Complex Number Question 10
Question 10 - 2024 (29 Jan Shift 2)
Let $\alpha, \beta$ be the roots of the equation $x^{2}-\sqrt{6} x+3=0$ such that $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Let $a, b$ be integers not divisible by 3 and $n$ be a natural number such that $\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^{\mathrm{n}}(\mathrm{a}+\mathrm{ib}), \mathrm{i}=\sqrt{-1}$. Then $\mathrm{n}+\mathrm{a}+\mathrm{b}$ is equal to
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Answer (49)
Solution
$x^{2}-\sqrt{6} x+6=0_{\Delta \beta}^{\alpha}$
$x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i)$
$\alpha=\sqrt{3}\left(e^{i \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(e^{-i \frac{\pi}{4}}\right)$
$\therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right)$
$=\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(e^{i 99 \frac{\pi}{4}}\right) \times \sqrt{2}$
$=3^{49}(-1+\mathrm{i})$
$=3^{\mathrm{n}}(\mathrm{a}+\mathrm{ib})$
$\therefore \mathrm{n}=49, \mathrm{a}=-1, \mathrm{~b}=1$
$\therefore \mathrm{n}+\mathrm{a}+\mathrm{b}=49-1+1=49$
