Question

lf $1,\omega ,{\omega }^2$ are cube roots of unity then the value $(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)(1-{\omega }^4+{\omega }^8)\dots ...$ (upto 2n factors) is?

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

Answer 1

The correct option is B $2^{2n}$

Use the result: $1+w+w^2=0$
$\therefore 1-w+w^2=-2w$
Also $(1-w^2+w^4)=(-2w^2)$
$(1-w^4-w^8)=-2w^4$
$\therefore$ the series becomes
$(-2w)(-2w^2)(-2w^4)(-2w^8)....2n$
Clubbing adjacent two terms, we get
$(+2^2{w}^3)\left(2^2{w}^{12}\right)...$2n terms
$=2^{2n}$(every 2 adjacent terms give $2^2$; we get n such terms)