Question

If $Q$ is real and $z_1,z_2$ are connected by $z_1^2+z_2^2+2z_1{z}_2\cos \theta =0$, then triangle with vertices $0,z_1$ and $z_2$ is

• A. equilateral
• B. right-angled
• C. isosceles
• D. None of these

Answer 1

Answer: (C)

isosceles

Given, $z_1^2+z_2^2+2z_1{z}_2\cos \theta =0$
$\Rightarrow {\left(\frac{z_1}{z_2}\right)}^2+2\left(\frac{z_1}{z_2}\right)\cos \theta +1=0$
$\Rightarrow {(\frac{z_1}{z_2}+\cos \theta )}^2=-(1-{\cos }^2\theta )=-{\sin }^2\theta$
$\Rightarrow \frac{z_1}{z_2}=-\cos \theta \pm i\sin \theta$
$\Rightarrow \left|\frac{z_1}{z_2}\right|=\sqrt{{(-\cos \theta )}^2+{\left(\sin \theta \right)}^2}=1$
$\Rightarrow \left|z_1\right|=\left|z_2\right|$
$\Rightarrow |z_1-0|=|z_2-0|$
Thus, triangle with vertices $0,z_1,z_2$ is isosceles.