(C) Given that $1, \omega, \omega^2$ are the cube roots of unity,we have $1+\omega+\omega^2=0$. The fourth roots of unity are $1, i, -1, -i$. Let $\al

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(C) Given that $1, \omega, \omega^2$ are the cube roots of unity,we have $1+\omega+\omega^2=0$. The fourth roots of unity are $1, i, -1, -i$. Let $\alpha = i$. Then $\alpha^2 = -1$ and $\alpha^3 = -i$. Substituting these into the expression $\alpha+\alpha \omega-\alpha^3 \omega^2$: $\alpha+\alpha \omega-\alpha^3 \omega^2 = i + i\omega - (-i)\omega^2$ $= i(1+\omega+\omega^2)$ Since $1+\omega+\omega^2=0$,the expression becomes $i(0) = 0$.

Class 11 Mathematics 4-1.Complex numbers De Moivre's theorem and Roots of unity

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If $1, \omega, \omega^2$ are the cube roots of unity and $1, \alpha, \alpha^2, \alpha^3$ are the fourth roots of unity in usual notation,then $\alpha+\alpha \omega-\alpha^3 \omega^2=$

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$3$
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$1$
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$0$
D
$-1$

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If $1, \omega, \omega^2$ are the cube roots of unity and $1, \alpha, \alpha^2, \alpha^3$ are the fourth roots of unity in usual notation,then $\alpha+\alpha \omega-\alpha^3 \omega^2=$

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