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Question: Let the complex numbers \(z_{1},z_{2}\) and \(z_{3}\)be the vertices of an equilateral triangle. Let...

Let the complex numbers z1,z2z_{1},z_{2} and z3z_{3}be the vertices of an equilateral triangle. Let z0z_{0} be the circumcentre of the

triangle, then z12+z22+z32=z_{1}^{2} + z_{2}^{2} + z_{3}^{2} =

A

z02z_{0}^{2}

B

−z02- z_{0}^{2}

C

3z023z_{0}^{2}

D

−3z02- 3z_{0}^{2}

Answer

3z023z_{0}^{2}

Explanation

Solution

Sol. Let r be the circum-radius of the equilateral triangle and ω\omega the cube root of unity.

Let ABC be the equilateral triangle with z1,z2z_{1},z_{2} and z3z_{3}as its vertices A, B and C respectively with circumcentre O′(z0).O'(z_{0}). The vectors O′A,O′B,O′CO'A,O'B,O'C are equal and parallel to OA′,OB′,OC′OA^{'},OB^{'},OC' respectively.

Then the vectors OA→′=z1−z0=reiθ\overset{\rightarrow}{OA}' = z_{1} - z_{0} = re^{i\theta}

⇒ OB→′=z2−z0=rei(θ+2π/3)=rωeiθ\overset{\rightarrow}{OB}' = z_{2} - z_{0} = re^{i(\theta + 2\pi/3)} = r\omega e^{i\theta}

⇒OC→′=z3−z0=rei(θ+4π/3)=rω2eiθ\overset{\rightarrow}{OC}' = z_{3} - z_{0} = re^{i(\theta + 4\pi/3)} = r\omega^{2}e^{i\theta}∴ z1=z0+reiθ,z2=z0+rωeiθ,z3=z0+rω2eiθz_{1} = z_{0} + re^{i\theta},z_{2} = z_{0} + r\omega e^{i\theta},z_{3} = z_{0} + r\omega^{2}e^{i\theta}Squaring and

adding, we get,

z12+z22+z32=3z02+2(1+ω+ω2)z0reiθ+(1+ω2+ω4)r2ei2θ=3z02,z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3z_{0}^{2} + 2(1 + \omega + \omega^{2})z_{0}re^{i\theta} + (1 + \omega^{2} + \omega^{4})r^{2}e^{i2\theta} = 3z_{0}^{2},

since 1+ω+ω2=0=1+ω2+ω4.1 + \omega + \omega^{2} = 0 = 1 + \omega^{2} + \omega^{4}.