Question

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

• A. ${\omega }^{\frac{1}{k}}$
• B. ${\omega }^{-1}$
• C. ${\omega }^{n-k}$
• D. ${\omega }^{\frac{n}{k}}$

Answer 1

Answer: (C)

${\omega }^{n-k}$

The roots of $n^{th}$ roots of unity are $1,\omega ,{\omega }^2,...,{\omega }^{n-1},{\omega }^n=1$

ωns a group with respect to multiplication .

Let $G=\{1,\omega ,{\omega }^2,...{\omega }^{n-1}\}$ be the group with respect to multiplication.

Identity element is $1$.

Thus for $k<n$,

${\omega }^k\cdot {\omega }^{n-k}={\omega }^{k+n-k}={\omega }^n=1$ (inverse property)

Thus inverse of ${\omega }^k$ is ${\omega }^{n-k}$