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