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(C) The given quadratic equation is $x^2-2x+4=0$. Since $\alpha$ and $\beta$ are the roots,we have $\alpha+\beta=2$ and $\alpha\beta=4$. Note that the

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$-2^{10}$

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(C) The given quadratic equation is $x^2-2x+4=0$. Since $\alpha$ and $\beta$ are the roots,we have $\alpha+\beta=2$ and $\alpha\beta=4$. Note that the roots of $x^2-2x+4=0$ are $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm i\sqrt{3}$. In polar form,$1 \pm i\sqrt{3} = 2(\cos(\frac{\pi}{3}) \pm i\sin(\frac{\pi}{3})) = 2e^{\pm i\pi/3}$. Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$. Then $\alpha^9 = (2e^{i\pi/3})^9 = 2^9 e^{i3\pi} = 2^9(\cos(3\pi) + i\sin(3\pi)) = 2^9(-1) = -2^9$. Similarly,$\beta^9 = (2e^{-i\pi/3})^9 = 2^9 e^{-i3\pi} = 2^9(\cos(-3\pi) + i\sin(-3\pi)) = 2^9(-1) = -2^9$. Therefore,$\alpha^9+\beta^9 = -2^9 - 2^9 = -2 	imes 2^9 = -2^{10}$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^9+\beta^9$ is equal to

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