Let p , q ∈ ℝ and ( 1 - 3 i ) 200 = 2 199 ( p + i q ) , i = - 1 . Then, p + q + q 2 and p - q + q 2 are roots of the equation.

Step-by-step solution

Given: ( 1 - 3 i ) 200 = 2 199 ( p + i q ) ⇒ 1 2 - 3 2 i 200 = 1 2 p + i q ⇒ cos π 3 - i sin π 3 200 = 1 2 p + i q ⇒ cos 200 π 3 - i sin 200 π 3 = 1 2 p + i q ⇒ cos 201 π - π 3 - i sin 201 π - π 3 = 1 2 p + i q ⇒ - cos π 3 - i sin π 3 = 1 2 p + i q ⇒ 2 - 1 2 - i 3 2 = p + i q ⇒ - 1 - i 3 = p + i q So, p = - 1 ,   q = - 3 α = p + q + q 2 = 2 - 3 β = p - q + q 2 = 2 + 3 So, α + β = 4 α · &#94