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In the multiplicative group of $n{ th } $ roots of unity, the inverse of $\omega^{k}$ is $(k<n)$

\[\begin{array}{1 1}(1)\omega^{1/k}&(2)\omega^{-1}\\(3)\omega^{n-k}&(4)\omega^{n/k}\end{array}\]

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The roots of nth roots of unity are
$1, \omega,\omega^2.........\omega^{n-1}$ and $\omega^n=1$
$G= \{1, \omega,\omega^2,........ \omega^{n-1}\}$ is a group with respect to multiplication .
Identify element is 1.
$\omega^k. \omega^{n-k} =\omega^{k+n-k}=\omega^n =1$
The inverse of $\omega^k$ is $\omega^{n-k}$
Hence 3 is the correct answer.
answered May 22, 2014 by meena.p

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