Question

If ω be complex nth root of unity and r is an integer not divisible by n, then the sum of the rth powers of the nth roots of unity is:
 $
  {\text{A}}{\text{. 0}} \\
  {\text{B}}{\text{. 1}} \\
  {\text{C}}{\text{. }}\omega \\
  {\text{D}}{\text{. n}} \\
 $

Answer 1

Hint: In order to find the sum of the rth powers of the nth roots of unity, we use the concept of nth roots of unity, their general form. We also need to apply the concept of geometric progression as the terms in the sequence are a multiplication of previous terms.

Complete step-by-step answer:
Given Data,
ω is the complex nth root of unity
r is an integer and is not divisible by n
To find,
Sum of rth powers of nth roots of unity.

Given that the nth root of unity is ω.
Let us consider the nth root of unity is $ {\omega _{\text{i}}} $ , where i = 0, 1, 2 …….

Now the sum of these nth roots of unity will be (it should start with unity 1):
 $ {\text{1 + }}{\omega _1} + {\omega _2} + {\omega _3} + .......... + {\omega _{{\text{n - 1}}}} = \dfrac{{1\left( {1 - {{\left( \omega \right)}^2}} \right)}}{{1 - \omega }} $
Where, $ {\omega _{\text{i}}} = {{\text{e}}^{{\text{i}}\dfrac{{2\pi {\text{k}}}}{{\text{n}}}}} $
Therefore we can say,
 $ {\omega _1} = {{\text{e}}^{{\text{i}}\dfrac{{2\pi }}{{\text{n}}}}} $ , $ {\omega _2} = {{\text{e}}^{{\text{i}}\dfrac{{4\pi }}{{\text{n}}}}} $ …….. Are in G.P with a ratio $ \omega = {{\text{e}}^{{\text{i}}\dfrac{{2\pi }}{2}}} $

Now we express the equation in the rth power, it becomes
 $ {\text{1 + }}{\omega _1}^{\text{r}} + {\omega _2}^{\text{r}} + {\omega _3}^{\text{r}} + .......... + {\omega ^{\text{r}}}_{{\text{n - 1}}} = \dfrac{{1\left( {1 - {{\left( {{\omega ^{\text{r}}}} \right)}^2}} \right)}}{{1 - {\omega ^{\text{r}}}}} $
Given that r is not divisible by n, that means r ≠ kn, where k is some constant.
That implies, $ {\omega ^{\text{r}}} \ne 1 $
 $
   \Rightarrow \dfrac{{1 - {{\left( {{\omega ^{\text{n}}}} \right)}^{\text{r}}}}}{{1 - {\omega ^{\text{r}}}}} \\
  {\text{But }}{\omega ^{\text{2}}} = 1 \\
  {\text{Thus }}\dfrac{{1 - 1}}{{1 - {\omega ^{\text{r}}}}} = 0 \\
 $

Therefore the sum of the rth powers of the nth roots of unity is zero.
Option A is the correct answer.

Note – In order to solve this type of question the key is to know the concept of roots of unity. They are a complex number. Also the concept and formula of geometric progression.
We say the last term is entirely zero because the numerator is equal to zero, making the entire term zero. Carefully understanding the question is the most important part of solving this type of problem.