Let p , q R and ( 1 - 3 i ) 200 = 2 199 ( p + i q ) , i = - 1 . The p + q + q 2 and p - q + q 2 are roots of the equation [2023]

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Solution

Q.
Let $p, q \in R$ and ${(1 - \sqrt{3} i)}^{200} = 2^{199} (p + i q), i = \sqrt{- 1} .$ The $p + q + q^{2}$ and $p - q + q^{2}$ are roots of the equation [2023]

1 $x^{2} - 4 x + 1 = 0$

2 $x^{2} + 4 x + 1 = 0$

3 $x^{2} - 4 x - 1 = 0$

4 $x^{2} + 4 x - 1 = 0$

Ans.
(1)

$Given: {(1 - \sqrt{3} i)}^{200} = 2^{199} (p + i q)$

$\Rightarrow {(2 e^{- i π / 3})}^{200} = 2^{199} (p + i q)$

$\Rightarrow 2^{200} {(\cos \frac{π}{3} - i \sin \frac{π}{3})}^{200} = 2^{199} (p + i q)$

$2 (\cos \frac{200 π}{3} - i \sin \frac{200 π}{3}) = p + i q$

$∴ p = - 1, q = - \sqrt{3}$

$\Rightarrow p + q + q^{2} = - 1 - \sqrt{3} + 3 = 2 - \sqrt{3}$

$and p - q + q^{2} = - 1 + \sqrt{3} + 3 = 2 + \sqrt{3}$

$Sum of roots = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$

$Product of roots = (2 + \sqrt{3}) (2 - \sqrt{3}) = 4 - 3 = 1$

$Required quadratic equation is x^{2} - 4 x + 1 = 0$

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