<s>[INST] 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?
 [/INST] You can write $z=e^{it}=\cos t+i\sin t$. Then $z^2+\overline z^2=2\cos 2t$.
So we need to solve $\cos2t=\pm\frac12$. Thus $2t=\pm\frac\pi3+2k\pi$
or $2t=\pm\frac{2\pi}3+2k\pi$, that is $t=\pm\frac\pi6+k\pi$ or
$t=\pm\frac\pi3+k\pi$, so that $z=e^{it}=\frac12(\pm\sqrt3\pm i)$
or $\frac12(\pm1\pm i\sqrt3)$.
</s>