Question

If $1,a_1,a_2,.a_{n-1}$, are $n$ roots of unity, then the value of $(1-a_1)(1-a_2)(1-a_3).(1-a_{n-1})$ is equal to ?

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

Answer 1

The correct option is C $n$

Given $1,a_1,a_2,--------,a_{n_1}n$ roots of unity.

As $1$ is the root of $x^n-1=0$ than $x-1$ is a factor of $x^n-1=0$

As , $x-1$ is a factor of $x^n-1=0$ then $n-1$ divides $x^n-1=0$

$\Rightarrow x-1)x^n+o{x}^{n-1}+o{x}^{n-2}+-------+ox-1(x^{n-1}+x^{n-2}+-----+x^2+x+1)$

$x^n-x^{n-1}$

_________________

$x^{n-1}+o{n}^{n-2}$

$x^{n-1}-x^{n-2}$

__________________

$x^{n-2}+o{x}^{n-3}$

.

.

.

.

__________________

$x-1$

$x-1$

______________________

$0$

$\therefore \frac{x^n-1}{x-1}=x^{n-1}{x}^{n-2}+---------+x^2+x+1$ ------------(1)

As $1,a_1,a_2,................................,a_{n-1}$ arev n roots of unity

$(x^2-1)=(x-1)(x-a_1)(x-a_2)$ -------------(x-a_{n-1})

$\frac{\left(x^n\right)-1}{x-1}=(n-a_1)(x-a_2)-------(x-a_{n-1})$

From equation (1) we get

$x^{n-1}+x^{n-2}+---------+x+x+1=(x-a_1)(x-a_2)----------(x-a_{n-1})$

substitute $x=1$ we get

$11+1+-----------+1+=(1-a_2)(1-a)---------(1-a_{n-1})$

$\therefore (1-a_1)(1-a_2)-------(1-a_{n-1}=n)$