Zeroes of a Polynomial (and its Derivative)

**roninpro** · January 23rd 2011, 07:03 PM

I am working on the following problem:

Let $P(z)$ be a holomorphic polynomial of degree at least 1 so that $P(z)\ne 0$ in the upper half plane $\mathbb{R}_+^2 = \{z\in \mathbb{C}: \text{Im }z>0\}$ . Prove that $P'(z)\ne 0$ on $\mathbb{R}_+^2$ .

I wrote

$\displaystyle P'(z)=\sum_{j=1}^n \frac{n_j}{z-z_j}P(z)$

where $z_j$ are the roots of the polynomial and $n_j$ are the multiplicity. Then, taking the real and complex parts, I have

$\displaystyle \text{Re }P'(z)&=n_j\sum_{j=1}^n \frac{x-x_j}{(x-x_j)^2+(y-y_j)^2}\text{Re }P(z)$
$\displaystyle \text{Im }P'(z)&=n_j\sum_{j=1}^n \frac{-y+y_j}{(x-x_j)^2+(y-y_j)^2}\text{Im }P(z)$

Since $y_j<0$ by assumption, if we input $z=x+iy$ where $y>0$ , then all of the terms in sum the complex part will be negative, so $\text{Im }P'(z)\ne 0$ and therefore $P(z)\ne 0$ . But there is a little bit of a problem with this: it is possible that $\text{Im }P(z)=0$ . If this is the case, then we must have $\text{Re }P(z)\ne 0$ . However, I do not see how to guarantee that $\text{Re }P'(z)\ne 0$ . It almost seems possible to construct a counterexample to the problem with this in mind.

Does anybody have any suggestions?

**tonio** · January 23rd 2011, 07:27 PM

Originally Posted by roninpro

I am working on the following problem:

Let $P(z)$ be a holomorphic polynomial of degree at least 1 so that $P(z)\ne 0$ in the upper half plane $\mathbb{R}_+^2 = \{z\in \mathbb{C}: \text{Im }z>0\}$ . Prove that $P'(z)\ne 0$ on $\mathbb{R}_+^2$ .

I wrote

$\displaystyle P'(z)=\sum_{j=1}^n \frac{n_j}{z-z_j}P(z)$

where $z_j$ are the roots of the polynomial and $n_j$ are the multiplicity. Then, taking the real and complex parts, I have

$\displaystyle \text{Re }P'(z)&=n_j\sum_{j=1}^n \frac{x-x_j}{(x-x_j)^2+(y-y_j)^2}\text{Re }P(z)$
$\displaystyle \text{Im }P'(z)&=n_j\sum_{j=1}^n \frac{-y+y_j}{(x-x_j)^2+(y-y_j)^2}\text{Im }P(z)$

Since $y_j<0$ by assumption, if we input $z=x+iy$ where $y>0$ , then all of the terms in sum the complex part will be negative, so $\text{Im }P'(z)\ne 0$ and therefore $P(z)\ne 0$ . But there is a little bit of a problem with this: it is possible that $\text{Im }P(z)=0$ . If this is the case, then we must have $\text{Re }P(z)\ne 0$ . However, I do not see how to guarantee that $\text{Re }P'(z)\ne 0$ . It almost seems possible to construct a counterexample to the problem with this in mind.

Does anybody have any suggestions?

An idea: use the Gauss-Lucas theorem (Gauss) , whose proof is pretty

easy and beautiful.

Tonio

**roninpro** · January 23rd 2011, 07:41 PM

Thanks for the theorem.

I was actually thinking about a geometric interpretation of the sums above, trying to locate $z=x+iy$ in relation to those roots. I had some idea that the expressions were about taking averages somehow, but it wasn't really clear. Now I see the connection.

In any case, I would still like to know if my work above is salvageable. Any thoughts would be appreciated.

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