Advertisement Remove all ads

Advertisement Remove all ads

Advertisement Remove all ads

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`

Advertisement Remove all ads

#### Solution

ω is a complex cube root of unity.

∴ ω^{3} = 1 and 1 + ω + ω^{2} = 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)`

= ω^{2 }= R.H.S.

Concept: Cube Root of Unity

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads