Complex Numbers Question?

Question

aα^2 + bα + c = 0, where a,b,c € R and a € C. Show that a(CONJUGATE)α^2 + b(CONJUGATE)α + c = 0

Answer 1

Suppose alpha = p+iq the its conjugate = p-iq.

Now a(p+iq)^2 + b(p+iq) +c =0=0+i0.

a(p^2 +2ipq - q^2) + bp+ibq +c =0,

so ap^2-aq^2+bp+c=0 and 2apq+bq =0 comparing real and inagnary parts.Hence

ap^2-aq^2+bp+c=0 - i(2apq+bq) =0,

So ap^2 - 2iapq - aq^2 +b(p-iq) +c=0,

a(p-iq)^2 +b(p-iq)+c=0. which was to be proved.

Answer 2

No calculations needed: just notice that α is the root of a quadratic equation with real coefficients, so it's conjugate must be a root of the same equation.