Question

Show that the complex numbers $z_1,z_2$ and the origin form an equilateral triangle only if $z_1^2+z_2^2-z_1{z}_2=0$.

Answer 1

$OA=OB$ ........... (i)
$\therefore$ Coni method
$\frac{z_1-0}{z_2-0}=\frac{OA}{OB}{e}^{\frac{2\pi }{3}i}$
$\Rightarrow \frac{z_1}{z_2}=(cos\frac{2\pi }{3}+isin\frac{2\pi }{3})$ {from (i)}
$\Rightarrow \frac{z_1}{z_2}=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$
$\Rightarrow (\frac{z_1}{z_2}+\frac{1}{2})=\frac{i\sqrt{3}}{2}$
squaring both sides,
$\Rightarrow \frac{z_1^2}{z_2^2}+\frac{1}{4}+\frac{z_1}{z_2}=-\frac{3}{4}$
$\Rightarrow \frac{z_1^2}{z_2^2}+\frac{z_1}{z_2}+1=0$
$\Rightarrow z_1^2+z_1{z}_2+z_2^2=0$
$\therefore z_1^2+z_2^2+z_1{z}_2=0$
Ans: 1