Question

If $a_1,a_2,...,a_{n-1}$ are the nth roots of unity, then the value of $(1-a_1)(1-a_2)...(1-a_{n-1})$ is equal to

• A. $\sqrt{3}$
• B. $\frac{1}{2}$
• C. $n$
• D. $0$

Answer 1

Answer: (C)

Since, $1,a_1,a_2,....,a_{n-1}$ are the nth root of unity.
$\therefore$ $x^n-1=(x-1)(x-a_1)...(x-a_{n-1})$
$\Rightarrow$ $\frac{x^n-1}{x-1}=(x-a_1)(x-a_2)...(x-a_{n-1})$
$\therefore$ $x^{n-1}+x^{n-2}+...+x^2+x+1$
$=(x-a_1)(x-a_2)...(x-a_{n-1})$
Put $x=1$ , we get $(1-a_1)(1-a_2)...(1-a_{n-1})=1+1+...+n$ times $=n$