Question

If $\omega$ is a cube root of unity and ‘n’ is a positive integer satisfying $1+{\omega }^n+{\omega }^{2n}=0$ then , ‘n’ is of the type :

• A. $3m$
• B. $3m+1$
• C. $3m+2$
• D. $Noneofthese$

Answer 1

The correct option is B $3m+1$

Since we know that,
$w^{(3m+1)}=w^{3m}w=w$ and $w^{2(3m+1)}=w^{6m}{w}^2=w^2$
So for $n=3m+1$, we have, $1+w^n+w^{2n}=1+w+w^2=0$
So $n=3m+1$ satisfies the equation.
If we substitute $n=3m$,
We get,
$1+{\omega }^{3m}+{\omega }^{6m}=3\ne 0$
$n=3m+3$ would lead to the same results as option A.
If $n=2m+3$, we get
$L.H.S=1+{\omega }^{2m+3}+{\omega }^{4m+6}$
The value of this expression depends on the value of $m$, hence option D does not satisfy the expression