Question

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

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

Answer 1

The correct option is C ${\omega }^{n-k}$

Since $\omega$ is the $n^{th}$ root of unity, hence ${\omega }^n=1$.
Now inverse of ${\omega }^k$ where $k<n$ is
$=\frac{1}{{\omega }^k}$
$=\frac{{\omega }^n}{{\omega }^k}$
$={\omega }^{n-k}$.
Hence inverse of ${\omega }^k$
$={\omega }^{-k}$
$={\omega }^{n-k}$.