Complex numbers question

Question

Suppose a complex number z satisfies the equation
2z^2 - (3+8i)z - (m+4i) = 0
Given that m is a real constant and that one of the two solutions is real, find the two solutions of the above complex quadratic equation. (Hint: The real solution is easy to find. To find the other solution, use the fact that the two solutions z1, z2 of the quadratic equation az^2 + bz+c = 0 satisfy z1+z2 = -b/a.)

Thank you

Answer 1

2z²-(3+8i)z-(m+4i)=0, and if one of the roots z₁=x where x is real, then

2x²-3x-8ix-m-4i=0. Therefore -8x-4=0 because there is no imaginary component remaining, making x=-½ and z₁=-½. The quadratic in x becomes 2/4+3/2-m=0, making m=2

Rewrite the equation: z²-½(3+8i)-(1+2i)=0.

The sum of the roots z₁+z₂=½(3+8i) so z₂=½(3+8i)+½=2+4i.

The product of the roots is -1-2i=-½(2+4i)=z₁z₂, confirming the solution.

The roots are -½ and 2+4i.