If $1, w, w^{2}, . . . w^{n - 1}$ are the $n^{t h}$ roots of unity, then $(1 - w) (1 - w^{2}) . . . (1 - w^{n - 1}) i s$

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Solution

The correct option is A $0$
we know that,

$1 + w + w^{2} = 0$

$\Rightarrow w + w^{2} = - 1$ , $w^{3} = 1$

Given,

$(1 - w) (1 - w^{2}) (1 - w^{3}) . . . . . . . (1 - w^{n - 1})$

$= (1 - w - w^{2} + w^{3}) (1 - w^{3}) (1 - w^{4}) . . . . . . . (1 - w^{n - 1})$

$= (1 + 1 + 1) (1 - 1) (1 - w) . . . . . . . . . . . (1 - w^{n})$

$= 0$

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