If 1, omega, omega^2, omega^3, dots, omega^{n | vedclass

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(C) Since $1, \omega, \omega^2, \dots, \omega^{n-1}$ are the $n^{th}$ roots of unity,they are the roots of the equation $x^n - 1 = 0$.Therefore,we can

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$n$

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(C) Since $1, \omega, \omega^2, \dots, \omega^{n-1}$ are the $n^{th}$ roots of unity,they are the roots of the equation $x^n - 1 = 0$.Therefore,we can write the polynomial as:$x^n - 1 = (x - 1)(x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$Dividing both sides by $(x - 1)$,we get:$\frac{x^n - 1}{x - 1} = (x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$Using the geometric series formula,the left side simplifies to:$x^{n-1} + x^{n-2} + \dots + x + 1 = (x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$Now,taking the limit as $x 	o 1$ or simply substituting $x = 1$ on both sides:$1^{n-1} + 1^{n-2} + \dots + 1 + 1 = (1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1})$Since there are $n$ terms on the left side,the sum is $n$.Thus,$(1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1}) = n$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

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

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