1. ## complex numbers.pt2

Problem:

Is it possible to determine the real constant a so that
$\displaystyle z=2+i$
is a root for the equation
$\displaystyle z^3+az^2+13z-10=0$

I started out like so that if p(b)=0, where p(z) is a polynomial of a degree n,
then p(z)=(z-b)*f(z) where f(z) is a polynomial of a degree n-1

meaning that

$\displaystyle z^3+az^3+13z-10=(z-(2+1))(z^2+dz+e)$

but apparently that was wrong..

Sorry for all the questions, but I have an exam coming up this monday

2. Originally Posted by Greenb
Problem:

Is it possible to determine the real constant a so that
$\displaystyle z=2+1$
is a root for the equation
$\displaystyle z^3+az^3+13z-10=0$

I started out like so that if p(b)=0, where p(z) is a polynomial of a degree n,
then p(z)=(z-b)*f(z) where f(z) is a polynomial of a degree n-1

meaning that

$\displaystyle z^3+az^3+13z-10=(z-(2+1))(z^2+dz+e)$

but apparently that was wrong..

Sorry for all the questions, but I have an exam coming up this monday
Do you mean $\displaystyle z = 2 + {\color{red}i}$ is the root?

Just substitute into the cubic, simplify and equate the real and imaginary parts of the result to zero. Then solve for $\displaystyle a$.

3. This is what I got:

$\displaystyle (z-(2+i))(z^2+dz+e)=$
$\displaystyle =z^3+dz^2+ez-2z^2-2dz-2e-iz^2-idz-ie$

somehow I think I need the values for d and e to proceed..

How do I "equate the real and imaginary parts of the result to zero" seeing that I should leave the left side intact?

It must contain some mistakes.
You want $\displaystyle 2+\color{red}i$ to be a root of $\displaystyle z^3+az^{{\color{red}2}}+13z-10$.
If I have guessed correctly, there is a simple answer for $\displaystyle a$.

5. Ah didn't see that I had typed it wrong Anyway, yes a=-6 but I really can't see how.

6. Originally Posted by Greenb
a=-6 but I really can't see how.
Indeed, that is what I got.
$\displaystyle \left( {2 + i} \right)^3 + a\left( {2 + i} \right)^2 + 13\left( {2 + i} \right) - 10 = \left( {3a + 18} \right) + i\left( {4a + 24} \right)$

7. I thought way more complicated and was actually totally off the track hehe. Thanks for clearing it up for me, I now see that mr fantastic also meant this solution, I was trying to fit in his advice into my solution ^^