Let be the roots of the equation x 2 - 6 x + 3 = 0 such that. Let a , b be integers not divisible by 3 and n be a natural number such that 99 + 98 = 3 n ( a + i b ) , ? i = - 1 . Then n + a + b is equal to ______ . [2024]

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Solution

Q.
Let $α, β$ be the roots of the equation $x^{2} - \sqrt{6} x + 3 = 0$ such that $Im (α) > Im (β) .$ Let $a, b$ be integers not divisible by 3 and $n$ be a natural number such that $\frac{α^{99}}{β} + α^{98} = 3^{n} (a + i b), i = \sqrt{- 1} .$ Then $n + a + b$ is equal to ______ . [2024]

Ans.
(49)

Roots of the equation $x^{2} - \sqrt{6} x + 3 = 0$ are

$x = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2} \Rightarrow x = \frac{\sqrt{6} \pm \sqrt{6 i}}{2}$

$∵$ $Im (α) > Im (β)$

$∴$ $α = \frac{\sqrt{6} + \sqrt{6 i}}{2}, β = \frac{\sqrt{6} - \sqrt{6 i}}{2}$

$\Rightarrow α^{2} = \frac{2 \cdot \sqrt{6} \cdot \sqrt{6} \cdot i}{4}, β^{2} = \frac{- 2 \cdot \sqrt{6} \cdot \sqrt{6 i}}{4}$

$α^{2} = 3 i, β^{2} = - 3 i \Rightarrow \frac{α}{β} = i$

Now, $\frac{α^{99}}{β} + α^{98} = 3^{n} (a + i b)$

$\Rightarrow α^{98} (\frac{α}{β} + 1) = 3^{n} (a + i b) \Rightarrow {(3 i)}^{49} (i + 1) = 3^{n} (a + i b)$

$\Rightarrow 3^{49} \cdot i (i + 1) = 3^{n} (a + i b) \Rightarrow 3^{49} \cdot (- 1 + i) = 3^{n} (a + i b)$

$\Rightarrow n = 49, a = - 1, b = 1$ $∴$ $n + a + b = 49$

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