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 other root?
I did a lot of work on hand and plugging this into the quadratic formula and with wolfram alpha but it doesn't simplify itself easily at all....anyone have any help to offer? Thanks!
On
Let the other root be $u+iv$ where $u,v$ are real
Using Vieta's formula $\displaystyle u+iv+4-5i=-a-8i\iff -a-8i=u+4-i(5-v)$
Comparing the real & the imaginary parts,
$\displaystyle u+4=-a\iff u=-a-4$ and $5-v=8\iff v=-3$
Again, $\displaystyle(u+iv)(4-5i)=-39+bi\iff -(a+4+3i)(4-5i)=-(39-ib)$
Multiply & compare the real & the imaginary parts
Since $$(4-5i)^2+(a+8i)(4-5i)+(-39+bi)=0$$$$\iff (4a-8)+(-5a+b-8)i=0,$$ we have $$4a-8=-5a+b-8=0\iff a=2,b=18.$$
Also, you'll have $$z^2+(2+8i)z+(-39+18i)=\{z-(4-5i)\}\{z-(-6-3i)\}.$$ So, what you want is $z=-6-3i$.
You can find this transformation by setting $\{z-(4-5i)\}\{z-(c+di)\}.$