Finding Roots of Complex Polynomials

[jf623]

Started my first year of Electronic Engineering a few months back and I'm already struggling with the mathematics. I have been to this forum a number of times over the last year and finally decided to join just 10 minutes ago!

Homework Statement

Find the roots of P(x) given that two roots are x=1, x=j

Homework Equations

P(x) = x^5 -9x^4 +24x^3 -24x^2 +23x -15
(where j^2 = -1)

The Attempt at a Solution

I tried polynomial division initially, but I was unsure of how to handle the imaginary terms, so then I tried using the method of undetermined coefficients and ended up with P(x) = (x-1)(x-j)(x^3 -8x^2 +jx^2 +15x -j8x +j15). Now I am unsure of how to factor or find the roots of this cumbersome third degree complex number.

 

[Zryn]

One of the roots is x = j, and P(x) has no imaginary components (j's) in it. What does this tell you?

 

[Apphysicist]

The nice thing is that given a real polynomial, you can determine how many positive, real and negative, real roots it has using Descarte's Rule of Signs. And you know it has to have five roots since it is of the fifth degree.

The coefficients of your polynomial are: +1,-9,+24,-24,+23,-15. Since they flip "sign" 5 times, you know there are 5, 3, or 1 positive real roots...[you're given at least one positive real root]

If you replace each x with -1, that means each odd-powered x will change the sign of the coefficient, thus: -1,-9,-24,-24,-23,-15. Since this flips "sign" 0 times, you know there are 0 negative real roots.

If you haven't read up on complex numbers, their conjugates, and their place in real polynomials, I would suggest you do so, as that's what Zryn is getting at. [Hint: there's a reason why the possible numbers of real roots (5,3,1) are separated by two each time.]

After that, it should be fairly simple to divide by the factors you've found to reduce the polynomial to a solvable quadratic.

 

[jf623]

Now I feel even more lost than I was before. I looked up Descarte's Rule of Signs as I have never encountered it before, and I am unsure as to how knowing the signs of the roots will help me determine what they are. Polynomials of this sort are fairly new to me, complex numbers are not. I know there are 5 roots and at least 1 is imaginary. The only thing that springs to mind is finding the other real roots by inspection, is this what you are nodding at?

 

[Zryn]

Looking closely at the polynomial you will notice that there is no j components, but we definitely know that 1 of the roots is x = j, so why isn't there a liberal spreading of j's throughout the equation?

For example, given (x - y)(x - 1) = x^2 - x - yx + 1, and that the roots of the equation are clearly x = y and x = 1, when we look at our P(x) and know that x = j is definitely one of the roots, why do we not see any j's in the P(x) equation, the same as we see y's in this example equation?

 

[eumyang]

@jf623: Don't worry about not having encountered Descartes' Rule of Signs. I never encountered it either when I was a student.

When you say "polynomial division," do you mean polynomial long division or synthetic division? I hope you meant the latter, as it is less cumbersome.

What the the two posters here were trying to tell you is that the problem actually didn't give you two roots at the start, but three. To find the third given root, look into the Complex conjugate root theorem.

BTW, if you are using synthetic division, in the factored polynomial that you provided, I would leave the like terms grouped. Instead of
$$P(x) = (x-1)(x-j)(x^3 -8x^2 +jx^2 +15x -j8x +j15)$$
I would write
$$P(x) = (x-1)(x-j)[x^3 -(8-j)x^2 + (15 - j8)x +j15]$$

 

[jf623]

Ok now I've slept I think I understand a little more. By elimination, there must be 3 positive real roots by using Descartes Rule, and therefore 2 complex roots? That there are no imaginary terms in the polynomial means that the complex conjugate of the one root I'm given must also be another root! So now I have P(x)=(x-1)(x-j)(x+j)(Ax^2 +Bx +C), I will try undetermined coeffs again because I am familiar with it. I have never heard of synthetic division either.

 

[Dick]

Ok now I've slept I think I understand a little more. By elimination, there must be 3 positive real roots by using Descartes Rule, and therefore 2 complex roots? That there are no imaginary terms in the polynomial means that the complex conjugate of the one root I'm given must also be another root! So now I have P(x)=(x-1)(x-j)(x+j)(Ax^2 +Bx +C), I will try undetermined coeffs again because I am familiar with it. I have never heard of synthetic division either.

Multiply the (x-1)(x-j)(x+j) out and you'll get a real polynomial. Now divide P(x) by it. You don't have to do it 'synthetically'. Any way will do.

 

[jf623]

Yes I did that too Dick, it all seems to work and I can confirm the factors are (x-1)(x-3)(x-5)(x-j)(x+j). Thank you all for your kind help.

 

[Outlined]

If all your coefficients are real then for any root holds the complex conjugate is also a root. In your case $$j$$ is a root so $$\bar{j}$$ is a root too.

Next step: Devide $$x^{5} -9x^{4} +24x^{3} -24x^{2} +23x -15$$ by
$$(x - 1)(x - j)(x + j) = (x - 1)(x^{2} + 1) = x^{3} - x^{2} + x - 1$$.