Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4+y^4$$. I am stuck here. Is it wrong approach? is there an easier one?

Firstly, $z=x+yi$, where $x$ and $y$ are reals.
Thus, it should be $$|x^2-y^2|=\frac{1}{2}$$ and $$x^2+y^2=1.$$
Finally, we obtain $$\left\{\pm\frac{\sqrt3}{2}\pm\frac{1}{2}i,\pm\frac{1}{2}\pm\frac{\sqrt3}{2}i\right\}$$
