Hi, I'm stuck on this one. I don't know why though. The answer in the book is 8.

if $\displaystyle P(z) = z^3 + 2z^2 - 6z + a$ and $\displaystyle P(1 - i) = 0$, then a is equal to?

A) 4 B) -6 C) 8 D) 6 E) 5

So the first thing I saw was that there is a conjugate zero also so it looks like this

$\displaystyle P(z) = (z - 1 - i) (z - 1 + i) (z - z_{3})$

So then I thought i would expand out just the z's, not the z^2s etc. and let them equal the z term in the original expression, -6z

$\displaystyle z(-1 + i) + z(-z_{3}) + z(-1 - i) + z(-z_{3}) + z(-1 - i) + z(-1 + i) = -6z$

the z cancels out

$\displaystyle -1 + i - z_{3} -1 - i -z_{3} - 1 - i - 1 + i = -6$

$\displaystyle -4 - 2z_{3} = -6$

$\displaystyle -2z_{3} = -2$

$\displaystyle z_{3} = 1$

So then I now have (z - 1 - i)(z - 1 + i)(z - 1) and I will just expand our the constants to get the a

But this doesn't seem to work. I won't bother writing it down because I have done it about 5 times and I can't seem to get the answer. Does anyone know a method to get this one out?

David.