Question

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

• A. $2$
• B. $3$
• C. $6$
• D. $12$

Answer 1

Answer: (B)

$3$

${(1+{\omega }^4)}^n={(1+{\omega }^8)}^n$

${(1+\omega )}^n={(1+{\omega }^2)}^n$

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

For $n=2$

${\left({\omega }^2\right)}^2=\omega \ne {\omega }^2$

For $n=3$

${\omega }^6={\omega }^3=1$

$\therefore$ Least value of n is $3$