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If ω is a complex cube root of unity, show that (a+bω+cω2)c+aω+bω2=ω2 - Mathematics and Statistics

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Sum

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`

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Solution

ω is a complex cube root of unity.
∴ ω3 = 1 and 1 + ω + ω2 = 0

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, therefore omega^4 = omega]`

= `(omega^2("c" + "a"omega + "b"omega^2))/("c" + "a"omega + "b"omega^2)`

= ω2 = R.H.S.

Concept: Cube Root of Unity
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 3 Complex Numbers
Exercise 3.3 | Q 1. (iii) | Page 42
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