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Let \( a, b \in R \) and the roots \( \alpha, \beta \) of the equation \( z^{2}+a z+b=0 \) be complex. If the origin, \( \alpha \) and \( \beta \) represent the vertices of an equilateral triangle on the Argand plane, thenimage

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If z1, z2, z3 forms an equilateral triangle then

z12+ z2+ 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

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