Complex Number Question 8
Question 8 - 2024 (29 Jan Shift 1)
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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Answer (13)
Solution
$$ \begin{aligned} & \alpha^{6}+\alpha^{4}+\beta^{4}-5 \alpha^{2} \ & =\alpha^{4}(\alpha-2)+\alpha^{4}-5 \alpha^{2}+(\beta-2)^{2} \ & =\alpha^{5}-\alpha^{4}-5 \alpha^{2}+\beta^{2}-4 \beta+4 \ & =\alpha^{3}(\alpha-2)-\alpha^{4}-5 \alpha^{2}+\beta-2-4 \beta+4 \ & =-2 \alpha^{3}-5 \alpha^{2}-3 \beta+2 \ & =-2 \alpha(\alpha-2)-5 \alpha^{2}-3 \beta+2 \ & =-7 \alpha^{2}+4 \alpha-3 \beta+2 \ & =-7(\alpha-2)+4 \alpha-3 \beta+2 \ & =-3 \alpha-3 \beta+16=-3(1)+16=13 \end{aligned} $$
