If z1 and z2 are two roots of the equation z^2 + az + b= 0, z being a complex number. Further assume that the origin, z1 and z2 form an equilateral triangle. Then (1) a^2 = b (2) a^2 = 2b (3) a^2 = 3b (4) a^2 = 4b

Given z² + az + b= 0,

z₁+z₂ = -a (sum of roots)

z₁z₂ = b (product of roots)

z₂ = z₁(cos π/3 + i sin π/3)

= z₁(½ +i√3/2)

2z₂-z₁ = i√3z₁

(2z₂-z₁ )² = -3z₁²

z₁²+z₂² = z₁z₂

So a²-2b = b

a² = 3b

Hence option (3) is the answer.