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If |z1|+|z2|=1a n dz1+z2+z3=0 then the a...

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`

A

`3sqrt(3//4)`

B

`sqrt(3//4)`

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
A
`|z_(1)|=|z_(2)|=|z_(3)|=1`
Hence, the cirumcenter of triangle is origin. Also, centroid `(z_(1)+z_(2)+z_(3))//3=0`, which coincides with the circumcentrer. So the triangle is equilateral . Since radius is 1, length of side is `a=sqrt(3)`. Therefore, the area of the triangle is `(sqrt3//4)a^(2)=(3sqrt3//4)`.
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