If $1,\ a_{1},\ a_{2},\ldots,a_{n - 1}$ are the $n$ roots of unity, then the value of $\left( 1 - a_{1} \right)\left( 1 - a_{2} \right)\left( 1 - a_{3} \right)\ldots(1 - a_{n - 1})$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
As, $1,\ a_{1},\ a_{2},\ \ldots,\ a_{n - 1}$ are $n$th roots unity
$\Rightarrow \ \ \ \left( x^{n} - 1 ight) = (x - 1)\left( x - a_{1} ight)\left( x - a_{2} ight)\ldots(x - a_{n - 1})$
$\Rightarrow \ \ \frac{x^{n} - 1}{x - 1} = \left( x - a_{1} ight)\left( x - a_{2} ight)\ldots(x - a_{n - 1})$
$	herefore\ \ \ \ x^{n - 1} = x^{n - 2} + ...x^{2} + x + 1$
$= \left( x - a_{1} ight)\left( x - a_{2} ight)\ldots(x - a_{n - 1})$
$\left\lbrack as\ \frac{x^{n} - 1}{x - 1} = x^{n - 1} + x^{n - 2} + ... + x + 1 ightbrack$Putting $x = 1$, we get
$1 + 1 + \ldots n$ times $= \left( 1 - a_{1} ight)\left( 1 - a_{2} ight)\ldots\left( 1 - a_{n - 1} ight)$
$\Rightarrow \ \ \ \left( 1 - a_{1} ight)\left( 1 - a_{2} ight)\ldots\left( 1 - a_{n - 1} ight) = n$

Q. If $1,\ a_{1},\ a_{2},\ldots,a_{n - 1}$ are the $n$ roots of unity, then the value of $\left( 1 - a_{1} \right)\left( 1 - a_{2} \right)\left( 1 - a_{3} \right)\ldots(1 - a_{n - 1})$ is equal to

Explanation

Reviewed by Dr. Abubacker

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