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If $\omega$ is the $n$th root of unity then

\[\begin{array}{1 1}(1)1+\omega^{2}+\omega^{4}+\dots=\omega+\omega^{3}+\omega^{5}+\dots&(2)\omega^{n}=0\\(3)\omega^{n}=1&(4)\omega=\omega^{n-1}\end{array}\]

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1 Answer

Given $\omega$ is the $n^{th}$ root of unity
$\omega ^n=1$
Hence 3 is the correct answer.
answered May 15, 2014 by meena.p
 

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