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Math Help - A theorem of Gauss

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

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

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


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

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

    \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 p(z)\neq 0 . So if z is a root of p'(z) but not of p(z) we get:

    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:

    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}}} = \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 z wrt a_1,\ldots,a_n (please do pay attention

    closely to the last expression's indexes).

    Tonio
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