Question

If ${(1+{\omega }^2)}^m={(1+{\omega }^4)}^m$, where $\omega$ is an imaginary cube root of unity, then least positive integral value of $m$ is:

• A. greater then $1$
• B. greater then $2$
• C. greater then $3$
• D. greater then $4$

Answer 1

The correct option is A greater then $1$

Given,

${(1+w^2)}^m={(1+w^4)}^m$

We know that $1+w+w^2=0$ and $w^3=1$

$1+w^2=-w$ and $1+w^4=1+w=-w^2$

$\therefore {(-w)}^m={\left({-w}^2\right)}^m$

$\therefore m>1$