If 1, a1, a2, ldots, an-1 are the nth roots of unity, then (1-a1)(1-a2)(1-a3) ldots(1-an-1)=

Question

If 1, a1, a2, ldots, an-1 are the nth roots of unity, then (1-a1)(1-a2)(1-a3) ldots(1-an-1)= (A) n+1 (B) n (C) n-1 (D) None of these

Tardigrade Admin · Accepted Answer

Let √[n]1=x ; ∴ xn=1 ; ∴ xn-1=0 ∴ xn-1=(x-1)(x-a1)(x-a2) ldots(x-an-1) ∴ (x-a1)(x-a2)(x-a3) ldots(x-an-1) =(xn-1/x-1)=(1-xn/1-x) =1+x+x2+ ldots,+xn-1 . Putting x=1, we get (1-a1)(1-a2)(1-a3) ldots(1-an-1)=n

Q. If $1, a_{1}, a_{2}, \dots, a_{n - 1}$ are the $n$ th roots of unity, then $(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) \dots (1 - a_{n - 1}) =$

$n + 1$

$n$

$n - 1$

None of these

Q. If 1,a1​,a2​,…,an−1​ are the nth roots of unity, then (1−a1​)(1−a2​)(1−a3​)…(1−an−1​)=

n+1

n

n−1

None of these

Let n1​=x;∴xn=1; ∴xn−1=0 ∴xn−1=(x−1)(x−a1​)(x−a2​)…(x−an−1​) ∴(x−a1​)(x−a2​)(x−a3​)…(x−an−1​) =x−1xn−1​=1−x1−xn​ =1+x+x2+…,+xn−1. Putting x=1, we get (1−a1​)(1−a2​)(1−a3​)…(1−an−1​)=n

Q. If $1, a_{1}, a_{2}, \dots, a_{n - 1}$ are the $n$ th roots of unity, then $(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) \dots (1 - a_{n - 1}) =$

$n + 1$

$n$

$n - 1$