ABCD is a rhombus in the Argand plane. If the affixes of the vertices are `z_(1),z_(2),z_(3)` and `z_(4)` respectively, and `angleCBA=pi//3`, then

Options

A. `z_(1)+omegaz_(2)+omega^(2)z_(3)=0`
B. `z_(1)-omegaz_(2)-omega^(2)z_(3)=0`
C. `omegaz_(1)+z_(2)+omega^(2)z_(3)=0`
D. `omega^(2)z_(1)+omegaz_(2)+z_(3)=0`

Answer

A

Explanation

Since the diagonala of a rhombus bisect each other.
Therefore, `BC=BA`. Also `angleCBA=pi/3`.

`therefore bar(BA)` is obtained by rotating `bar(BC)` through `pi//3`.
`rArr bar(BA)=bar(BC)e^(ipi//3)`.
`rArr (z_(1)-z_(2))=(z_(3)-z_(2))(cospi//3)+isinpi//3)`
`rArr (z_(1)-z_(2))=(z_(3)-z_(2))(-omega^(2)) [therefore omega^(2)=-1/2-isqrt(3)/2]`
`z_(1)-z_(2)(1+omega^(2))+z_(3)omega^(2)=0` `[therefore 1+omega^(2)=-omega]`