Solving polynominals with complex numbers

Question

I have been given this question. Where one root is given $z=1+i$ I need to solve for $a$ and $b$ for the polynominal

$$z^2-aiz+b=0$$

Give that both root are real

P.s i just dont know how to solve this as the Complex conjugate root therom doesent work in this context (there is an $i$ in the coefficient)... is there any other method i should use?

Answer 1

Just multiply out:

$ (x - z) (x - 1 - i) = x^2 - (z + 1 + i) x + z (1 + i) \\ $

You see that $z + 1 + i = a i$ and $z (1 + i) = b$. This is two (complex) linear equations, you get a relation between $z$, $a$ and $b$.

Answer 2

$$(1+i)^2-2ai(1+i)+b=0\iff a+b=0,-a+2=0$$ or $$a=2,b=-2.$$

The other root is $-1+i$.

Answer 3

Sub in $z=1+i$ $\to(1+i)^2-ai(1+i)+b=0\to2i-ai+a+b$ $\iff2-a=0$ and $a+b=0$