What is the locus of $z^2+\bar{z}^2=2$? I have to prove that it's the equation of an equilateral hyperbola
$$z^2+z^{-2}=2$$
I try this 
$z^2+z^{-2} +2 = 2+2$
$(z^2+1/z^2 +2 ) = 4$
$(z+1/z)^2=4$
$ z+1/z= 2 $
$ x+yi + 1/(x+yi) = 2 $
$ ((x+yi)^2+1)/(x+yi)=2$
$ x^2+2xyi-y^2+1=2x+2yi$
$ x^2-2x+1 +2xyi -y^2=2yi $
$(x-1)^2 +2xyi-y^2= 2yi$
I don't know how to end it.

Hint One can rewrite this quickly using polar coordinates: For $z = r e^{i \theta}$, we have
$$2 = z^2 + \bar z^2 = (r e^{i\theta})^2 + (r e^{-i\theta})^2 = r^2 (e^{2 i \theta} + e^{-2 i \theta}) = r^2 \cdot 2 \cos 2 \theta .$$
Now, divide both sides by $2$, apply the double angle identity for $\cos$, and rewrite in terms of the components $x, y$ of $z = x + i y$.

 Doing so gives $$1 = r^2 (\cos^2 \theta - \sin^2 \theta) = (r \cos \theta)^2 - (r \sin \theta)^2 = x^2 - y^2 .$$

