# complex numbers polynomial

• February 18th 2011, 06:09 PM
kumquat
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)
• February 18th 2011, 07:02 PM
Ackbeet
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?
• February 18th 2011, 07:02 PM
Prove It
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.
• February 18th 2011, 07:14 PM
kumquat
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".
• February 18th 2011, 07:15 PM
Ackbeet
It is a coefficient: of the $z^{0}$ term!
• February 18th 2011, 07:57 PM
kumquat
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
• February 18th 2011, 08:10 PM
Prove It
But you know it's a factor, so the remainder must be 0...
• February 18th 2011, 08:16 PM
kumquat
so the end part (b+8+(2a+b-4)i)/(z-1-i), MUST equal zero?

-----> b+8+(2a+b-4)i = 0
• February 18th 2011, 08:21 PM
kumquat
ignore previous post
• February 18th 2011, 08:22 PM
Prove It
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 $\displaystyle a,b$.
• February 18th 2011, 08:24 PM
kumquat
ok ive solved it and know what i did right/wrong, i just abandoned a couple of easy solving techniques along the way