Let $a$ and $b$ be real numbers. The complex number $4 - 5i$ is a root of the quadratic \[z^2 + (a + 8i) z + (-39 + bi) = 0.\]What is the ot

There seems to be 2 different questions here.

I will answer the first one.

Let  a  and  b  be real numbers. The complex number    4 - 5i     is a root of the quadratic

\(z^2 + (a + 8i) z + (-39 + bi) = 0\).

I am told that one solution is z=4-5i

so

\((4-5i)^2 + (a + 8i) (4-5i) + (-39 + bi) = 0\\ \text{expand and simplify and we get}\\ (4a-8)+(b-5a-8)i=0\\ so\\ 4a-8=0\\ a=2\\ b-5a-8=0\\ b=18\\ \text{so the equation becomes}\\ z^2 + (2 + 8i) z+ (-39 + 18i) = 0\\ \)

Now I will use the quadratic formula to solve this.

\(z^2 + (2 + 8i) z+ (-39 + 18i) = 0\\ z = {-(2 + 8i) \pm \sqrt{\triangle} \over 2}\\ (-1-4i)\pm\frac{\sqrt{\triangle}}{2}=4-5i \qquad \text{I'm not sure if it is + or - so i will keep it together}\\ \pm\frac{\sqrt{\triangle}}{2}=5-i\\ \text{So the solutions are}\\ z=-1-4i\pm (5-i)\\ z=-1-4i+ (5-i) \qquad z=-1-4i- (5-i)\\ z=4-5i \qquad \qquad \quad \qquad z=-6-3i\\~\\ \text{So the other root is } -6-3i\)

I already answered you (A full day before you edited your question) 

You have not responded to my answer in any way.

What is wrong with you?  

Are you just totally rude?

I even read your mind and answered the question that you wanted to ask, rather than the garbled nonsense that you wrote.