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If ω is a complex cube root of unity, show that abωcωcaωbωa+bω+cω2c+aω+bω2 = ω2

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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

Sum
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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"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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Chapter 1: Complex Numbers - Exercise 1.4 [Page 20]

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