<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] Use exponential form: since $|z|=1$, $z=\mathrm e^{i\theta}$, so the second equation can be rewritten as 
$$1=\bigl|z^2+\bar z^2\bigr|^2=\bigl(z^2+\bar z^2\bigr)^2=\bigl(\mathrm e^{2i\theta}+\mathrm e^{-2i\theta}\bigr)^2=\mathrm e^{4i\theta}+\mathrm e^{-4i\theta}+2=2\cos4\theta+2,$$
so finally we have to solve
$$\cos4\theta=-\frac12\iff 4\theta\equiv =\pm\frac{2\pi}3\pmod{2\pi}\iff\theta\equiv\pm\frac\pi6\mod{\frac\pi2}.$$
This corresponds the $8$ different complex numbers in $\mathbf U$:*
$$z=\mathrm e^{\pm\tfrac{k\pi}6},\quad k=1,2,4,5.$$
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