Question

If the complex numbers $z_1,z_2,z_3$ represent the the vertices of an equilateral triangle such that $|z_1|=|z_2|=|z_3|$ , then prove that $z_1+z_2+z_3=0$

Answer 1

Let $z_1,z_2,z_3$ represent the vertices $A_1,A_2,A_3$ of $\Delta A_1{A}_2{A}_3$ respectively and let O be the origin.
Then $z_1=\overrightarrow{O{A}_1},z_2=\overrightarrow{O{A}_2},z_3=\overrightarrow{O{A}_3}$
Since $|z_1{|}^3=|z_2{|}^3=|z_3{|}^3$
we have $O{A}_1=O{A}_2=O{A}_3$ This show that O is the circum centre of the $\Delta A_1{A}_2{A}_3$ .
$\angle A_{1}O{A}_2=\angle A_{2}O{A}_3=\angle A_{3}O{A}_1=\frac{2\pi }{3}$
Hence we can write
$z_2=z_1{e}^{2\pi i/3}=z_1\omega and{z}_3=z_1{e}^{4\pi i/3}=z_3{\omega }^2$
Hence $z_1+z_2+z_3=z_1(1+\omega +{\omega }^2)=0$