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If α and β are the complex cube root of unity, show that α4 + β4 + α−1β−1 = 0 - Mathematics and Statistics

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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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Cube Root of Unity
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Chapter 1: Complex Numbers - Exercise 1.4 [Page 20]

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