JEE Mains · Maths · STD 11 - 4.1 complex nubers
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 \(\mathrm{n}+\mathrm{a}+\mathrm{b}\) is equal to ...........
- A \(49\)
- B \(42\)
- C \(45\)
- D \(59\)
Answer & Solution
Correct Answer
(A) \(49\)
Step-by-step Solution
Detailed explanation
\( x^2-\sqrt{6} x+6=0 \) \( x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i) \) \( \alpha=\sqrt{3}\left(e^{i \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(e^{-i \frac{\pi}{4}}\right) \)…
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