Let p,q∈R and (1−3i)200=2199(p+iq), i=−1 then p+q+q2, and p−q+q2 are roots of the equation.

(1−3i)200=2199(p+iq)(12−32i)200=p2+iq2(cosπ3−isinπ3)200=p2+iq2−12−i32=p2+iq2p=−1,q=−3p+q+q2=2−3 p−q+q2=2+3Q.Eqn is x2−4x+1=0

Let p,q∈R and (1−3i)200=2199(p+iq), i=−1 then p+q+q2, and p−q+q2 are roots of the equation.

(1−3i)200=2199(p+iq)(12−32i)200=p2+iq2(cosπ3−isinπ3)200=p2+iq2−12−i32=p2+iq2p=−1,q=−3p+q+q2=2−3 p−q+q2=2+3Q.Eqn is x2−4x+1=0

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

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

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

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

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

answer is B.

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Detailed Solution

${(1 - \sqrt{3} i)}^{200} = 2^{199} (p + i q)$
${(\frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{200} = \frac{p}{2} + \frac{i q}{2}$
${(\cos \frac{π}{3} - i \sin \frac{π}{3})}^{200} = \frac{p}{2} + \frac{i q}{2}$
$\frac{- 1}{2} - \frac{i \sqrt{3}}{2} = \frac{p}{2} + \frac{i q}{2}$
$p = - 1, q = - \sqrt{3}$
$p + q + q^{2} = 2 - \sqrt{3} p - q + q^{2} = 2 + \sqrt{3}$
$Q . E q^{n}$ is $x^{2} - 4 x + 1 = 0$