In this question, Ideas for parameterizing this curve in the complex plane and calculating its length by (numerical) contour integration?, I plotted the imaginary part of $$\tanh(\ln(1+Z(t)^2))$$ and conjectured in The Hyperbolic Tangent of the Logarithm of One Plus The Square of The Hardy Z Function that there is an analytic curve, either via circle inversion or actually more likely, inversion with respect to an ellipse, that will show that the imaginary part of $Y(t)$ is equal to the real part $Y(t)$ transformed via elliptical inversion around the roots . I would like to find the complex plane-curve which passes thru the singular parts of the curve $Y(t)$ , which appear in the image as the white points on the zero-lines . I believe the point where the real and imaginary curves vanish when $Z(t)= \pm i\times\sqrt{(2)}$ are the foci of the ellipse. There are 4 points that need to be on the ellipse, but half of them are symmetric, same with the foci. Are these foci and points on the curve enough to uniquely define the ellipse of inversion?
This graph shows the curves where the imaginary and real parts vanish.. the hyperbolas and the vertical curved line are the vanishing imaginary parts, the lemniscate curve is the vanishing real part.
This graph shows the circle inversion of the imaginary part of Y: I think the appropriately placed elliptical inversion would show that the curves match exactly
