If ( 3 + i ) ^ 100 = 2 ^ 99 ( p + iq ) , then p and q are roots of...

$( \sqrt{3} + i{)}^{100} = {2}^{99} ( p + iq )$

${2}^{100}{\left(cos\dfrac{\pi}{6} + isin\dfrac{\pi}{6}\right)}^{100} = {2}^{99}\left(p + iq\right)$

${2}^{100}{e}^{i\dfrac{50\pi}{3}} = {2}^{99}\left(p + iq\right)$

$2\left(cos\dfrac{50\pi}{3} + isin\dfrac{50\pi}{3}\right) = p + iq$

$2\left(\cos \left(17\pi - \dfrac{\pi}{3}\right) + isin\left(17\pi - \dfrac{\pi}{3}\right)\right) = p + iq$

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

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

$p = - 1 \text{ and } q = \sqrt{3}$

$p$ and $q$ Roots of the equation
${x}^{2} - ( \sqrt{3} - 1 ) x - \sqrt{3} = 0$