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Question

Let the complex numbers z1,z2 and z3 be the vertices of an equilateral triangle. Let z0 be the circumcentre of the triangle, then z21+z22+z23=


A

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C

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Solution

The correct option is C


Let r be the circum radius of the equilateral triangle and ω the cube root of unity.

Let ABC be the equilateral triangle with z1,z2 and z3 as its vertices A, B and C respectively with circumcentre O(z0). The vectors OA,OB,OC are equal and parallel to OA,OB,OC respectively.

Then the vectors ¯¯¯¯¯¯¯¯¯¯OA = z1z0 = reiθ

¯¯¯¯¯¯¯¯¯¯OB = z2z0 = re(θ+2π3) = rωeiθ

¯¯¯¯¯¯¯¯¯¯OC = z3z0 = rei(θ+4π3) = rω2eiθ

z1 = z0+reiθ,z2 = z0+rωeiθ, z3 = z0+rω2eiθ

Squaring and adding

z21+z22+z23=3z20+2(1+ω+ω2)z0reiθ+(1+ω2+ω4)r2ei2θ

=3z20, Since 1+ω+ω2 = 0 = 1+ω2+ω4


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