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(C) Given the equation $x^2-x+2=0$, the roots are $\alpha, \beta = \frac{1 \pm \sqrt{1-8}}{2} = \frac{1 \pm i\sqrt{7}}{2}$.Since $\operatorname{Im}(\a

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$13$

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(C) Given the equation $x^2-x+2=0$, the roots are $\alpha, \beta = \frac{1 \pm \sqrt{1-8}}{2} = \frac{1 \pm i\sqrt{7}}{2}$.Since $\operatorname{Im}(\alpha) > \operatorname{Im}(\beta)$, we have $\alpha = \frac{1 + i\sqrt{7}}{2}$ and $\beta = \frac{1 - i\sqrt{7}}{2}$.Note that $\alpha + \beta = 1$ and $\alpha \beta = 2$.Also, $\alpha^2 = \alpha - 2$.Then $\alpha^4 = (\alpha-2)^2 = \alpha^2 - 4\alpha + 4 = (\alpha-2) - 4\alpha + 4 = -3\alpha + 2$.And $\alpha^6 = \alpha^2 \cdot \alpha^4 = (\alpha-2)(-3\alpha+2) = -3\alpha^2 + 2\alpha + 6\alpha - 4 = -3(\alpha-2) + 8\alpha - 4 = 5\alpha + 2$.Similarly, $\beta^4 = -3\beta + 2$.Substituting these into the expression $\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$:$= (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2)$$= 5\alpha - 3\alpha - 3\beta + 6 - 5\alpha + 10$$= -3\alpha - 3\beta + 16$$= -3(\alpha + \beta) + 16$$= -3(1) + 16 = 13$.

Let $\alpha, \beta$ be the roots of the equation $x^2-x+2=0$ with $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Then $\alpha^6+\alpha^4+\beta^4-5 \alpha^2$ is equal to

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