If `theta` is real and `z_1, z_2` are connected by `z1 2+z2 2+2z_1z_2costheta=0,` then prove that the triangle formed by vertices `O ,z_1a n dz_2` is isosceles.

Explanation

we have,
`z_(1)^(2)+z_(2)^(2) +2z_(1)z_(2) cos theta=0`
` Rightarrow (z_(1)/z_(2))^(2) +2z_(1)/z_(2) cos theta+1=0`
` Rightarrow z_(1)/z_(2) = ( - 2 cos theta +-sqrt(3cos^(2) theta-4))/2`
`= - cos theta +- isin theta`
`Rightarrow |z_(1)/z_(2)|=sqrt((-cos theta)^(2)+(+-sintheta)^(2))=1`
`Rightarrow |z_(1)|=|z_(2)|`
`Rightarrow OP = OQ`
` Rightarrow Delta OPQ` is isosceles