Question

If $1, a_{1}, a_{2}, \ldots, a_{n-1}$ are the $n$th roots of unity, then $\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right) \ldots\left(1-a_{n-1}\right)=$

• A. $n+1$
• B. $n$
• C. $n-1$
• D. None of these

Answer 1

Answer: (B)

Let $\sqrt[n]{1}=x ; \therefore x^{n}=1 ; $
$\therefore x^{n}-1=0$
$\therefore x^{n}-1=(x-1)\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n-1}\right) $
$\therefore \left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right) \ldots\left(x-a_{n-1}\right) $
$=\frac{x^{n}-1}{x-1}=\frac{1-x^{n}}{1-x}$
$=1+x+x^{2}+\ldots,+x^{n-1} .$
Putting $x=1$, we get
$\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right) \ldots\left(1-a_{n-1}\right)=n$