Question

In the multiplicative group of ${n^{th}}$ roots of unity, the inverse of ${\omega ^k}$ is $\left( {k < n} \right):$
A ${\omega ^{\dfrac{1}{k}}}$
B ${\omega ^{ - 1}}$
C ${\omega ^{n - k}}$
D ${\omega ^{\dfrac{n}{k}}}$

Answer 1

Complex numbers are the numbers that are expressed in the form of $a + ib$where, a $a$and $b$are real numbers and $i$ is an imaginary number. Hence, a complex number is a simple representation of addition of two numbers, i.e., real number and an imaginary number. One part of it is purely real and the other part is purely imaginary.

Complete step by step answer:
Let us write the given data:
In the multiplicative group of ${n^{th}}$ roots of unity, we need to find the inverse of ${\omega ^k}$ i.e.,
The roots of ${n^{th}}$roots of unity are:
$1,\omega ,{\omega ^2},.....,{\omega ^{n - 1}},{\omega ^n} = 1$
Where,$\omega$ is a ${n^{th}}$ roots of unity.
Let, G = \left\\{ {1,\omega ,{\omega ^2},.....,{\omega ^{n - 1}}} \right\\} be the group with respect to the multiplication, hence the identity element is 1.
Now inverse of ${w^k}$, where $k < n$ we have:
$\Rightarrow {\omega ^k} \cdot {\omega ^{n - k}} = {\omega ^{k + n - k}}$
With respect to inverse property, we get:
$\Rightarrow {\omega ^k} \cdot {\omega ^{n - k}} = {\omega ^n}$
$\Rightarrow {\omega ^k} \cdot {\omega ^{n - k}} = 1$
Hence, inverse of ${w^k}$ is
$\Rightarrow {\omega ^{n - k}}$

So, the correct answer is “Option C”.

Note: We must note that the numbers which are not real are imaginary numbers. When we square an imaginary number, it gives a negative result. Therefore, the square of the imaginary number gives a negative value. We must know that $0$ is a real number and real numbers are part of complex numbers. Therefore, $0$ is also a complex number and can be represented as $0 + 0i$.