Q99 / 180 · KCET 2011 · Mathematics
If $\omega$ is an imaginary cube root of unity, then the value of $\left(1-\omega+\omega^{2}\right) \cdot\left(1-\omega^{2}+\omega^{4}\right) \cdot\left(1-\omega^{4}+\omega^{8}\right) \cdot \ldots$
(2n factors) is
- Correct answerA$2^{2 n}$
- B$2^{n}$
- C1
- D0
Solution
Given, $\quad \omega^{3}=1,1+\omega+\omega^{2}=0$
Now,
$\left(1-\omega+\omega^{2}\right) \cdot\left(1-\omega^{2}+\omega^{4}\right) \cdot\left(1-\omega^{4}+\omega^{8}\right)$.
$\left(1-\omega^{8}+\omega^{16}\right) \ldots 2 n$ factors $=\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega\right)$
$\cdot\left(1-\omega+\omega^{2}\right) \cdot\left(1-\omega^{2}+\omega\right) \ldots .2 n$ factors
[from Eq. (i)]
$=(-\omega-\omega) \cdot\left(-\omega^{2}-\omega^{2}\right) \cdot(-\omega-\omega) \cdot\left(-\omega^{2}-\omega^{2}\right)$
... $2 n$ factors [from Eq. (i)]
$=(-2 \omega) \cdot\left(-2 \omega^{2}\right) \cdot(-2 \omega) \cdot\left(-2 \omega^{2}\right) \ldots 2 n$ factors $=(-2)^{2 n}(\omega)^{n}\left(\omega^{2}\right)^{n}$ $=(-1)^{2 n}(2)^{2 n} \omega^{3 n}$ $=2^{2 n} \cdot\left(\omega^{3}\right)^{n}=2^{2 n}(1)^{n}=2^{2 n} \quad$ [from Eq. (i)]
Hence, the correct answer is option: 'A'.