Question

Suppose  α + iβ is a solution of the polynomial equation a₄z⁴ + i a₃z³ + a₂z² + i a₁z + a₀ = 0 where α, β, a_i ∈ R ∀ i ∈ {0, 1, 2, 3, 4}. Which one of the following must also be a solution?

(A)  -α - βi

(B)  α - βi

(C)  -α + βi

(D)  β  +  αi

Answer 1

Correct option (C)  -α + βi

Let z = a be a real root. Then

αa² + a + a = 0 ...(1)

Let α = p + iq. Then 

(p + iq)a² + a + p − iq = 0

⇒ pa² + a + p = 0 and a2q − q = 0 

 Therefore,

a = ± 1 (since q ≠ 0)

From Eq. (1),

α ± 1 + (bar)α = 0, also |a| = 1