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

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(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,$\alpha_1 \cos n 	heta + \alpha_2 \cos (n-1) 	heta + \ldots + \alpha_n \cos 	heta = -1$.

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