If `|z_1|+|z_2|=1a n dz_1+z_2+z_3=0` then the area of the triangle whose vertices are `z_1, z_2, z_3` is `3sqrt(3)//4` b. `sqrt(3)//4` c. `1` d. `2`

Options

A. `(3sqrt(3))/4`
B. `sqrt(3)/4`
C. 1
D. 2

Answer

A

Explanation

Let `z_(1),z(2),z_(3)` be the affixes of points A,B and C respectively in the Argand plane. We have,
`|z_(1)|=|z_(2)|=|z_(3)|`
`rArr OA=OB=OC`
`rArr` O(origin) is the circumcenter of `triangleABC`.
Also, `z_(1)+z_(2)+z_(3)=0`
`rArr` Centroid of `triangleABC` is also the origin.
`therefore triangleABC` is equilateral.
`therefore z_(2)=z_(1)e^(i2pi//3)=z_(1)omega` and `z_(3)=z_(1)e^(i4pi//3)=z_(1)omega^(2)`
`rArr AB=|z_(1)omega-z_(1)|=|z_(1)||omega-1|=|-3/2+(isqrt(3))/(2)|=sqrt(3)`
Hence, Area of `triangleABC=sqrt(3)/4 (AB)^(2)=(3sqrt(3))/(4)` sq. units.