Complex Number Roots of quadratics with missing terms

Question

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

How would I go about finding $a$ and $b$ or finding the other root? I'm totally lost on this problem. I'm currently a student in pre-calculus. Thanks in advance!

Answer 1

Substitute $z=4-5i$ into your equation and separate real and imaginary parts. Equate both parts to zero to get the value of $a$ and $b$.

\begin{align*} (4-5i)^2+(a+8i)(4-5i)-39+bi=0\\ \implies (-9+4a+40-39)+i(-40-5a+32+b)=0\end{align*}

Answer 2

$$4-5i+z_2=-a-8i,$$ which gives $$z_2=-a-4-3i.$$ Thus, $$(-a-4-3i)(4-5i)=-39+bi,$$ which gives $a=2$ and $b=8$.