(A) We know that $1+\omega+\omega^{2}=0$ and $\omega^{3}=1$. Given expression: $(1+\omega)(1+\omega^{2})(1+\omega^{4})(1+\omega^{8})$ $= (1+\omega)(1+

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(A) We know that $1+\omega+\omega^{2}=0$ and $\omega^{3}=1$. Given expression: $(1+\omega)(1+\omega^{2})(1+\omega^{4})(1+\omega^{8})$ $= (1+\omega)(1+\omega^{2})(1+\omega)(1+\omega^{2})$ $= [(1+\omega)(1+\omega^{2})]^{2}$ $= [1+\omega^{2}+\omega+\omega^{3}]^{2}$ $= [1+(\omega^{2}+\omega)+1]^{2}$ $= [1+(-1)+1]^{2}$ $= [1]^{2} = 1$.

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,then $(1+\omega)(1+\omega^{2})(1+\omega^{4})(1+\omega^{8})$ is equal to

KCET 2007 All KCET

A
$1$
B
$0$
C
$\omega^{2}$
D
$\omega$

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4-1.Complex numbers

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If $1, \omega, \omega^{2}$ are the cube roots of unity,then $(1+\omega)(1+\omega^{2})(1+\omega^{4})(1+\omega^{8})$ is equal to

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