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On a textbook, I've arrived at the following function:

$\displaystyle \phi(z)=\log{\frac{|z-\sqrt{(z²-1})|}{2}}$

and it says that the formula has a simple interpretation: the level curves of $\phi(z)$ are the ellipses with foci $-1, 1$. I know the problem is reduced to proving $|z-\sqrt{(z²-1})| = k$ is an ellipse, $k$ constant, but I don't know why this is true. So, my question is: Why is this an ellipse?

Aloizio Macedo
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1 Answers1

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Without caring about choosing a branch of the square root etc., when $z$ traverses a level curve of $\phi$, then

$$w = z-\sqrt{z^2-1}$$

traverses a circle. So let's identify the inverse mapping of $z \mapsto z - \sqrt{z^2-1}$:

$$\begin{align} w = z - \sqrt{z^2-1} &\iff \sqrt{z^2-1} = z-w\\ &\Rightarrow z^2-1 = z^2-2zw+w^2\\ &\iff 2zw = w^2+1\\ &\iff z = \frac12\left(w+\frac1w\right). \end{align}$$

Now write $w = r e^{i\varphi}$ to see that that maps circles to ellipses.

Daniel Fischer
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  • Sorry, but it isn't quite clear to me that $w+\frac{1}{w}$ maps circles to ellipses. – Aloizio Macedo Nov 26 '13 at 15:16
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    $$\begin{align}w+\frac1w &= re^{i\varphi} + \frac{1}{re^{i\varphi}} = r(\cos\varphi + i\sin\varphi) + \frac1r(\cos\varphi - i\sin\varphi)\&= \left(r+\frac1r\right)\cos\varphi + i\left(r-\frac1r\right)\sin\varphi\end{align}$$ Do you recognise a parametrisation of an ellipse there? – Daniel Fischer Nov 26 '13 at 15:19
  • Right! Thank you, I see now. – Aloizio Macedo Nov 26 '13 at 15:20