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.

  • $\begingroup$ 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. $\endgroup$ Commented Mar 7, 2023 at 20:49

1 Answer 1


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 $|z_1+z_2+z_3|^4=4|z_1+z_2+z_3|^2\implies |z_1+z_2+z_3|=2$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .