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

2200(cosπ3−isinπ3)200=2199(p+iq) 2(−12−i32)=p+iq ⇒p=−1,q=−3 Let α=p+q+q2=−1−3+3=2−3 β=p−q+q2=−1+3+3=2+3 Req x2−[α+β]x+αβ=0⇒ 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

2200(cosπ3−isinπ3)200=2199(p+iq) 2(−12−i32)=p+iq ⇒p=−1,q=−3 Let α=p+q+q2=−1−3+3=2−3 β=p−q+q2=−1+3+3=2+3 Req x2−[α+β]x+αβ=0⇒ 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

$2^{200} {(\cos \frac{π}{3} - i \sin \frac{π}{3})}^{200} = 2^{199} (p + i q) 2 (\frac{- 1}{2} - \frac{i \sqrt{3}}{2}) = p + i q \Rightarrow p = - 1, q = - \sqrt{3} L e t α = p + q + q^{2} = - 1 - \sqrt{3} + 3 = 2 - \sqrt{3} β = p - q + q^{2} = - 1 + \sqrt{3} + 3 = 2 + \sqrt{3} R e q x^{2} - [α + β] x + α β = 0 \Rightarrow x^{2} - 4 x + 1 = 0$