Question

If $1,\omega ,{\omega }^2$ be the three cube roots of unity, then $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)...$ to $2n$ factors $=$

• A. $1$
• B. $2^{2n}$
• C. $2^{4n}$
• D. None of these

Answer 1

The correct option is A $1$

We have $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)...$ to $2n$ factors

$=(1+\omega )(1+{\omega }^2)(1+{\omega }^3.\omega )(1+{\omega }^6.{\omega }^2)...$ $2n$ factors

$=(1+\omega )(1+{\omega }^2)(1+\omega )(1+{\omega }^2)...$ to $2n$ factors $[\because {\omega }^3={\omega }^6=...=1]$

$=[(1+\omega )(1+\omega )...$ to $n$ factors $\left]\right[(1+{\omega }^2)(1+{\omega }^2)...$ to $n$ factors $]$

$={(1+\omega )}^n{(1+{\omega }^2)}^n={\left[(1+\omega )(1+{\omega }^2)\right]}^n$

$={[1+\omega +{\omega }^2+{\omega }^3]}^n={(0+1)}^n=1$

$[\because 1+\omega +{\omega }^2=0,{\omega }^3=1]$