Question

If $1, \omega, \omega^2, \omega^3, \ldots \ldots, \omega^{n-1}$ are the $n, n$th roots of unity, then $(1-\omega)\left(1-\omega^2\right) \ldots\left(1-\omega^{n-1}\right)$ equals

(a) 0

(b) 1

(c) n

(d) $n^2$

Answer 1

SOLUTION : Since, $1, \omega, \omega^2, \omega^3, \ldots, \omega^{n-1}$ are $n, n$th roots of unity, therefore, we have the identity

$(x-1)(x-\omega)\left(x-\omega^2\right) \ldots\left(x-\omega^{n-1}\right)=x^n-1 \\$

$\Rightarrow \quad(x-\omega)\left(x-\omega^2\right) \ldots\left(x-\omega^{n-1}\right)=\frac{x^n-1}{x-1} \\$

$=x^{n-1}+x^{n-2}+\ldots+x+1$

Putting $x=1$ on both sides, we get

$(1-\omega)\left(1-\omega^2\right) \ldots\left(1-\omega^{n-1}\right)=n$

So, The correct option will be (c).