polynomial with complex roots....

**Showcase_22** · September 30th 2008, 10:26 AM

Right. This is really getting on my nerves.

The question is:

Let $f=a_{n}X^n+a_{n-1}X^{n-1}+.........+a_0$ be a polynomial where the coefficientsn $a_0,.......,a_n$ are real and $a_n \neq 0.$

i). Show that if $\alpha$ is a complex root of f then it's conjugate $\overline {\alpha}$ is also a root of f.

Would this just be that the product of a complex number and it's conjugate produce a real number. Since this polynomial has real coefficients then there can't be a complex number on it's own (or some of the coefficients would be complex)?

ii) Suppose that the degree n is odd. Show that f must have at least one real root.

Would the solution to this be that a polynomial with a degree n has n roots. If the polynomial is odd then there would be $\frac {n-1}{2}$ pairs of complex numbers and their conjugates and one root left over. Since this root is on it's own, it must be real to avoid complex coefficients.

I do not know how to prove this though

.

Please help!

P.S: Sorry for two threads in one night, there's no gentle introduction on a maths course!

**Plato** · September 30th 2008, 01:59 PM

Do you know that ${(\overline{\alpha}})^n={\overline{\alpha^n}}$ ?
Do you know that $\overline{z+w}=\overline{z}+\overline{w}$ ?

You can use both of those in the proof.

**Showcase_22** · September 30th 2008, 11:21 PM

I know both those facts, so thanks for putting me on the right lines!

So would I prove part i) by:

$f(\alpha)=a_{n}(\alpha)^{n}+a_{n-1}(\alpha)^{n-1}+.......+a_{n-(n-1)}(\alpha)^{n-(n-1)}+a_{n-n}(\alpha^{n-n}$

$f(\alpha)=a_{n}(\alpha)^{n}+a_{n-1}(\alpha)^{n-1}+.......+a_{1}(\alpha)^{1}+a_{0}$

$f(\alpha)=a_{n}\alpha^{n}+a_{n-1}\alpha^{n-1}+.......+a_{1}\alpha^{1}+a_{0}$

I get that far (I managed to use the first thing you wrote but just for $\alpha$ ). I could do the same for $\overline {\alpha}$ but it's hard to try and prove that $f(\alpha)$ and $f(\overline {\alpha})$ are equal to 0.

This is one of my ideas to prove it is equal to 0:

since we know $f(\alpha)=a_{n}\alpha^{n}+a_{n-1}\alpha^{n-1}+.......+a_{1}\alpha^{1}+a_{0}$ and $f(\overline{\alpha})=a_{n}\overline{\alpha}^{n}+a_ {n-1}\overline{\alpha}^{n-1}+.......+a_{1}\overline{\alpha}^{1}+a_{0}$

we could find $f(\overline{\alpha})+f(\alpha)$ which we might be able to cancel and get somewhere. I could use the second thing you wrote if I did it this way. If the sum of them both does come out to 0 since $\alpha=a+bi$ and $\overline {\alpha}=a-bi$ (ie. the sign of the real part is the same) then $f(\alpha)\neq f(\overline{\alpha}$ . The only possible values of $f(\alpha)$ and $f(\overline{\alpha})$ are therefore 0.

Since the function is the same they are both roots of the function.

However, it now becomes a challenge to prove that $f(\alpha)+f(\overline{\alpha})=0$ . The complex parts would cancel out giving a sequence like:

$f(\alpha)+f(\overline{\alpha})=a_{n}(2a)^n+a_{n-1}(2a)^(n-1)......$

(i would put in more but I have to dash off to my test in a minute).

Since I would end up with this, $f(\overline{\alpha})+f(\alpha)$ could equal 0. Unfortunately, I get this far and i'm not really sure what to do!

Is this the right method? It makes sense to me so far but i'm not gifted with on overabundance of logic!

P.S: I'm sorry for such a long post, ideas kept coming to me as I kept writing!

**Showcase_22** · October 1st 2008, 02:59 AM

Sorry to double post but I don't want to confuse anyone reading this (this is a new idea).

From the question we know that $f(\alpha)=0+0i$ .

If $\alpha=a+bi$ then $f(a+bi)=0+0i$ .

$\overline {\alpha}=a-bi=a+(-b)i$ . Therefore $f(a+(-b)i)=0+0i$ as well.

I think that's a good way to do part one. What does everyone else think?

**bobak** · October 1st 2008, 03:27 AM

Originally Posted by Showcase_22

Sorry to double post but I don't want to confuse anyone reading this (this is a new idea).

From the question we know that $f(\alpha)=0+0i$ .

If $\alpha=a+bi$ then $f(a+bi)=0+0i$ .

$\overline {\alpha}=a-bi=a+(-b)i$ . Therefore $f(a+(-b)i)=0+0i$ as well.

I think that's a good way to do part one. What does everyone else think?

Well you have shown no working, but assuming your applying Plato hints accurately that is correct.

Bobak

**Showcase_22** · October 1st 2008, 03:39 AM

oh right. Well I was just going to write that!

So I need to prove that $f(a-bi)=0$ .

$f(a+bi)=f(\alpha)=a_{n}\alpha^n+a_{n-1}\alpha^{n-1}+.......+a_0=0$

$f(a-bi)=f(\overline{\alpha})=a_{n}\overline{\alpha}^n+ a_{n-1}\overline{\alpha}^{n-1}+.......+a_0=0$

So how do I prove the second one (I know i'm beating a bit around the mulberry bush but i'm really starting to get confused).

**bobak** · October 1st 2008, 03:42 AM

Originally Posted by Showcase_22

oh right. Well I was just going to write that!

So I need to prove that $f(a-bi)=0$ .

$f(a+bi)=f(\alpha)=a_{n}\alpha^n+a_{n-1}\alpha^{n-1}+.......+a_0=0$

$f(a-bi)=f(\overline{\alpha})=a_{n}\overline{\alpha}^n+ a_{n-1}\overline{\alpha}^{n-1}+.......+a_0=0$

So how do I prove the second one (I know i'm beating a bit around the mulberry bush but i'm really starting to get confused).

you need to prove that $f(\overline{\alpha}) = \overline{f(\alpha)}$ . This is quiet simple. Look at what plato wrote again.

Bobak

**Plato** · October 1st 2008, 03:45 AM

$f(z) = \sum\limits_{k = 0}^n {a_k z^k } \Rightarrow \,f(\alpha ) = \sum\limits_{k = 0}^n {a_k \alpha ^k } = 0$

$0 = \overline {\sum\limits_{k = 0}^n {a_k \alpha ^k } } = \sum\limits_{k = 0}^n {a_k \overline {\alpha ^k } } = \sum\limits_{k = 0}^n {a_k \overline \alpha ^k } = f\left( {\overline \alpha } \right)$

**Showcase_22** · October 1st 2008, 03:52 AM

Hang on a second, so $\alpha^k=\overline{\alpha^k}=\overline{\alpha}^k$ ??

(I'm trying to figure out what plato did :s)

what about when k=1? Then you would have $\alpha=\overline{\alpha}$ and they're different?

I know all of this is probably just a stupid question but it just seems really strange.

**bobak** · October 1st 2008, 04:01 AM

Originally Posted by Showcase_22

Hang on a second, so $\alpha^k=\overline{\alpha^k}=\overline{\alpha}^k$ ??

(I'm trying to figure out what plato did :s)

what about when k=1? Then you would have $\alpha=\overline{\alpha}$ and they're different?

I know all of this is probably just a stupid question but it just seems really strange.

nope Plato never implied that individual terms were equal. as the sum is equal to zero so is the conjugate of the sum. for example $2-2+3i - 3i = 0$ and so does $2-2-3i +3i$ .

Bobak

**Showcase_22** · October 1st 2008, 04:15 AM

as the sum is equal to zero so is the conjugate of the sum.

Do you get this bit from the proof Plato posted?

I have to admit i'm still a little lost with this :s

**bobak** · October 1st 2008, 05:23 AM

We know that $f( \alpha ) = 0$ so therefore $\overline{f( \alpha )} = 0$

however $\overline{f( \alpha )} = \overline{a_n \alpha^n + a_{n-1} \alpha^{n-1} + \cdots a_{1} \alpha + a_0}$

Now you just apply the hint Plato gave you in turn.

$\overline{f( \alpha )} = \overline{a_n \alpha^n} + \overline{a_{n-1} \alpha^{n-1}} + \cdots \overline{a_{1} \alpha} + \overline{a_0}$

$\overline{f( \alpha )} = a_n \overline{\alpha}^n + a_{n-1} \overline{\alpha}^{n-1} + \cdots a_{1}\overline{ \alpha} + \overline{a_0} \ \ = f( \overline{\alpha} )$

Does that makes sense ?

Bobak

**Showcase_22** · October 1st 2008, 09:02 AM

OMG! I actually understand that!

Thanks Bobak, you are a legend!!

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