If alpha_1, alpha_2, alpha_3, ldots, alpha_n | vedclass

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

Question

(C) Given the equation $\alpha_1 + \alpha_2 z + \alpha_3 z^2 + \ldots + \alpha_n z^{n-1} + z^n = 0$. Since $z = \cos 	heta + i \sin 	heta = e^{i 	h

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) Given the equation $\alpha_1 + \alpha_2 z + \alpha_3 z^2 + \ldots + \alpha_n z^{n-1} + z^n = 0$. Since $z = \cos 	heta + i \sin 	heta = e^{i 	heta}$,we have $z^n = \cos n 	heta + i \sin n 	heta$. Substituting $z$ into the equation: $\alpha_1 + \alpha_2 e^{i 	heta} + \alpha_3 e^{i 2 	heta} + \ldots + \alpha_n e^{i (n-1) 	heta} + e^{i n 	heta} = 0$. Multiply the entire equation by $e^{-i n 	heta}$: $\alpha_1 e^{-i n 	heta} + \alpha_2 e^{-i (n-1) 	heta} + \ldots + \alpha_n e^{-i 	heta} + 1 = 0$. Taking the real part of this equation: $\alpha_1 \cos (-n 	heta) + \alpha_2 \cos (-(n-1) 	heta) + \ldots + \alpha_n \cos (-	heta) + 1 = 0$. Since $\cos (-x) = \cos x$,this simplifies to: $\alpha_1 \cos n 	heta + \alpha_2 \cos (n-1) 	heta + \ldots + \alpha_n \cos 	heta + 1 = 0$. Therefore,$\alpha_1 \cos n 	heta + \alpha_2 \cos (n-1) 	heta + \ldots + \alpha_n \cos 	heta = -1$.

Similar Questions

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^{4}+x^{3}+x^{2}+x+1=0$,then $\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}$ is equal to

If $1, \omega, \omega^2, \ldots, \omega^8$ are the roots of the equation $x^9-1=0$,then $\sum_{r=1}^8 \left(\omega^r\right)^{99} =$

If $(\sqrt{3}+i)^8-(\sqrt{3}-i)^8=\alpha+i \beta$,then $\alpha-\frac{\sqrt{3}}{2} \beta=$

If $z = \sin \theta - i \cos \theta,$ then for any integer $n$

$\sum_{k=1}^6 \left[ \sin \left(\frac{2 \pi k}{7}\right) - i \cos \left(\frac{2 \pi k}{7}\right) \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module