Let p, q ∈ ℝ and ( 1 - 3 i ) ^ 200 = 2 ^ 199 ( p + iq ) , i = - 1 ....

Given:

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

$⇒ {\left(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\right)}^{200} = \dfrac{1}{2}\left(p + iq\right)$

$⇒ {\left(\cos \left(\dfrac{\pi}{3}\right) - isin\left(\dfrac{\pi}{3}\right)\right)}^{200} = \dfrac{1}{2}\left(p + iq\right)$

$⇒ \left(\cos \left(\dfrac{200\pi}{3}\right) - isin\left(\dfrac{200\pi}{3}\right)\right) = \dfrac{1}{2}\left(p + iq\right)$

$⇒ \left(\cos \left(201\pi - \dfrac{\pi}{3}\right) - isin\left(201\pi - \dfrac{\pi}{3}\right)\right) = \dfrac{1}{2}\left(p + iq\right)$

$⇒ \left(- \cos \left(\dfrac{\pi}{3}\right) - isin\left(\dfrac{\pi}{3}\right)\right) = \dfrac{1}{2}\left(p + iq\right)$

$⇒ 2\left(- \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right) = p + iq$

$⇒ - 1 - i\sqrt{3} = p + iq$

So,

$p = - 1, q = - \sqrt{3}$

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

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

So,

$\alpha + \beta = 4$

$\alpha \cdot \beta = 1$

Required equation is

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