Q48 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics
$n$ is a positive integer and not a multiple of 3 . If $\omega$ is non real cube root of unity, then $\omega^n+\omega^{2 n}$ is equal to
- Correct answerA-1
- B3
- C-3
- D1
Solution
If $n$ is not a multiple of 3 , then $n=3 K+1$ or
$n=3 K+2$
where $K \in I$
Case (I) : When, $n=3 K+1, K \in I$
$\begin{aligned}
& \omega^n+\omega^{2 n}=\omega^{3 K+1}+\omega^{2(3 K+1)} \\
& =\omega^{3 k} \cdot \omega+\omega^{6 K} \cdot \omega^2 \\
& \quad\left[\because 1+\omega+\omega^2=0 \text { and } \omega^3=1\right] \\
& =\left(\omega^3\right)^K \cdot \omega+\left(\omega^3\right)^{2 K} \cdot \omega^2 \\
& =\omega+\omega^2=-1
\end{aligned}$
Case (II): When, $n=3 K+2, K \in I$
$\omega^n+\omega^{2 n}=\omega^{3 K+2}+\omega^{2(3 K+2)}$
$\begin{aligned}
& =\omega^{3 K} \cdot \omega^2+\omega^{6 K} \cdot \omega^4 \\
& =\left(\omega^3\right)^K \cdot \omega^2-1\left(\omega^3\right)^{2 K} \cdot \omega \\
& =\omega^2+\omega=-1
\end{aligned}$
In both cases, $\omega^n+\omega^{2 n}=-1, \forall n \notin$ multiple of 3 .
Hence, the correct answer is option: 'A'.