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Question
If ω is a complex cube root of unity, show that (a + b) + (aω + bω2) + (aω2 + bω) = 0
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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) + (aω + bω2) + (aω2 + bω)
= a + b + aω + bω2 + aω2 + bω
= (a + aω + aω2) + (b + bω + bω2)
= a(1 + ω + ω2) + b(1 + ω + ω2)
= a(0) + b(0)
= 0
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