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Could we find two complex numbers such as $z_1$ and $z_2$, where both of them are roots of unity (de Moivre numbers) and also when we define

$$s = z_1 + z_2$$

$s$ is also a root of unity?

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If $|z_1|=|z_2|=1$ and $|z_1+z_2|=1$, then $$1=|z_1+z_2|^2=(z_1+z_2)(\bar{z}_1+\bar{z}_2)=z_1\bar{z}_1+z_2\bar{z}_2+z_1\bar{z}_2+\bar{z}_1z_2.$$ That is, $$1=|z_1|^2+|z_2|^2+2\Re(z_1\bar{z}_2)=1+1+2\Re(z_1\bar{z}_2).$$ Hence, $$\Re\left(\frac{z_1}{z_2}\right)=\Re(z_1\bar{z}_2)=-\frac12.$$ However, $$\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}=\frac{1}{1}=1.$$ That is, $$\frac{z_1}{z_2}=\frac{-1\pm\sqrt{3}i}{2}=\left(\frac{1\pm\sqrt{3}i}{2}\right)^2.$$ This means $$(z_1,z_2)=(u\alpha,u\alpha^{-1})$$ for some complex number $u$ such that $|u|=1$ and $\alpha=\frac{1\pm\sqrt{3}i}{2}$; in this case, $z_1+z_2=u$. If you want $z_1$ and $z_2$ be roots of unity, then it automatically follows that $u$ is also a root of unity.

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  • $\begingroup$ How did you conclude that Re(z1/z2) = Re(z1 * z2_)? (4th equality) $\endgroup$ – Jigsaw Oct 20 '19 at 17:24
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    $\begingroup$ If $|z|=1$, then $\bar{z}=1/z$. $\endgroup$ – Batominovski Oct 20 '19 at 17:24
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Yes. You can take $z_1=\frac12+\frac{\sqrt3}2i$ and $z_1=\frac12-\frac{\sqrt3}2i$. Then $s=1$

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Suppose the difference of their arguments is $\;\theta_1-\theta_2=\varphi$. Using Al-Kashi's theorem, we have $$|z_1+z_2|=1+1-2\cos\varphi=2(1-\cos\varphi),$$ $$\text{so }\hskip6em|z_1+z_2|=1\iff\cos\varphi=-\frac12\iff\varphi=\pm\frac{2\pi}3.\hskip11em$$

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