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?
 A: 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
$$ z=\cos t+i\sin t=\pm\frac{\sqrt3}{2}\pm\frac{1}{2}i,z=\pm\frac{1}{2}\pm\frac{\sqrt 3}{2}i. $$
A: 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 $$\left\{\pm\frac{\sqrt3}{2}\pm\frac{1}{2}i,\pm\frac{1}{2}\pm\frac{\sqrt3}{2}i\right\}$$
A: 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 $z^2\pm \sqrt{3}z+1=0$ 
A: 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 $\frac12(\pm1\pm i\sqrt3)$.
A: 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 $\mathbf U$:*
$$z=\mathrm e^{\pm\tfrac{k\pi}6},\quad k=1,2,4,5.$$
