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Thread: A theorem of Gauss

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    MHF Contributor Bruno J.'s Avatar
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    A theorem of Gauss

    Let $\displaystyle p(z)$ be a complex polynomial of degree $\displaystyle >1$. Show that the zeroes of $\displaystyle p'(z)$ lie in the convex hull of the zeroes of $\displaystyle p(z)$.
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    Quote Originally Posted by Bruno J. View Post
    Let $\displaystyle p(z)$ be a complex polynomial of degree $\displaystyle >1$. Show that the zeroes of $\displaystyle p'(z)$ lie in the convex hull of the zeroes of $\displaystyle p(z)$.

    Spoiler:
    This is Gauss-Lucas Theorem. Highlights of the proof:


    If $\displaystyle p(z)=p'(z)=0$ , then $\displaystyle z=1\cdot z+\sum^n_{i=1}0\cdot a_1$ is a convex combination of the roots of $\displaystyle p(z)$ , otherwise:

    write $\displaystyle p(z)=a\prod\limits^n_{i=1}(z-a_i) \,,\,\,a,a_i\in\mathbb{C}$ , and take the logarithmic derivative of this:

    $\displaystyle \frac{p'(z)}{p(z)}=\sum^n_{i=1}\frac{1}{z-a_i}=\sum^n_{i=1}\frac{\overline{z}-\overline{a_i}}{|z-a_i|^2}$ , which is valid whenever $\displaystyle p(z)\neq 0$ . So if $\displaystyle z$ is a root of $\displaystyle p'(z)$ but not of $\displaystyle p(z)$ we get:

    $\displaystyle 0=\sum^n_{i=1}\frac{\overline{z}-\overline{a_i}}{|z-a_i|^2}\Longrightarrow \sum^n_{i=1}\frac{1}{|z-a_i|^2}\overline{z}=\sum^n_{i=1}\frac{1}{|z-a_i|^2}\overline{a_i}$ , and taking conjugates and dividing we finally get:

    $\displaystyle z=\frac{\displaystyle{\sum^n_{i=1}\left(\frac{1}{| z-a_i|^2}\right)a_i}}{\displaystyle{\sum^n_{i=1}\fra c{1}{|z-a_i|^2}}}$ $\displaystyle = \frac{\displaystyle{\sum^n_{i=1}\frac{1}{|z-a_i|^2}}}{\displaystyle{\sum^n_{k=1}\frac{1}{|z-a_k|^2}}}\,a_i$ , and it's easy now to see these are barycentric coordinates of $\displaystyle z$ wrt $\displaystyle a_1,\ldots,a_n$ (please do pay attention

    closely to the last expression's indexes).

    Tonio
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