sqrt{(x-a)^2+y^2}}{2a}\\ \nu = \text{sign}(y)~ \text{arccos}\frac{\sqrt{(x+a)^2+y^2}-\sqrt{(x-a)^2+y^2}}{2a} $$

Note the sign(y) is needed to give the correct sign on the whole $xy$-plane.


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