If `A(z_(1))` and `B(z_(2))` are two points in the Argand plane such that `z_(1)^(2)+z_(2)^(2)+z_(1)z_(2)=0`, then `triangleOAB`, is

Options

A. equilateral
B. isosceles with `angleAOB=pi/2`
C. isosceles with `angleAOB=(2pi)/3`
D. isosceles with `angleAOB=pi/4`

Answer

C

Explanation

We have,
`z_(1)^(2)+z_(2)^(2)+z_(1)z_(2)=0`
`rArr (z_(1)-omegaz_(2))(z_(1)-omega^(2)z_(2))=0`
`rArr z_(1)=omegaz_(2)` or `z_(1)=omega^(2)z_(2)`
`rArr z_(1)=omegaz_(2)` or `z_(1)=omega^(2)z_(2)`
`rArr z_(1)=e^(i2pi//3)z_(2)` or `z_(1)=e^(4ipi//3)z_(2)`
`rArr z_(1)=e^(i2pi//3)z_(2)` or `z_(1)=e^(-2ipi//3)z_(2)` `[therefore e^(i4pi//3)=e^(-i2pi//3)]`
Now, `z_(1)=e^(i2pi//3)z_(2)`

`rArr bar(OA)` is obtained by rotating `bar(OB)` through `2pi//3` in anticlockwise sense.
`rArr` OA=OB and `angleBOA=2pi//3`
`rArr triangleOAB` is an isosceles triangle.
Similarly, `z_(1)=z_(2)e^(-2pi//3)` implies that `triangle OAB` is an isosceles triangle.