Question

If $\omega$ is a cube root of unity and $n$ is a positive integer satisfying $1 + {\omega ^n} + {\omega ^{2n}} = 0$ then, $n$ is of the type:
A.$3m$
B.$3m + 1$
C.$3m + 2$
D.None of these

Answer 1

Here, we will substitute each option in the given equation and try to prove that the left hand side is equal to the right hand side. The option which will satisfy this criterion will be the required answer.

Formula Used:
We will use the following formulas:
1.${\left( {{a^m}} \right)^n} = {a^{m \times n}}$
2.$1 + \omega + {\omega ^2} = 0$
3. ${a^{m + n}} = {a^m} \cdot {a^n}$

Complete step-by-step answer:
According to the question, $\omega$ is a cube root of unity.
Hence, this means that,
$\sqrt[3]{\omega } = 1$
Cubing both sides, we get
$\Rightarrow {\omega ^3} = 1$…………………….$\left( 1 \right)$
Now, $n$ is a positive integer satisfying $1 + {\omega ^n} + {\omega ^{2n}} = 0$.
We have to find that $n$ is of the type of which of the given options.
Now, we will solve this question, by substituting every option in the place of $n$
$3m$
Now, substituting $n = 3m$ in the equation $1 + {\omega ^n} + {\omega ^{2n}} = 0$, we get
$1 + {\omega ^{3m}} + {\omega ^{2\left( {3m} \right)}} = 0$
Now, let $m = 1$
$\Rightarrow 1 + {\omega ^3} + {\omega ^6} = 0$
This can also be written as:
Using the property ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$, we get
$\Rightarrow 1 + {\omega ^3} + {\left( {{\omega ^3}} \right)^2} = 0$
Hence, substituting ${\omega ^3} = 1$ from $\left( 1 \right)$, we get,
$\Rightarrow 1 + 1 + {\left( 1 \right)^2} = 3 \ne 0$
Since, LHS $\ne$ RHS
Hence, option A is rejected.
$3m + 1$
Now, substituting $n = 3m + 1$ in the equation $1 + {\omega ^n} + {\omega ^{2n}} = 0$, we get
$1 + {\omega ^{3m + 1}} + {\omega ^{2\left( {3m + 1} \right)}} = 0$
$\Rightarrow 1 + {\omega ^{3m}} \cdot \omega + {\omega ^{6m + 2}} = 0$
Using the property ${a^{m + n}} = {a^m} \cdot {a^n}$ , we get
$\Rightarrow 1 + {\omega ^{3m}} \cdot \omega + {\omega ^{6m}} \cdot {\omega ^2} = 0$
Now, let $m = 1$
$\Rightarrow 1 + {\omega ^3} \cdot \omega + {\omega ^6} \cdot {\omega ^2} = 0$
This can also be written as:
$\Rightarrow 1 + {\omega ^3} \cdot \omega + {\left( {{\omega ^3}} \right)^2} \cdot {\omega ^2} = 0$
Hence, substituting ${\omega ^3} = 1$ from $\left( 1 \right)$, we get,
$\Rightarrow 1 + \left( 1 \right) \cdot \omega + {\left( 1 \right)^2} \cdot {\omega ^2} = 0$
$\Rightarrow 1 + \omega + {\omega ^2} = 0$
Hence, LHS $=$ RHS
We know that $1 + \omega + {\omega ^2} = 0$ because the sum of three cube roots of unity is 0.
Hence, option B is the correct answer.
But, we will check this for option C as well:
$3m + 2$
Now, substituting $n = 3m + 2$ in the equation $1 + {\omega ^n} + {\omega ^{2n}} = 0$, we get
$1 + {\omega ^{3m + 2}} + {\omega ^{2\left( {3m + 2} \right)}} = 0$
$\Rightarrow 1 + {\omega ^{3m}} \cdot {\omega ^2} + {\omega ^{6m + 4}} = 0$
Using the property ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$, we get
${a^{m + n}} = {a^m} \cdot {a^n}$)
$\Rightarrow 1 + {\omega ^{3m}} \cdot {\omega ^2} + {\omega ^{6m}} \cdot {\omega ^4} = 0$
Now, let $m = 1$
$\Rightarrow 1 + {\omega ^3} \cdot {\omega ^2} + {\omega ^6} \cdot {\omega ^4} = 0$
This can also be written as:
$\Rightarrow 1 + {\omega ^3} \cdot {\omega ^2} + {\left( {{\omega ^3}} \right)^2} \cdot {\omega ^4} = 0$
Hence, substituting ${\omega ^3} = 1$ from $\left( 1 \right)$, we get,
$\Rightarrow 1 + \left( 1 \right) \cdot {\omega ^2} + {\left( 1 \right)^2} \cdot {\omega ^4} = 0$
$\Rightarrow 1 + {\omega ^2} + {\omega ^4} \ne 0$
Hence, LHS $\ne$ RHS
Also, this part completely depends on the value of $m$
Hence, option C is also rejected.
Therefore, if $\omega$ is a cube root of unity and $n$ is a positive integer satisfying $1 + {\omega ^n} + {\omega ^{2n}} = 0$ then, $n$ is of the type $3m + 1$
Hence, option B is the required answer.

Note: Let us assume the cube root of unity or 1 as:
$\sqrt[3]{1} = z$
Cubing both sides,
$\Rightarrow 1 = {z^3}$
Or
$\Rightarrow {z^3} - 1 = 0$
Now, using the formula $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$, we get
$\Rightarrow \left( {z - 1} \right)\left( {{z^2} + z + 1} \right) = 0$
Therefore, either $\left( {z - 1} \right) = 0$
$\Rightarrow z = 1$
Or, $\left( {{z^2} + z + 1} \right) = 0$
Comparing with $\left( {a{x^2} + bx + c} \right) = 0$
Here, $a = 1$, $b = 1$ and $c = 1$
Now, we know that the formula of determinant, $D = {b^2} - 4ac$.
Hence, for $\left( {{z^2} + z + 1} \right) = 0$,
$D = {\left( 1 \right)^2} - 4 \times 1 = 1 - 4 = - 3$
Now, Using quadratic formula,
$z = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Here, $a = 1$, $b = 1$, $c = 1$ and $D = - 3$
$\Rightarrow z = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}$
This can be written as:
$\Rightarrow z = \dfrac{{ - 1 \pm \sqrt 3 i}}{2}$
Therefore, the three cube roots of unity are:
$1$, $\dfrac{{ - 1}}{2} + \dfrac{{\sqrt 3 i}}{2}$ and $\dfrac{{ - 1}}{2} - \dfrac{{\sqrt 3 i}}{2}$
Now, according to the property, the sum of these three cube roots of unity will be equal to 0.
We know that,
$1 + \omega + {\omega ^2} = 0$
Here, $\omega$ represents the imaginary cube roots.
$\Rightarrow 1 + \dfrac{{ - 1}}{2} + \dfrac{{\sqrt 3 i}}{2} + \dfrac{{ - 1}}{2} - \dfrac{{\sqrt 3 i}}{2} = 1 - 1 + 0 = 0$
Hence, proved
Therefore, due to this property; in this question we have assumed that $1 + \omega + {\omega ^2} = 0$ and we were able to find the required answer.