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Math Help - Zeroes of a Polynomial (and its Derivative)

  1. #1
    Senior Member roninpro's Avatar
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    Zeroes of a Polynomial (and its Derivative)

    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?
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  2. #2
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    Quote Originally Posted by roninpro View Post
    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
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  3. #3
    Senior Member roninpro's Avatar
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    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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