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Question
If α and β are the complex cube roots of unity, show that α2 + β2 + αβ = 0.
Solution
α and β are the complex cube roots of unity.
∴ α = `(-1 + "i"sqrt(3))/2 and beta = (-1 - "i"sqrt(3))/2`
∴ αβ = `((- 1 + "i"sqrt(3))/2)((-1 - "i"sqrt(3))/2)`
= `((-1)^2 - (isqrt(3))^2)/4`
= `(1 - (-1)(3))/4` ...[∵ i2 = – 1]
= `(1 + 3)/4`
∴ αβ = 1
Also, α + β = `(-1 + "i"sqrt(3))/2 + (-1 - "i"sqrt(3))/2`
= `(-1 + "i"sqrt(3) - 1 - "i" sqrt(3))/2`
= `(-2)/2`
∴ α + β = – 1
L.H.S. = α2 + β2 + αβ
= α2 + 2αβ + β2 + αβ – 2αβ ...[Adding and subtracting 2αβ]
= (α2 + 2αβ + β2) – αβ
= (α + β)2 – αβ
= (– 1)2 – 1
= 1 – 1
= 0 = R.H.S.
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