Need help with Proof and conjugate roots theorem

[type_speed]

Need Assistance w/ existence of conjugate roots in polynomial

Prove:
given that f(x) = Anx^n + An-1X^n-1+ ...A1X + A0 , and An does not = 0
=> if f(x) has a root of the form (A+Bi), then it must have a root of the form (A-iB). (complex roots)

So far I've come up with the conjugate roots theorem to help me where it states that if a function f(x) is given with real coefficients then the conjugate must exist also. My professor suggested I try the Remainder Theorem??

 

[driscoll79]

What exactly are you stuck on? By inspection you know that if (A+Bi) divides your polynomial then (A-Bi) must also divide your polynomial because your polynomial does not have any complex coefficients. So, I'd recommend doing either a direct or indirect proof - though it seems as if a direct proof will be an easier way to go. Assume that (A+Bi) divides your polynomial. It follows then that:

$$A_{n}x^n + A_{n-1}x^{n-1} + ... + A_{1}x + A_{0} = 0 (mod (A+Bi))$$

The rest should fall out from there...

 

[type_speed]

Appreciate the fast reply, Well first I'm stuck on why my professor told me to use the Remainder Theorem where if a polynomial f(x) is divided by x-c then f(x) is the remainder. Second, I'm not quite sure I can prove that the conjugate exists with only the given statement of the polynomial function not having any complex coefficients. Third, I doubt that our professor would assign such a easy problem for homework. (He only assigned one problem)

What does (mod(A+Bi)) mean?

I am assuming that the roots are (x-(A+Bi))= 0 and (x+(A-Bi))= 0? Right?

Tell me what you think. Regards, Chris

 

[KoGs]

Mod is just a fancy way of saying remainder really. Example 4 mod 2 = 0.

 

[KoGs]

Let's look at a simple polynomial as an example.

Say 0 = x^2 + 2x -3
Your solution would be (x+3)(x-1) = 0.
So your roots are x= -3, and x = 1.

That means something * (x+3) gives you x^2 + 2x -3.
And something * (x-1) gives you x^2 + 2x -3.

In this case we know what the something is, as we put in all the numbers.