Questions & Answers

Question

Answers

A)$2$

B)$3$

C)$5$

D)$6$

Answer
Verified

${\omega ^3} - 1 = 0$ and use the formula $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ to find the least positive values of n.

Since it is given that ω is cube root of unity so it satisfies the following condition,

$ \Rightarrow $ ${\omega ^3} - 1 = 0$$ \Rightarrow {\omega ^3} = 1$ --- (i)

On using the formula $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$, we get-

$ \Rightarrow \left( {\omega - 1} \right)\left( {{\omega ^2} + \omega + 1} \right) = 0$

Now it is given that $\omega \ne 1$ so it follows that $\left( {{\omega ^2} + \omega + 1} \right) = 0$ -- (ii)

From this we can also write-

$ \Rightarrow 1 + {\omega ^2} = - \omega $ --- (iii)

It is also given that ${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$--- (iv)

Then we can write ${\omega ^4} = {\omega ^3}\omega $ and we know from equation (i) that ${\omega ^3} = 1$

So ${\omega ^4} = \omega $

Then we can write,

\[ \Rightarrow 1 + {\omega ^4} = 1 + \omega \]

From eq. (ii) substituting the value of $1 + \omega $ , we get

$ \Rightarrow 1 + {\omega ^4} = - {\omega ^2}$ --- (v)

On substituting the values of eq. (iii) and (v) in eq. (iv), we get

$ \Rightarrow {\left( { - \omega } \right)^n} = {\left( { - {\omega ^2}} \right)^n}$

We have to find the least positive value of n.

So we also write the above equation as-

$ \Rightarrow 1 = \dfrac{{{{\left( { - {\omega ^2}} \right)}^n}}}{{{{\left( { - \omega } \right)}^n}}}$ -- (vi)

On simplifying we get-

$ \Rightarrow 1 = {\left( {\dfrac{{{\omega ^2}}}{\omega }} \right)^n}$

$ \Rightarrow 1 = {\omega ^n}$ -- (vii)

Now from eq. (i) and eq. (vii), it is clear that-

$ \Rightarrow {\omega ^3} = {\omega ^n} = 1$

Since the base of the raised powers is same so the powers raised will also be equal to each other which means ${a^x} = {a^y} \Rightarrow x = y$

So on applying this in the obtained equation, we get-

$ \Rightarrow n = 3$

So the least positive value of n such that${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$ is 3.

For n=$2$ ,

$ \Rightarrow 1 = {\left( {\dfrac{{{\omega ^2}}}{\omega }} \right)^2}$

On solving we get,

$ \Rightarrow $ ${\omega ^4} = {\omega ^2}$ but we already proved that \[{\omega ^4} = \omega \] and $\omega \ne {\omega ^2}$ so n=$2$ is not possible

For n=$3$ ,we get

$ \Rightarrow 1 = {\left( {\dfrac{{{\omega ^2}}}{\omega }} \right)^3}$

On solving we get,

${\omega ^3} = 1$ which is given is the condition of the cube root of unity so the least positive values for n=$3$.