If 1, alpha_1, alpha_2, ldots, alpha_{n-1} ar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Since $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$ are the $n$th roots of unity,they are the roots of the equation $x^n - 1 = 0$. Thus,we can wri

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Since $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$ are the $n$th roots of unity,they are the roots of the equation $x^n - 1 = 0$. Thus,we can write $x^n - 1 = (x - 1)(x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_{n-1})$. Dividing both sides by $(x - 1)$,we get $(x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_{n-1}) = \frac{x^n - 1}{x - 1}$. Substituting $x = -1$ into the equation,we get $(-1 - \alpha_1)(-1 - \alpha_2) \ldots (-1 - \alpha_{n-1}) = \frac{(-1)^n - 1}{-1 - 1}$. Factoring out $(-1)$ from each of the $(n-1)$ terms on the left side,we have $(-1)^{n-1}(1 + \alpha_1)(1 + \alpha_2) \ldots (1 + \alpha_{n-1}) = \frac{(-1)^n - 1}{-2}$. Since $n$ is an even natural number,$(-1)^n = 1$. Therefore,$(-1)^{n-1}(1 + \alpha_1)(1 + \alpha_2) \ldots (1 + \alpha_{n-1}) = \frac{1 - 1}{-2} = 0$. Since $(-1)^{n-1} 
eq 0$,it follows that $(1 + \alpha_1)(1 + \alpha_2) \ldots (1 + \alpha_{n-1}) = 0$.

If $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$ are the $n$th roots of unity and $n$ is an even natural number,then $(1+\alpha_1)(1+\alpha_2) \ldots (1+\alpha_{n-1})=$

Similar Questions

If $\omega$ is a complex cube root of unity,then $\sin \left[\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right]=$

One of the values of $(\sqrt{3}-i)^{\frac{2}{5}}$ is

If $Z_r = \cos \left(\frac{\pi}{2^r}\right) + i \sin \left(\frac{\pi}{2^r}\right)$ for $r = 1, 2, 3, \ldots$,then the product $Z_1 Z_2 Z_3 \ldots \infty$ is equal to:

The product of the distinct $(2n)^{\text{th}}$ roots of $1+i\sqrt{3}$ is equal to:

$(\sqrt{3}+i)^{10}+(\sqrt{3}-i)^{10}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module