Question

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

Answer 1

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 - (i sqrt (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.