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?

Let $z=\cos t+i\sin t$. Then $|z^2+\bar{z}^2|=1$ becomes
$$2|\cos(2t)|=1 $$
or
$$ \cos(2t)=\pm\frac{1}{2}$$
So
$$ \cos t=\pm\sqrt{\frac{1+\cos(2t)}{2}}=\pm\frac{\sqrt3}{2}\text{ or }\pm\frac{1}{2}$$
and 
$$ \sin t=\pm\sqrt{\frac{1-\cos(2t)}{2}}=\pm\frac12\text{ or }\pm\frac{\sqrt3}{2}. $$
Thus
$$ z=\cos t+i\sin t=\pm\frac{\sqrt3}{2}\pm\frac{1}{2}i,z=\pm\frac{1}{2}\pm\frac{\sqrt 3}{2}i. $$
