Question

If $\omega \left( { \ne 1} \right)$ be the cube root of unity and ${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$, the least +ve value of n is
(A) 2
(B) 3
(C) 4
(D) 5

Answer 1

Hint: Solve the above equation using $1 + \omega + {\omega ^2} = 0$ and ${\omega ^{3x}} = 1$ conditions.

Complete step-by-step answer:
We know that, if $\omega$ is the cube root of unity then,
$\Rightarrow 1 + \omega + {\omega ^2} = 0$ $\to$ (1)
$\Rightarrow {\omega ^{3x}} = 1$ (for all x as positive integer) $\to$ (2)
And it is given that,
$\Rightarrow {\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$ $\to$ (3)
We, can write ${\omega ^4} = \omega \cdot {\omega ^3}$ and we know that ${\omega ^3} = 1$ (from equation 2)
$\Rightarrow$ ${\omega ^4} = \omega$
Putting the value of ${\omega ^4}$ in equation (3) we get,
$\Rightarrow {\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + \omega } \right)^n}$
From equation (1) we get,
$\Rightarrow \left( {1 + {\omega ^2}} \right) = -\omega$
$\Rightarrow 1 + \omega = - {\omega ^2}$
So, putting the value of $1 + {\omega ^2}$ and $1 + \omega$ in equation (4) we get,
$\Rightarrow {\left( { - \omega } \right)^n} = {\left( { - {\omega ^2}} \right)^n}$
On solving above equation we get,
$\Rightarrow {\left( { - 1} \right)^n}{\omega ^n} = {\left( { - 1} \right)^n}{\omega ^{2n}}$
$\Rightarrow {\omega ^n} = {\omega ^{2n}}$
$\Rightarrow {\omega ^{2n-n}} = 1$
$\Rightarrow {\omega ^{n}} = 1$
Clearly we can see that n=3, is the least value of n satisfying the above equation.
Hence B is the correct option.

Note: Whenever we face these types of problems, remember the properties of the cube root of unity and try to simplify the given expression, it will lead us to the answer.