# Finding two roots of unity which their sum is also root of unity

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?

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.
• If $|z|=1$, then $\bar{z}=1/z$. – Batominovski Oct 20 '19 at 17:24
Yes. You can take $$z_1=\frac12+\frac{\sqrt3}2i$$ and $$z_1=\frac12-\frac{\sqrt3}2i$$. Then $$s=1$$
Suppose the difference of their arguments is $$\;\theta_1-\theta_2=\varphi$$. Using Al-Kashi's theorem, we have $$|z_1+z_2|=1+1-2\cos\varphi=2(1-\cos\varphi),$$ $$\text{so }\hskip6em|z_1+z_2|=1\iff\cos\varphi=-\frac12\iff\varphi=\pm\frac{2\pi}3.\hskip11em$$