1. ## complex numbers polynomial

the question is "If z = 1+i is a zero of the polynomial z^3 + az^2 + bz + 10 - 6i, find the constants 'a' and 'b' given that they are real."

now becuase 1+i is a solution i am going to assume 1-i is a solution becuase the polynomial has real co-efficients. so (z - 1 - i)(z - 1 + i) = z^2 - 2z + 2 as a quadratic factor

im suspecting there could be a problem with the question becuase on my CAS calculator 1+i is certainly a solution but 1-i is NOT (substituting the answers from the back of the book a=6, b=-8)

2. If the coefficients are ALL real, then complex roots come in conjugate pairs. But, if you introduce even one complex coefficient, all bets are off. You do know that (z-1-i) is a factor of the polynomial. Why not try either polynomial long division or synthetic division?

3. If the coefficients are real, then complex roots always occur as complex conjugates. You are correct there. But the coefficients are not all real (you have -6i at the end).

I would start by long dividing your cubic by the linear factor z - 1 - i, to get the quadratic factor.

4. ok i'll try dividing by the given factor i didnt realise that there was a complex co-efficient i thought that only applied to the terms with a "z".

5. It is a coefficient: of the $z^{0}$ term!

6. ok i divided the cubic and the remaining quadratic is z^2 + (1 + i + a)z + a + b + (a + 2)i + (b + 8 + (2a + b - 4)i)/z - 1 - i

ive verified this is correct on the CAS calculator but i dont know how to get the answer, the "quadratic" has 4 terms

7. But you know it's a factor, so the remainder must be 0...

8. so the end part (b+8+(2a+b-4)i)/(z-1-i), MUST equal zero?

-----> b+8+(2a+b-4)i = 0

9. ignore previous post

10. Correct, but what to do becomes more obvious if you write it as $\displaystyle b + 8 + (2a + b - 4)i = 0 + 0i$, because now you equate real and imaginary parts to get two equations to solve simultaneously for $\displaystyle a,b$.

11. ok ive solved it and know what i did right/wrong, i just abandoned a couple of easy solving techniques along the way