Question

If $\omega $ (not equal to $1$) is a cube root of unity and ${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$, then the least positive value of $n$ is
1) $2$
2) $3$
3) $5$
4) $6$

Answer 1

Hint: First we have to find the value of $\omega $, by finding the cube root of $1$. Then we will substitute the suitable value of the cube root of $1$ in the condition given to us in the question. Then, on performing suitable conditions we will get a suitable expression that helps us to get the least positive value of $n$, which we are asked for in the question.

Complete answer:
Given, $\omega $ is a cube root of unity.
So, we can write it as, $\omega = \sqrt[3]{1}$
Now, cubing both sides, we get,
$ \Rightarrow {\omega ^3} = 1$
Now, taking all the terms on the left hand side, we get,
$ \Rightarrow {\omega ^3} - 1 = 0$
We can write $1$ as ${1^3}$.
So, doing this, we get,
$ \Rightarrow {\omega ^3} - {1^3} = 0$
Using the formula, ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$, we get,
$ \Rightarrow \left( {\omega - 1} \right)\left( {{\omega ^2} + \omega + 1} \right) = 0$
$ \Rightarrow \omega - 1 = 0$ or ${\omega ^2} + \omega + 1 = 0$
$ \Rightarrow \omega = 1$ or ${\omega ^2} + \omega + 1 = 0$
In the question it is given that, $\omega \ne 1$.
Therefore, ${\omega ^2} + \omega + 1 = 0$
By applying the quadratic roots formula for this equation, we get,
$ \Rightarrow \omega = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4.1.1} }}{{2.1}}$
$ \Rightarrow \omega = \dfrac{{ - 1 \pm \sqrt {1 - 4} }}{2}$
$ \Rightarrow \omega = - \dfrac{1}{2} \pm \dfrac{{\sqrt { - 3} }}{2}$
Now, we know, $\sqrt { - 1} = i$
Replacing this value, gives us,
$ \Rightarrow \omega = - \dfrac{1}{2} \pm \dfrac{{\sqrt { - 1} .\sqrt 3 }}{2}$
$ \Rightarrow \omega = - \dfrac{1}{2} \pm i\dfrac{{\sqrt 3 }}{2}$
Therefore, we get, two values of $\omega $, that are,
$\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ and $\omega = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$
Now, if we take, $\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ and squaring both sides, we get,
\[ \Rightarrow {\omega ^2} = {\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^2}\]
\[ \Rightarrow {\omega ^2} = {\left( { - \dfrac{1}{2}} \right)^2} - 2.\dfrac{1}{2}.i\dfrac{{\sqrt 3 }}{2} + {\left( {i\dfrac{{\sqrt 3 }}{2}} \right)^2}\]
Now, opening the brackets and simplifying, we get,
$ \Rightarrow \Rightarrow {\omega ^2} = \dfrac{1}{4} - i\dfrac{{\sqrt 3 }}{2} + {i^2}\dfrac{3}{4}$
We know, $i = \sqrt { - 1} \Rightarrow {i^2} = - 1$.
Substituting this value, we get,
$ \Rightarrow {\omega ^2} = \dfrac{1}{4} - i\dfrac{{\sqrt 3 }}{2} - \dfrac{3}{4}$
$ \Rightarrow {\omega ^2} = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$
This is our second cube root of unity.
Therefore, we can say that, $\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ and ${\omega ^2} = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$
Therefore, we are given that, in the question,
${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$
Therefore, we can write it as,
$ \Rightarrow {\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {{\left( {{\omega ^2}} \right)}^2}} \right)^n}$
Now, substituting the value of ${\omega ^2}$, gives us,
$ \Rightarrow {\left( {1 + \left( { - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)} \right)^n} = {\left( {1 + {{\left( { - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)}^2}} \right)^n}$
Now, opening the brackets and simplifying, we get,
$ \Rightarrow {\left( {1 + - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)^n} = {\left( {1 + \left( {{{\left( { - \dfrac{1}{2}} \right)}^2} - 2.\left( { - \dfrac{1}{2}} \right)i\dfrac{{\sqrt 3 }}{2} + {{\left( { - i\dfrac{{\sqrt 3 }}{2}} \right)}^2}} \right)} \right)^n}$
$ \Rightarrow {\left( {\dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)^n} = {\left( {1 + \left( {\dfrac{1}{4} + i\dfrac{{\sqrt 3 }}{2} + {i^2}\dfrac{3}{4}} \right)} \right)^n}$
Simplifying the terms, we get,
$ \Rightarrow {\left( {\dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)^n} = {\left( {1 + \left( {\dfrac{1}{4} + i\dfrac{{\sqrt 3 }}{2} - \dfrac{3}{4}} \right)} \right)^n}$
[Using, ${i^2} = - 1$]
$ \Rightarrow {\left( {\dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)^n} = {\left( {1 + \left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)} \right)^n}$
$ \Rightarrow {\left( {\dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)^n} = {\left( {1 - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^n}$
$ \Rightarrow {\left( {\dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)^n} = {\left( {\dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^n} - - - \left( 1 \right)$
Now, we already found,
$\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$
Multiplying, both sides by $ - 1$, gives us,
$ \Rightarrow - \omega = \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$
And, ${\omega ^2} = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$
Again, multiplying both sides by $ - 1$, we get,
$ \Rightarrow - {\omega ^2} = \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$
Now, substituting these values $\left( 1 \right)$, gives us,
$ \Rightarrow {\left( { - \omega } \right)^n} = {\left( { - {\omega ^2}} \right)^n}$
$ \Rightarrow {\omega ^n} = {\omega ^{2n}}$
Now, dividing both sides by ${\omega ^n}$, we get,
$ \Rightarrow 1 = \dfrac{{{\omega ^{2n}}}}{{{\omega ^n}}}$
$ \Rightarrow {\omega ^n} = 1$
We found that, $\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ and ${\omega ^2} = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$.
Therefore, we can write,
${\omega ^3} = \omega .{\omega ^2}$
Substituting the values, gives us,
$ \Rightarrow {\omega ^3} = \left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right).\left( { - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)$
Using, the formula, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$, we get,
$ \Rightarrow {\omega ^3} = {\left( { - \dfrac{1}{2}} \right)^2} - {\left( {i\dfrac{{\sqrt 3 }}{2}} \right)^2}$
$ \Rightarrow {\omega ^3} = \dfrac{1}{4} - {i^2}\dfrac{3}{4}$
We know, ${i^2} = - 1$.
So, using that, we get,
$ \Rightarrow {\omega ^3} = \dfrac{1}{4} + \dfrac{3}{4}$
$ \Rightarrow {\omega ^3} = 1$
We earlier had, ${\omega ^n} = 1$.
Therefore, the least value of $n$, for which, ${\omega ^n} = 1$ is $3$, as ${\omega ^3} = 1$.

Therefore, the correct option is 2.

Note:
We could have also done this problem by using the property of cube roots of unity, $\omega $. The properties of $\omega $ gives us directly the equations like $\left( {1 + {\omega ^2}} \right)$ and many more directly. And also the values like ${\omega ^3} = 1$. So, if these properties are known, one can directly solve them.