Question

If $\omega (\ne 1)$ is a complex cube root of unity and ${\left(1+{\omega }^4\right)}^n={\left(1+{\omega }^8\right)}^n$, then the least positive integral value of $n$ is

• A. 2
• B. 3
• C. 6
• D. 12

Answer 1

Answer: (B)

As ${\omega }^4=\omega ,{\omega }^8={\omega }^2$, we get
$(1+\omega {)}^n={\left(1+{\omega }^2\right)}^n$
$\Rightarrow {\left(-{\omega }^2\right)}^n=(-\omega {)}^n$
$\Rightarrow {\omega }^n=1$
$\therefore n=3$