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

Given:

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

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

$\Rightarrow {\left(\cos \left(\frac{\mathrm{\pi }}{3}\right)-i\sin \left(\frac{\mathrm{\pi }}{3}\right)\right)}^{200}=\frac{1}{2}\left(p+iq\right)$

$\Rightarrow \left(\cos \left(\frac{200\mathrm{\pi }}{3}\right)-i\sin \left(\frac{200\mathrm{\pi }}{3}\right)\right)=\frac{1}{2}\left(p+iq\right)$

$\Rightarrow \left(\cos \left(201\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)-i\sin \left(201\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right)=\frac{1}{2}\left(p+iq\right)$

$\Rightarrow \left(-\cos \left(\frac{\mathrm{\pi }}{3}\right)-i\sin \left(\frac{\mathrm{\pi }}{3}\right)\right)=\frac{1}{2}\left(p+iq\right)$

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

$\Rightarrow -1-i\sqrt{3}=p+iq$

So,

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

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

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

So,

$\alpha +\beta =4$

$\alpha ·\beta =1$

Required equation is

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