If the complex numbers $z_{1}$ and $z_{2}$ and the origin form an isosceles triangle with vertical angle $\frac{2 π}{3}$ , then show that $z_{1}^{2} + z_{2}^{2} + z_{1} z_{2} = 0$ .

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Solution

$O A = O B$ ........... (i)
$∴$ Coni method
$\frac{z_{1} - 0}{z_{2} - 0} = \frac{O A}{O B} e^{\frac{2 π}{3} i}$
$\Rightarrow \frac{z_{1}}{z_{2}} = (c o s \frac{2 π}{3} + i s i n \frac{2 π}{3})$ {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,
$\Rightarrow \frac{z_{1}^{2}}{z_{2}^{2}} + \frac{1}{4} + \frac{z_{1}}{z_{2}} = - \frac{3}{4}$
$\Rightarrow \frac{z_{1}^{2}}{z_{2}^{2}} + \frac{z_{1}}{z_{2}} + 1 = 0$
$\Rightarrow z_{1}^{2} + z_{1} z_{2} + z_{2}^{2} = 0$
$∴ z_{1}^{2} + z_{2}^{2} + z_{1} z_{2} = 0$
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