# Thread: Zeroes of a Polynomial (and its Derivative)

1. ## Zeroes of a Polynomial (and its Derivative)

I am working on the following problem:

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

I wrote

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

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

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

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

Does anybody have any suggestions?

2. Originally Posted by roninpro
I am working on the following problem:

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

I wrote

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

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

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

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

3. Thanks for the theorem.

I was actually thinking about a geometric interpretation of the sums above, trying to locate $\displaystyle z=x+iy$ 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.