Prove that the level curves of $\log\frac{|z-\sqrt{z^2-1}|}{2}$ are ellipses

Question

In a book it was mentioned that the level curves of the complex function $\phi (z) = \log\frac{|z-\sqrt{z^2-1}|}{2}$ is an ellipse with foci $1$ and $-1$. I can not establish a relation between the ellipse equation in the complex plane $|z-a|+|z-b|=c$ and the function $\phi(z)$. Is the change of variable possible here? (for example I tried $z = \sinh\omega$ but it did not work).

Can anyone help me to prove it?

Thanks in advance.

Answer 1

You can just ignore $\log$ and $2$ in the fraction, and concentrate on $|z-\sqrt{z^2-1}| = r$ for $r>0$.

Let $w = z-\sqrt{z^2-1}$. Then $z - w = \sqrt{z^2-1}$ and $(z-w)^2 = z^2 - 1$. It follows that $z = \frac{w^2+1}{2w}.$ We have

\begin{align} \left| \frac{w^2+1}{2w} + 1 \right| + \left| \frac{w^2+1}{2w} - 1 \right| &= \left| \frac{(w+1)^2}{2w}\right| + \left| \frac{(w-1)^2}{2w}\right|\\ &= \frac{(w+1)(\overline w + 1)}{2|w|} + \frac{(w-1)(\overline w - 1)}{2|w|}\\ &= \frac{w\overline w + w + \overline w + 1}{2|w|} + \frac{w\overline w - w - \overline w + 1}{2|w|}\\ &= \frac{|w|^2 + 1}{|w|} \end{align}

and thus if $|w|$ is constant, so is $\left| \frac{w^2+1}{2w} + 1 \right| + \left| \frac{w^2+1}{2w} - 1 \right|$. Therefore, solutions of $|z-\sqrt{z^2-1}|=r$ lie on an ellipse.

However, it seems the contours are actually not the whole ellipses.

Answer 2

You know that $|z - \sqrt{z^2 - 1}| = c$, where c is a real number. This defines some level curve of $\phi(z)$. Try substituting $z = x + iy$ and seeing whether this leads to the equation of an ellipse.