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Question
If ω is a complex cube root of unity, show that `("a" + "b"ω + "c"ω^2)/("c" + "a"ω + "b"ω^2)` = ω2
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"ω + "c"ω^2)/("c" + "a"ω + "b"ω^2)`
= `("a"ω^3 + "b"ω^4 + "c"ω^2)/("c" + "a"ω + "b"ω^2)` ...[∵ ω3 = 1, ∴ ω4 = ω]
= `(ω^2("a"ω + "b"ω^2 + "c"))/("c" + "a"ω + "b"ω^2)`
= `(ω^2("c" + "a"ω + "b"ω^2))/("c" + "a"ω + "b"ω^2)`
= ω2
= R.H.S.
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