If `z_(1),z_(2)` are vertices of an equilateral triangle with `z_(0)` its centroid, then `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=`

Options

A. `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1)`
B. `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=2(z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1))`
C. `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)+z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1)=0`
D. None of these

Answer

A

Explanation

Let ABC be the given equilateral triangle such that the affixes of A,B,C are `z_(1),z_(2)` and `z_(3)` respectively.

Clearly, `bar(AC)` is obtained by rotating `bar(AB)` through `pi//3` in anticlockwise sense.
`therefore bar(AC)=bar(A)Be^(ipi//3) rArr z_(3)-z_(1)=(z_(2)-z_(1))e^(ipi//3)`...........(i)
Similarly, we have
`bar(BA)=bar(BC)e^(ipi//3) rArr z_(1)-z_(2)=(z_(3)-z_(2))e^(ipi//3)`..............(ii)
From (i) and (ii), we get
`(z+(3)-z_(1))/(z_(1)-z_(2)) = (z_(2)-z_(1))/(z_(3)-z_(2))`
`rArr z_(3)^(2)-z_(1)z_(3)-z_(2)z_(3)+z_(1)z_(2)=z_(1)z_(2)-z_(1)^(2)-z_(2)^(2)+z_(1)z_(2)`
`rArr z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=z_(1)z_(2)-z_(1)-z_(2)^(2)+z_(1)z_(2)`
`rArr z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1)`