If $\omega$ is an imaginary cube root of unity, $n$ is a positive integer but not a multiple of 3, then the value of $1 + \omega^{n} + \omega^{2n}$ is

Question

Dr. Abubacker · Accepted Answer

(c)
Since, $n$ is not a multiple of 3, therefore$n = 3m + 1,\ n = 3m + 2$, where $m$ is a positive integer.
For $n = 3m + 1$,
$1 + \omega^{n} + \omega^{2n} = 1 + \omega^{3m + 1} + \omega^{2(3m + 1)}$
$= 1 + \omega^{3m}\omega + \left( \omega^{3} ight)^{2m}\omega^{2} = 1 + \omega + \omega^{2} = 0$
Similarly, for $n = 3m + 2$
$	herefore\ \ \ \ 1 + \omega^{n} + \omega^{2n} = 1 + \omega^{3m + 2} + \omega^{2(3m + 2)}$
$= 1 + \omega^{3m}.\omega^{2} + \left( \omega^{3} ight)^{2m}.\omega^{3}.\omega = 0$
$\lbrack\because\ \ \omega^{3} = 1brack$

Q. If $\omega$ is an imaginary cube root of unity, $n$ is a positive integer but not a multiple of 3, then the value of $1 + \omega^{n} + \omega^{2n}$ is

Explanation

Reviewed by Dr. Abubacker

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