​​Let α, β be the roots of the equation x^2 − √6x + 3 = 0 such that Im(α) > Im(β).

Question

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 \(\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^{n}(a+i b), i=\sqrt{-1}\). Then \(n+a+b\) is equal to ____.

Answer 1

Correct answer: 49

\(\mathrm{x}^{2}-\sqrt{6} \mathrm{x}+6=0_{\rightarrow \beta}^{\rightarrow\alpha}\)

\(\mathrm x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i)\)

\(\alpha=\sqrt{3}\left(\mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}\right)\)

\(\therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right)\)

\(=\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(\mathrm{e}^{\mathrm{i} 99 \frac{\pi}{4}}\right) \times \sqrt{2}\)

\(=3^{49}(-1+\mathrm{i})\)

\(=3^{\mathrm{n}}(\mathrm{a}+\mathrm{ib})\)

\(\therefore \mathrm{n}=49, \mathrm{a}=-1, \mathrm{~b}=1\)

\(\therefore \mathrm{n}+\mathrm{a}+\mathrm{b}=49-1+1=49\)