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If α and β are the complex cube root of unity, show that α2 + β2 + αβ = 0

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Question

If α and β are the complex cube root of unity, show that α2 + β2 + αβ = 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`

∴ α + β = `(-1 + "i"sqrt(3))/2 + (-1 - "i" sqrt(3))/2`

= `(-1 + "i"sqrt(3) - 1 - "i"sqrt(3))/2`

= `(-2)/2`

= – 1

and αβ = `((-1 + "i"sqrt(3))/2)((-1 - "i"sqrt(3))/2)`

= `((-1)^2 - ("i"sqrt(3))^2)/4`

= `(1 - 3"i"^2)/4`

= `(1 + 3)/4`

= 1 ..........[∵ i2 = – 1]

α2 + β2 + aβ = (α + β)2 – αβ

= (–  1)2 –  1

= 1 – 1

= 0

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Chapter 1: Complex Numbers - Exercise 1.4 [Page 20]

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