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

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(D) The product is given by $\omega \cdot \omega^2 \cdot \omega^3 \cdots \omega^n = \omega^{1 + 2 + 3 + \cdots + n} = \omega^{n(n + 1)/2}$.For $n = 1$

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Both $B$ and $C$

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(D) The product is given by $\omega \cdot \omega^2 \cdot \omega^3 \cdots \omega^n = \omega^{1 + 2 + 3 + \cdots + n} = \omega^{n(n + 1)/2}$.For $n = 1$,the product is $\omega^1 = \omega$.For $n = 2$,the product is $\omega^{2(3)/2} = \omega^3 = 1$.For $n = 3$,the product is $\omega^{3(4)/2} = \omega^6 = 1$.For $n = 4$,the product is $\omega^{4(5)/2} = \omega^{10} = \omega^1 = \omega$.Since $\omega = \frac{-1 + i\sqrt{3}}{2}$,the value $-\frac{1 - i\sqrt{3}}{2} = \frac{-1 + i\sqrt{3}}{2} = \omega$. Thus,the product results in values $1$ and $\omega$.

If $\omega$ is a complex cube root of unity,then for a positive integral value of $n$,the product $\omega \cdot \omega^2 \cdot \omega^3 \cdots \omega^n$ will be:

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