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Logarithmic Differentiation

If 1, a1, a2,...

Question

If $1, a_{1}, a_{2}, . . . a_{n - 1}$ are the $n t h$ roots of unity, then the value of $(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) . . . (1 - a_{n - 1})$ is

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
$n$

Explanation for correct option:
Given,
$1, a_{1}, a_{2}, . . . a_{n - 1}$ are the $n t h$ roots of unity So,
$x^{n} - 1 = (x - 1) (x - a_{1}) . . . (x - a_{n - 1}) \frac{x^{n} - 1}{(x - 1)} = (x - a_{1}) (x - a_{2}) . . . (x - a_{n - 1}) x^{n - 1} + x^{n - 2} + . . . + x^{2} + x + 1 = (x - a_{1}) (x - a_{2}) . . . (x - a_{n - 1})$
Substitute $x = 1$
$1^{n - 1} + 1^{n - 2} + . . . + 1^{2} + 1 + 1 = (1 - a_{1}) (1 - a_{2}) . . . (1 - a_{n - 1}) = 1 + 1 + 1 + . . . + n t i m e s$
$\Rightarrow$ $R H S$ $= n$
$\Rightarrow$ $L H S$ $= 1^{n - 1} + 1^{n - 2} + . . . + 1^{2} + 1 + 1$
$\Rightarrow$ $L H S = 1 + 1 + . . . + 1 + 1 + 1$
$\Rightarrow$ $L H S = n$
$\Rightarrow$ $L H S = R H S$
Hence option (C) is correct.

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Similar questions

1, a_{1}, a_{2}, . a_{n - 1}

n

roots of unity, then the value of

(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) . (1 - a_{n - 1})

a_{1}, a_{2}, . . ., a_{n - 1}

n^{t h}

roots of unity then

\frac{1}{1 - a_{1}} + \frac{1}{1 - a_{2}} + \dots + \frac{1}{1 - a_{n - 1}}

1, a_{1}, a_{2} . . ., a_{n - 1}

n^{t h}

roots of unity, then

\frac{1}{1 - a_{1}} + \frac{1}{1 - a_{2}} + . . . + \frac{1}{1 - a_{n - 1}}

a_{1}, a_{2}, . . . ., a_{n - 1}

n t h

roots of unity then the value of

(1 - a_{1}) (1 - a_{2}) . . . (1 - a_{n - 1})

a_{1}, a_{2}, . . . . a_{n - 1}

n

th roots of unity then prove that
(i)

(1 - a_{1}) (1 - a_{2}) (1 - a_{n - 1}) . . . = n

1 + a_{1} + a_{2} + . . . . + a_{n - 1} = 0

\frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + . . . . \frac{1}{2 - a_{n - 1}} = \frac{(n - 2) 2^{n - 1} + 1}{2^{n} - 1}

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