Question

If it is given that \[\omega \] is the ${{n}^{th}}$ root of unity then:
A. $1+{{\omega }^{2}}+{{\omega }^{4}}+....=\omega +{{\omega }^{3}}+{{\omega }^{5}}+....$
B. ${{\omega }^{n}}=0$
C. ${{\omega }^{n}}=1$
D. ${{\omega }^{n}}={{\omega }^{n-1}}$

Answer 1

Hint: We here have been given that $\omega $ is the ${{n}^{th}}$ root of unity. We will then find the values of different powers of $\omega $ by keeping $\omega ={{\left( 1 \right)}^{\dfrac{1}{n}}}$ and then check for all the options using these values. As a result, we will obtain the correct and wrong options separately. Hence, we will get the required answer.

Complete step-by-step solution:
Here we have been given that $\omega $ is the ${{n}^{th}}$ root of unity, i.e. 1.
Thus, we can say that:
$\omega ={{\left( 1 \right)}^{\dfrac{1}{n}}}$
Now, if we raise the power ‘n’ on both the sides, we will get:
${{\left( \omega \right)}^{n}}={{\left( {{\left( 1 \right)}^{\dfrac{1}{n}}} \right)}^{n}}$ …..(i)
Now, we know that ${{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}$
Using this formula in equation (i) we get:
$\begin{align}
  & {{\left( \omega \right)}^{n}}={{\left( {{\left( 1 \right)}^{\dfrac{1}{n}}} \right)}^{n}} \\
 & \Rightarrow {{\omega }^{n}}={{\left( 1 \right)}^{\dfrac{n}{n}}} \\
 & \Rightarrow {{\omega }^{n}}={{\left( 1 \right)}^{1}} \\
 & \therefore {{\omega }^{n}}=1 \\
\end{align}$
Now we will check for the options.
Option-A:
Here, we have been given that:
$1+{{\omega }^{2}}+{{\omega }^{4}}+....=\omega +{{\omega }^{3}}+{{\omega }^{5}}+....$
Here, we have been given that $\omega $ is the ${{n}^{th}}$ root of unity. Thus, since we do not know the value of ‘n’, we cannot know the values of $\omega ,{{\omega }^{2}},{{\omega }^{3}}...$
Thus, the value of $1+{{\omega }^{2}}+{{\omega }^{4}}+....=\omega +{{\omega }^{3}}+{{\omega }^{5}}+....$ is non determinable.
Thus, this option is wrong.
Option-B:
Here, we have been given that:
${{\omega }^{n}}=0$
But, we have already established that ${{\omega }^{n}}=1$.
Thus, this option is wrong.
Option-C:
Here, we have been given that:
${{\omega }^{n}}=1$
We already established above that this is true.
Thus, this option is correct.
Option-D:
Here, we have been given that:
${{\omega }^{n}}={{\omega }^{n-1}}$
Now, if we try to find out the value of ${{\omega }^{n-1}}$, we will get:
$\begin{align}
  & {{\omega }^{n}}=1 \\
 & \Rightarrow {{\omega }^{n-1}}.\omega =1 \\
 & \Rightarrow {{\omega }^{n-1}}=\dfrac{1}{\omega } \\
 & \Rightarrow {{\omega }^{n-1}}={{\omega }^{-1}} \\
\end{align}$
Thus, this option is also wrong.
Hence, from all these observations we get that option (C) is the only correct option.
Thus, option (C) is the correct option.

Note: We here have been given $\omega $ as the ${{n}^{th}}$ root of unity that’s why option (A) is indeterminable. Normally, $\omega $ is the symbol of the cube root of unity and in that case, we have certain properties given as:
1. If $\omega $ is the cube root of unity, ${{\omega }^{2}}$ is also the cube root of unity.
2. $1+\omega +{{\omega }^{2}}=0$
3. ${{\omega }^{3}}=1$
Using these properties, option (A) will also come out as true.