# A theorem of Gauss

• Jun 7th 2010, 02:09 PM
Bruno J.
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 $\displaystyle p(z)$.
• Jun 7th 2010, 04:01 PM
tonio
Quote:

Originally Posted by Bruno J.
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