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?

Since $w+\overline{w}\in \mathbb{R}$ for each $w\in \mathbb{C}$
we have $z^2+ \overline{z}^2 =1$ and from $|z|=1$ we have $z\overline{z}=1$.
So $$(z+\overline{z})^2=  z^2+2z\overline{z} +\overline{z}^2 =3$$ 
and thus we have  $z+\overline{z} =\pm \sqrt{3}$
So, since $\overline{z} = {1\over z} $ we have $z^2\pm \sqrt{3}z+1=0$ 
