If z1, z2, z3 forms an equilateral triangle then
z12+ z22 + z32 = z1z2 + z2z3 + z1z3
Let z1 = 0, z2 = α, z3 = ß
\(\therefore\) z22+ z32 = z2z3
⇒ z22+ z32 + 2 z2z3 = z2z3 + 2z2z3
⇒ (z2 + z3)2 = 3z2z3
⇒ (α + ß)2 = 3αß---(1)
But given that α and ß are the roots of the equation
z2 + az + b = 0
∴ α + ß = -a
and α ß = b
∴ From (1), We obtain
(-a)2 = 3b
⇒ a2 = 3b
Alternative:
∵ origin α and ß are vertices of an equilateral triangle
∴ OB = OA and angle between OA & OB is 60°
∴ OB = OA ei60°
⇒ ß = α (cos 60° + isin60°)
⇒ ß = α(\(\frac12+i\frac{\sqrt3}2\))
since, α and ß are roots of equation
z2 + az + b = 0
∴ α + ß = -a and αß = b
now, α + ß = α + α(\(\frac12+i\frac{\sqrt3}2\))
= α(\(\frac32+i\frac{\sqrt3}2\))
and αß = α2(\(\frac12+i\frac{\sqrt3}2\))---(1)
(α +ß)2 = α2(\(\frac12+i\frac{\sqrt3}2\))2
= α2(\(\frac94-\frac34+2i\times\frac32\times\frac{\sqrt3}2\))
= α2(\(\frac64+\frac{3\sqrt3 i}2\))
= α2(\(\frac32+\frac{3\sqrt3 i}2\))
= 3α2(\(\frac12+i\frac{\sqrt3}2\))
= 3αß (From (1))
⇒ (-a)2 = 3b
⇒ a2 = 3b