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Question

If $a_{1}, a_{2}, . . . ., a_{n - 1}$ are the $n t h$ roots of unity then the value of $(1 - a_{1}) (1 - a_{2}) . . . (1 - a_{n - 1})$ is equal to

A

\sqrt{3}

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B

\frac{1}{2}

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C

n

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D

0

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Solution

The correct option is C $n$
Since, $1, a_{1}, a_{2}, . . . ., a_{n - 1}$ are the $n t h$ root of unity.
$∴ x^{n} - 1 = (x - 1) (x - a_{1}) . . . (x - a_{n - 1})$
$\Rightarrow \frac{x^{n} - 1}{x - 1} = (x - a_{1}) (x - a_{2}) . . . . (x - a_{n - 1})$
$∴ x^{n - 1} + x^{n - 2} + . . . + x^{2} + x + 1$
$= (x - a_{1}) (x - a_{2}) . . . . (x - a_{n - 1})$
Put $x = 1$ , we get
$(1 - a_{1}) (1 - a_{2}) . . . (1 - a_{n - t}) = 1 + 1 + . . . + n t i m e s$
$= n$ .

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