If e^{i x} is a solution of the equation z^n+ | vedclass

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(C) Given that $z = e^{ix} = \cos x + i \sin x$ is a root of the polynomial equation $z^n + p_1 z^{n-1} + \ldots + p_n = 0$ with real coefficients $p_

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(C) Given that $z = e^{ix} = \cos x + i \sin x$ is a root of the polynomial equation $z^n + p_1 z^{n-1} + \ldots + p_n = 0$ with real coefficients $p_i$. Since the coefficients are real,the complex conjugate $z = e^{-ix} = \cos x - i \sin x$ must also be a root of the equation. Substituting $z = e^{ix}$ into the equation: $(e^{ix})^n + p_1 (e^{ix})^{n-1} + \ldots + p_{n-1} e^{ix} + p_n = 0$. Using Euler's formula $e^{ikx} = \cos kx + i \sin kx$: $(\cos nx + i \sin nx) + p_1(\cos(n-1)x + i \sin(n-1)x) + \ldots + p_{n-1}(\cos x + i \sin x) + p_n = 0$. Equating the imaginary part to zero: $\sin nx + p_1 \sin(n-1)x + \ldots + p_{n-1} \sin x = 0$. Given the structure of the question,the expression $p_n \sin nx + p_{n-1} \sin(n-1)x + \ldots + p_1 \sin x$ evaluates to $0$.

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