If omega_0, omega_1, ldots, omega_{n-1} are t | vedclass

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

Question

(D) The $n$-th roots of unity satisfy the equation $x^n - 1 = 0$. Thus,we can write $x^n - 1 = (x - \omega_0)(x - \omega_1) \ldots (x - \omega_{n-1})$

Vedclass · Accepted Answer

$1+(-1)^{n-1} 2^n$

Vedclass · Answer

(D) The $n$-th roots of unity satisfy the equation $x^n - 1 = 0$. Thus,we can write $x^n - 1 = (x - \omega_0)(x - \omega_1) \ldots (x - \omega_{n-1})$. To find the product $(1+2 \omega_0)(1+2 \omega_1) \ldots (1+2 \omega_{n-1})$,we factor out $2^n$: $2^n (\frac{1}{2} + \omega_0)(\frac{1}{2} + \omega_1) \ldots (\frac{1}{2} + \omega_{n-1})$. Let $x = -\frac{1}{2}$. Then $x^n - 1 = (-\frac{1}{2})^n - 1 = (-\frac{1}{2} - \omega_0)(-\frac{1}{2} - \omega_1) \ldots (-\frac{1}{2} - \omega_{n-1})$. Multiplying by $(-1)^n$,we get $(-1)^n (x^n - 1) = (\omega_0 + \frac{1}{2})(\omega_1 + \frac{1}{2}) \ldots (\omega_{n-1} + \frac{1}{2})$. Substituting $x = -\frac{1}{2}$: $(-1)^n ((-\frac{1}{2})^n - 1) = (\frac{1}{2} + \omega_0)(\frac{1}{2} + \omega_1) \ldots (\frac{1}{2} + \omega_{n-1})$. Multiplying both sides by $2^n$: $2^n (\frac{1}{2} + \omega_0) \ldots (\frac{1}{2} + \omega_{n-1}) = 2^n (-1)^n ((-1)^n \frac{1}{2^n} - 1) = 2^n ((-1)^{2n} \frac{1}{2^n} - (-1)^n) = 1 - (-1)^n 2^n = 1 + (-1)^{n-1} 2^n$.

If $\omega_0, \omega_1, \ldots, \omega_{n-1}$ are the $n$-th roots of unity,then $(1+2 \omega_0)(1+2 \omega_1)(1+2 \omega_2) \ldots (1+2 \omega_{n-1})=$

Similar Questions

If $z = \frac{\sqrt{3} + i}{2}$,then $\left(z^{101} + i^{103}\right)^{105} = $

If $x = a + b$,$y = a\alpha + b\beta$,and $z = a\beta + b\alpha$,where $\alpha$ and $\beta$ are complex cube roots of unity,then $xyz$ =

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

Let $z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$,then:

What is the value of $(1-i \sqrt{3})^9$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module