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Question
If ω is a complex cube root of unity, then prove the following: (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
Also, 1 + ω2 = - ω, 1 + ω = - ω2
and ω + ω2 = – 1
L.H.S. = (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 = R.H.S.
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