Question

If 1,z1,z2,z3 be the nth roots of unity and w be a non real complex cube root of unity then the product (w-zr) can be equal to

Answer 1

Answer:

0 or 1 or -ω²

Step-by-step explanation:

Hi,

Let 1, z₁, z₂, z₃,....... z(n-1) be n roots of unity

We know that n roots of unity are the roots of the algebraic expression

xⁿ - 1 = 0

Thus,

xⁿ - 1 = (x - 1)(x - z₁)(x - z₂)......(x - z(n-1))

⇒ (xⁿ - 1 )/ (x - 1) = (x - z₁)(x - z₂)......(x - z(n-1)) --------(1)

Now, we know that 1, ω, ω² are cube roots of unity

such that ω³ = 1

Now, substitute x = ω in equation (1), we get

(ωⁿ - 1 )/ (ω - 1) = (ω - z₁)(ω - z₂)......(ω - z(n-1))----(2)

If suppose n is of form 3k, then ωⁿ = 1

Hence , the product will become 0.

Suppose n is of form 3k + 1,

then ωⁿ = ω, so the product will become 1.

Suppose n is of the form 3k + 2,

then ωⁿ = ω², hence the product would be equal to

(ω² - 1)/(ω - 1) = (ω + 1) = -ω² .

Hence, the product can take either 0 or 1 or -ω².

Hope, it helped !