# Writing a polynomial as a product of it's roots

• Apr 22nd 2013, 03:41 AM
Ant
Writing a polynomial as a product of it's roots
Let $\displaystyle P(z)$ be a polynomial. We're trying to show that we can write:

$\displaystyle P(z)=c(z-z_1)^{m_1}(z-z_2)^{m_2}...(z-z_{k})^{m_k}$.

If $\displaystyle P(z)$ has a root $\displaystyle z_1$ (and it does by the FTA) then we can write:

$\displaystyle P(z)=(z-z_1)^{m_1}P_1(z)$ for some entire function $\displaystyle P_1(z)$.

Next, in my notes it says "Clearly, $\displaystyle |P_1(z) \leq C(|z|+1)^n$

I really don't see how that is "clear", could anyone help?

Thanks.
• Apr 22nd 2013, 06:00 AM
majamin
Re: Writing a polynomial as a product of it's roots
Quote:

Originally Posted by Ant
Let $\displaystyle P(z)$ be a polynomial. We're trying to show that we can write:

$\displaystyle P(z)=c(z-z_1)^{m_1}(z-z_2)^{m_2}...(z-z_{m_k})^{m_k}$.

If $\displaystyle P(z)$ has a root $\displaystyle z_1$ (and it does by the FTA) then we can write:

$\displaystyle P(z)=(z-z_1)^{m_1}P_1(z)$ for some entire function $\displaystyle P_1(z)$.

Next, in my notes it says "Clearly, $\displaystyle |P_1(z) \leq C(|z|+1)^n$

I really don't see how that is "clear", could anyone help?

Thanks.

Did you mean to have subscript of k instead of m_k in the last factor? As in:

$\displaystyle P(z)=c(z-z_1)^{m_1}(z-z_2)^{m_2}...(z-z_{k})^{m_k}$

Are there further details for the z_i's? For instance, $\displaystyle z_i \leq 1$?
• Apr 22nd 2013, 06:09 AM
Ant
Re: Writing a polynomial as a product of it's roots
yeah I did, I'll edit that now. Thanks.

There are no further restrictions on the z_i's, they are the roots distinct roots of P(Z) in $\displaystyle \mathbb{C}$
• Apr 22nd 2013, 10:33 AM
Ant
Re: Writing a polynomial as a product of it's roots
Okay, I think I might have it.

I think $\displaystyle n$ is the degree of our polynomial $\displaystyle P(z)$ and so $\displaystyle P_1(z)$ is a polynomial of degree $\displaystyle n-m_1$.

Then, I think the inequality follows? Could anyone confirm this?
• Apr 22nd 2013, 09:14 PM
majamin
Re: Writing a polynomial as a product of it's roots
Quote:

Originally Posted by Ant
Okay, I think I might have it.

I think $\displaystyle n$ is the degree of our polynomial $\displaystyle P(z)$ and so $\displaystyle P_1(z)$ is a polynomial of degree $\displaystyle n-m_1$.

Then, I think the inequality follows? Could anyone confirm this?

Since $\displaystyle P(z)=c(z-z_1)..(z-z_k)$, wouldn't it make sense that deg(P(z))=k? Not sure how "n" fits into this. And I still think that would be too "handwavy" to say it follows thereafter. Regardless,

$\displaystyle |P(z)|=|c(z-z_1) ... (z-z_k)| \leq |c(z-z_1)||(z-z_2)...(z-z_k)| \leq C |(z-z_2)|...|(z-z_k)|$

The reason why I asked about the z_i earlier was because it would follow nicely that if for each z_i, $\displaystyle |z_i|\leq 1$ then

$\displaystyle C |(z-z_2)|...|(z-z_k)| \leq C (|z|+|z_2|)...(|z|+|z_k|) \leq C (|z| +1) ... (|z| +1) = C(|z|+1)^{k-1}$

Otherwise, if this restriction on z_i isn't as above, then more will have to be done to bound the polynomial. Thoughts?
• Apr 22nd 2013, 09:39 PM
majamin
Re: Writing a polynomial as a product of it's roots
Another attempt here, bypassing the need for $\displaystyle |z_i|\leq 1$:

let $\displaystyle z^* = \max\{|z_1|,|z_2|,...\}$ then $\displaystyle P_1(z) \leq C |z+z^*|...|z+z^*| = C|z-z^*|^{k-1} \leq C (|z|+|z^*|)^{k-1} \leq C (|z|+1)^n$

where $\displaystyle n = (k-1) \log{(|z|+z^*)}/\log{(|z|+1)}$