What is the locus of $z^2+\bar{z}^2=2$?

Question

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.

Answer 1

The locus of $$z^2+(\bar z)^2=2 \implies (x+iy)^2+(x-iy)^2=2 \implies x^2-y^2=1$$ which is a hyperbola.

Answer 2

Let $z = x+iy$ to get $x=\frac12(z+\bar z)$, $y=-\frac i2(z-\bar z)$, and

$$x^2-y^2=\frac14(z+\bar z)^2 + \frac14(z-\bar z)^2 =\frac12(z^2+\bar z^2)= 1$$

which shows that the locus is an equal-axis hyperbola.

Answer 3

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