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