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If ω is a complex cube root of unity, show that (a + b)2 + (aω + bω2)2 + (aω2 + bω)2 = 6ab - Mathematics and Statistics

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Question

If ω is a complex cube root of unity, show that (a + b)2 + (aω + bω2)2 + (aω2 + bω)2 = 6ab

Sum
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Solution

ω is a complex cube root of unity.

∴ ω3 = 1 and 1 + ω + ω2 = 0.

∴ ω + ω2 = – 1, 1 + ω = – ω2 and 1 + ω2 = – ω.

(a + b)2 + (aω + bω2)2 + (aω2 + bω)2 

= (a2 + 2ab + b2) + (a2ω2 + 2abω3 + b2ω4) + (a2ω4 + 2abω3 + b2ω2)

= a2 + 2ab + b2 + a2ω2 + 2ab + b2ω + a2ω + 2ab + b2ω2   ...[∵ ω4 = ω3 × ω = ω]

= 6ab + (a2 + a2ω2 + a2ω) + (b2 + b2ω + b2ω)

= 6ab +a2(1 + ω + ω2) + b2(1 + ω + ω2)

= 6ab + a2(0) + b2(0)

= 6ab

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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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