Question

If the complex numbers $z_1,z_2$ and the origin form an isosceles triangle with vertical angle $\frac{2\pi }{3}$ then $z_1^2+z_2^2+z_2{z}_2=0$.

• A. True
• B. False

Answer 1

The correct option is A True

$Letvertexisatorigin$
$\sin ce\text{triangle is isoscoles with vertex at origin}$
$\text{then}\left|z_1\right|=\left|z_2\right|$
$\sin cevertexangle=2\pi /3$
$hence$
$z_2=z_1(\cos 2\pi /3+i\sin 2\pi /3)$
$=z_1(-1/2+i\sqrt{3}/2)$
$z_1^2+z_2^2+z_1{z}_2$
$=z_1^2+{\left(z_1(-1/2+i\sqrt{3}/2)\right)}^2+z_1\left(z_1(-1/2+i\sqrt{3}/2)\right)$
$=z_1^2+z_1^2{(-1/2+i\sqrt{3}/2)}^2+z_1^2(-1/2+i\sqrt{3}/2)$
$=z_1^2(1+\frac{1}{4}-\frac{\sqrt{3}i}{2}-\frac{3}{4}-\frac{1}{2}+\frac{\sqrt{3}i}{2})$
$=z_1^2\left(0\right)$
$=0$
$Hence\text{the given statement is true}$