Let z 1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle, then ?

Hello student,
Given that z₁ and z₂ be two roots of the equation z² + az + b = 0, z being complex.
So, we have z₁ + z₂ = -a, z₁z₂ = b (sum and product of roots)
So, z₂ = z₁ ( cos 60° + i sin 60°)
= z₁ (1/2 + √3/2 i)
So, 2z₂ – z₁ = √3 i z₁
This gives, (2z₂ -z₁)² = -3z₁²
Hence, (z₁² + z₂²) = z₁z₂
So, a² – 2b = b
this gives a² = 3b.

As z1, z2 are roots of z2 + az + bz1 + z2 = –a and z1 z2 = bAgain 0, z1, z2 are vertices of an equilateral triangle.Therefore, 02 + z12 + z22 = 0z1 + z1z2 + 0z2 = 0z12 + z22 = z1z2(z1 + z2)2 = 3z1z2