The roots of the quadratic equation

Question

z^2+(4+i+qi)z+20=0 are w and w*
I) in the case where q is real, explain why q must be -1

Answer 1

clearly the roots of the given equation must be complex numbers, thus they must be conjugates of each other
let w = a+bi
then w* is a - bi

sum of roots = a+bi + a-bi = 2a
product or roots = (a+bi)(a-bi) = a^2 + b^2

but from the given equation:
sum of roots = -4-i-qi
product of roots = 20

a^2 + b^2 = 20, and
2a = -4-i-qi = -4 - (q+1)i
since a is the real part of w, 2a must be real
so in -4 - (q+1)i, the imaginary part must drop out, thus
q+1 = 0
q = -1

let's work backwards, suppose q = -1
then our equation simply becomes
z^2 + 4z+ 20 = 0
z^2 + 4z + 4 = -16
(z+2)^2 = -16
z+2 = ±4i
z = -2 ± 4i

so the sum of the roots = -2+4i + (-2 - 4i) = -4
product of roots = (-2+4i)(-2-4i)
= 4 - 16i^2 = 20

all is good!

Answer 2

Thank you so much