If 1, z_1, z_2, ldots, z_{n-1} are the nth ro | vedclass

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(C) The $n$th roots of unity are the roots of the equation $z^n - 1 = 0$. We can factorize $z^n - 1$ as: $z^n - 1 = (z - 1)(z - z_1)(z - z_2) \ldots (

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$n$

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(C) The $n$th roots of unity are the roots of the equation $z^n - 1 = 0$. We can factorize $z^n - 1$ as: $z^n - 1 = (z - 1)(z - z_1)(z - z_2) \ldots (z - z_{n-1})$. We also know that $z^n - 1 = (z - 1)(z^{n-1} + z^{n-2} + \ldots + z + 1)$. Equating the two expressions for $z^n - 1$: $(z - 1)(z^{n-1} + z^{n-2} + \ldots + z + 1) = (z - 1)(z - z_1)(z - z_2) \ldots (z - z_{n-1})$. Dividing both sides by $(z - 1)$ (for $z 
eq 1$): $z^{n-1} + z^{n-2} + \ldots + z + 1 = (z - z_1)(z - z_2) \ldots (z - z_{n-1})$. Taking the limit as $z 	o 1$ or simply substituting $z = 1$ in the polynomial identity: $1^{n-1} + 1^{n-2} + \ldots + 1 + 1 = (1 - z_1)(1 - z_2) \ldots (1 - z_{n-1})$. Since there are $n$ terms on the left side (from $z^0$ to $z^{n-1}$),the sum is $n$. Therefore,$(1 - z_1)(1 - z_2) \ldots (1 - z_{n-1}) = n$.

If $1, z_1, z_2, \ldots, z_{n-1}$ are the $n$th roots of unity,then the value of $(1-z_1)(1-z_2) \ldots (1-z_{n-1})$ is equal to

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