# Math Help - complex numbers

1. ## complex numbers

Hello,

I have the complex polynomial: $z^3+(3+i)z^2+(4-2i)z+(a-3i) = 0$

I need to determine the parameter a, so that the equation gets a real negative root.
I don't know any root of the polynomial, if i did it would have been easy.

Polynomial division?
Any ideas?

2. Substitute $z=-1$. You will see what $a$ must equal.

3. I do, care to elaborate on the science behind your methodology?

4. Originally Posted by Jones
I do, care to elaborate on the science behind your methodology?
There is no methodology here.
If $z=-1$ gives a zero result then $-1$ is negative root.
$\left( { - 1} \right)^3 + \left( {3 + i} \right)\left( { - 1} \right)^2 + \left( {4 - 2i}\right)\left( i \right) + \left( {a - 3i} \right) = 0$

Do you see what value of $a$ makes that true?

5. What Plato is saying is that this equation will have many different roots depending on what a is. And that you can make one root to be anything you want by choosing a correctly. In particular, to be sure that the equation has at least one real, negative, root, you can just choose z to be any real, negative, number you want.

for example, if you chose to make that root -2 instead of -1, you would have the equation $(2)^3+ (3+ i)(-2)^2+ (4- 2i)(-2)+ (a- 3i)= 0$ and solve that for a.

However, I see a difficulty with this. It is possible that the form (a- 3i) is supposed to indicate that a is a real number. If so, that should have been said, but if so then that makes the problem harder!

Assuming that a must be real, we can separate the equation into real and imaginary parts:
Real part: $z^3+ 3z^2+ 4z+ a= 0$
Imaginary part: $z^2- 2z- 3= 0$.
(I have used the fact that z is a real number in separating.)

Now you can solve that second, quadratic, equation for z and see what values of a allow a negative z that satifies both equations.

6. Originally Posted by Plato
There is no methodology here.
If $z=-1$ gives a zero result then $-1$ is negative root.
$\left( { - 1} \right)^3 + \left( {3 + i} \right)\left( { - 1} \right)^2 + \left( {4 - 2i}\right)\left( i \right) + \left( {a - 3i} \right) = 0$
Typographical error: That " $(4- 2i)(i)$" should be " $(4- 2i)(-1)$".

Do you see what value of $a$ makes that true?