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?

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- July 25th 2010, 10:01 AMJonescomplex 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? - July 25th 2010, 11:13 AMPlato
Substitute . You will see what must equal.

- July 25th 2010, 03:51 PMJones
I do, care to elaborate on the science behind your methodology?

- July 25th 2010, 05:29 PMPlato
- July 26th 2010, 02:27 AMHallsofIvy
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. - July 26th 2010, 02:28 AMHallsofIvy