If `z_(1),z_(2),z_(3)` are the vertices of an equilational triangle ABC such that `|z_(1)-i|=|z_(2)- i| = |z_(3)-i|,` then `|z_(1)+z_(2)+z_(3)|` equals to

Options

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

Answer

C

Explanation

`|z_(1)-i| = |z_(2) - p| = |z_(3) -i|`
Hence, `z_(1),z_(2),z_(3)` lie on the circle whose centre is I . Also given that the triangle is equilarteral. Hence, centroid and circumcentre conicides.
`therefore (z_(1) + z_(2) + z_(3))/(3) = i`
` rArr |z_(1) + z_(2) + z_(3)|=3`