# Let $z_1, z_2, z_3 \in\mathbb C$ such that $|z_1|=|z_2|=|z_3|=1$. If $z_1+z_2+z_3\neq0$ and ${z_1}^2+{z_2}^2+{z_3}^2=0$. Then find $|z_1+z_2+z_3|$.

Let $$z_1, z_2, z_3 \in\mathbb C$$ such that $$|z_1|=|z_2|=|z_3|=1$$. If $$z_1+z_2+z_3\neq0$$ and $${z_1}^2+{z_2}^2+{z_3}^2=0$$. Then find $$|z_1+z_2+z_3|$$.

My Approach: Let $$z_1=e^{i \theta _1}, z_2=e^{i \theta _2}, z_3=e^{i \theta _3}$$

Now $$(z_1+z_2+z_3)^2={z_1}^2+{z_2}^2+{z_3}^2+2(z_1z_2+z_2z_3+z_3z_1)\implies (z_1+z_2+z_3)^2=2(z_1z_2+z_2z_3+z_3z_1)$$.

$$\implies$$ $$|z_1+z_2+z_3|^2=2|(z_1z_2+z_2z_3+z_3z_1)|\implies$$ $$2 |(\cos\theta_1+\cos\theta_2+\cos\theta_3)+i(\sin\theta_1+\sin\theta_2+\sin\theta_3)|$$

$$\implies$$ $$|z_1+z_2+z_3|^2=2\bigg(\sqrt{3+2(\cos(\theta_1-\theta_2)+\cos(\theta_2-\theta_3)+\cos(\theta_3-\theta_1)})\bigg)$$

How to proceed further?

I know how to solve this question using other method. I want to know Can given question be solved using this method? If yes, How can i proceed further.

• Note that $|z_1^2| = |z_2^2| = |z_3^2| = 1$. Since they add to $0$, they must be some constant multiple of the roots of unity. Commented Mar 7, 2023 at 20:49

Finish off by computing $$|z_1z_2+z_2z_3+z_1z_3|^2=|z_1z_2|^2+|z_2z_3|^2+|z_1z_3|^2+2\Re(z_1z_2\overline{z_2z_3}+z_1z_2\overline{z_1z_3}+z_2z_3\overline{z_1z_3})=3+2\Re(z_1\overline{z_3}+z_2\overline{z_3}+z_2\overline{z_1})=3 +|z_1+z_2+z_3|^2-|z_1|^2-|z_2|^2-|z_3|^2=|z_1+z_2+z_3|^2,$$ and thus $$|z_1+z_2+z_3|^4=4|z_1+z_2+z_3|^2\implies |z_1+z_2+z_3|=2$$.