Given:
(1−3i)200=2199(p+iq)
⇒(21−23i)200=21(p+iq)
⇒(cos(3π)−isin(3π))200=21(p+iq)
⇒(cos(3200π)−isin(3200π))=21(p+iq)
⇒(cos(201π−3π)−isin(201π−3π))=21(p+iq)
⇒(−cos(3π)−isin(3π))=21(p+iq)
⇒2(−21−i23)=p+iq
⇒−1−i3=p+iq
So,
p=−1,q=−3
α=p+q+q2=2−3
β=p−q+q2=2+3
So,
α+β=4
α⋅β=1
Required equation is
x2−4x+1=0