Question

If $\omega$ is an imaginary cube root of unity, then the value of $(1-\omega+\omega^2)\cdot(1-\omega^2+\omega^4)\cdot(1-\omega^4+\omega^8)\dots$ (2n factors ) is

• A. $2^{2n}$
• B. $2^n$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

Given, $\omega^{3}=1,1+\omega+\omega^{2}=0$...(i)
Now,
$\left(1-\omega+\omega^{2}\right) \cdot\left(1-\omega^{2}+\omega^{4}\right) \cdot\left(1-\omega^{4}+\omega^{8}\right)$
$\left(1-\omega^{8}+\omega^{16}\right)...2 n$ factors
$=\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega\right) \cdot\left(1-\omega+\omega^{2}\right) \cdot\left(1-\omega^{2}+\omega\right)...2 n$ 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)]
$=(-2 \omega) \cdot\left(-2 \omega^{2}\right) \cdot(-2 \omega) \cdot\left(-2 \omega^{2}\right)... 2 n$ factors
$=(-2)^{2 n}(\omega)^{n}\left(\omega^{2}\right)^{n}$
$=(-1)^{2 n}(2)^{2 n} \omega^{3 n}$
$=2^{2 n} \cdot\left(\omega^{3}\right)^{n}=2^{2 n}(1)^{n}=2^{2 n}$ [from Eq .(i)]