If omega is a complex cube root of unity,then | vedclass

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(B) Given the expression: $P = (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors.Since $\omega^3 = 1$,we have $\omega^4 =

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(B) Given the expression: $P = (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors.Since $\omega^3 = 1$,we have $\omega^4 = \omega$ and $\omega^8 = \omega^2$.Thus,the expression becomes: $P = (1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) \dots$ to $2n$ factors.This can be grouped into $n$ pairs of $(1 + \omega)(1 + \omega^2)$.We know that $1 + \omega + \omega^2 = 0$,so $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.Therefore,$(1 + \omega)(1 + \omega^2) = (-\omega^2)(-\omega) = \omega^3 = 1$.Since there are $n$ such pairs,the product is $1^n = 1$.

If $\omega$ is a complex cube root of unity,then $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors is equal to:

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