`z_(1),z_(2),z_(3)` are the vertices of an equilateral triangle taken in counter clockwise direction. If its circumference is at the origin and `z_(1)=1+i`, then

Options

A. `z_(2)=z_(1)e^(i2pi//3), z_(3)=e^(pi//3)`
B. `z_(2)=z_(1)e^(i2pi//3),z_(3)=z_(1)e^(i4pi//3)`
C. `z_(2)=z_(1)e^(i4pi//3), z_(3)=z_(1)e^(i2pi//3)`
D. `z_(2)=z_(1)e^(ipi//3),z_(3)=z_(1)e^(i2pi//3)`

Answer

B

Explanation

Clearly,`bar(OB)` and `bar(OC)` are obtained by rotating `bar(OA)` through `2pi//3` and `4pi//3` respectively.

`therefore bar(O)B=bar(O)C=bar(O)Ae^(i4pi//3)`
`rArr z_(2)=z_(1)e^(i2pi//3)` and `z_(3)=z_(1)e^(i4pi//3)`