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

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

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

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 $(1+2 \omega_0)(1+2 \omega_1)(1+2 \omega_2) \ldots (1+2 \omega_{n-1})=$

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