Thread: complex numbers

Hello,

I have the complex polynomial:

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?

Substitute . You will see what must equal.

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

Originally Posted by Jones

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

There is no methodology here.
If gives a zero result then is negative root.


Do you see what value of makes that true?

There is no mystery there if you understand anything about this question.

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 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:
Imaginary part: .
(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.

Originally Posted by Plato

There is no methodology here.
If gives a zero result then is negative root.

Typographical error: That " " should be " ".

Do you see what value of makes that true?

There is no mystery there if you understand anything about this question.