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Question
If α and β are the complex cube root of unity, show that α4 + β4 + α−1β−1 = 0
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Solution
α and β are the complex cube roots of unity
∴ α = `(-1 + "i"sqrt(3))/2` and β = `(-1 - "i"sqrt(3))/2`
∴ αβ = `((-1 + "i"sqrt(3))/2)((-1 - "i"sqrt(3))/2)`
= `((-1)^2 - ("i"sqrt(3))^2)/4`
= `(1 - (-1)(3))/4`
= `(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
α4 + β4 + α−1β−1
= `α^4 + β^4 + 2α^2β^2 - 2α^2β^2 + 1/"αβ"` ...[Adding and subtracting 2α2β2]
= `(α^2 + β^2)^2 - 2α^2β^2 + 1/"αβ"`
= `[(α + β)^2 - 2αβ]^2 - 2(αβ)^2 + 1/"αβ"`
= `[(-1)^2 - 2(1)]^2 - 2(1)^2 + 1/1`
= (1 – 2)2 – 2 + 1
= (– 1)2 – 1
= 1 – 1
= 0
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