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Motivated by a geometric proof of the Fundamental Theorem of Algebra we ask:

Is there a geometric proof for the Gauss-Lucas theorem? Since we are working on a half plane, can one imagine a possible proof of the above fact with some methods in hyperbolic geometry?

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  • $\begingroup$ The Gauss-Lucas theorem has nothing particular to do with half-planes, it concerns any convex subset of the plane. $\endgroup$
    – abx
    Jun 19, 2016 at 13:19
  • $\begingroup$ @abx Yes but an equivalent formulation is the following: If all roots of a polynomial lie at a half plane, then all roots of its derivative lie at the same half plane(See the Ahlfors book, complex analysis). $\endgroup$ Jun 19, 2016 at 14:24
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    $\begingroup$ @abx I don't see the point of your objection - convex subsets of the plane have a lot to do with half-planes! $\endgroup$
    – dvitek
    Mar 22, 2017 at 21:43
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    $\begingroup$ @abx actually, it does: I recently gave a proof of GL via the numerical range of a matrix, which involves half-planes (see my answer below for details). $\endgroup$ Feb 8, 2020 at 17:43

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Good question! I think the following article may qualify as a "yes":

Arnaud Ch´eritat, Yan Gao, Yafei Ou, Lei Tan. "A refinement of the Gauss-Lucas theorem (after W. P. Thurston)". 2015. https://hal.archives-ouvertes.fr/hal-01157602/document

(An essential observation here is that you can make the Gauss-Lucas theorem explicitly about the behavior half-planes in $\mathbb{C}$. )

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    $\begingroup$ Thank you very much for your very perfect and interesting answer. $\endgroup$ Feb 23, 2017 at 21:11
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I think the problem with that is that the theorem is so easy (close to trivial really) that there's no need for any tools. If we want to work in a zero free half plane, then the following simple argument settles matters immediately: Assume, for convenience, that $p\not=0$ on $\mathbb C^+ =\{ z: \textrm{Im}\, z> 0\}$. If we write $p(z)=c\prod (z-a_j)$, then $$ \frac{p'(z)}{p(z)} = \sum \frac{1}{z-a_j} . $$ Since $\textrm{Im}\, 1/(z-a)<0$ for $z\in\mathbb C^+$ if $\textrm{Im}\, a\le 0$, it follows that $p'/p$ has no zeros in $\mathbb C^+$, either.

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    $\begingroup$ Thank you for your interesting answer.But I search for a proof with a serious usage og a riemanian metric or some thing in the upper half plane as the main object of poin care plane. I guess that yourmethod is similar to the book of Ahlfors $\endgroup$ Jun 19, 2016 at 18:37
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    $\begingroup$ @AliTaghavi Why do you want or hope that something with a direct proof should have a proof using some complicated tools? Not everything to do with a half-plane is to do with hyperbolic space $\endgroup$
    – Yemon Choi
    Jun 19, 2016 at 20:54
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I gave a simple geometric proof of Bocher's theorem (a generalization of the Gauss-Lucas theorem) in a paper in Computational Methods and Function Theory in 2015. You can find a pdf at my webpage linked here. The paper is the one titled

  • "Level curve configurations and conformal equivalence of meromorphic functions"

It is a long paper (48 pages), but the proof of Bocher's theorem is entirely contained on page 6.

The idea is as follows. Suppose that a rational function $R(z)$ has all of its zeros contained in one disk $D_1$ on the Riemann sphere, and all of its poles on another disk $D_2$, and that $D_1\cap D_2=\emptyset$. Bocher's theorem says that there will be no critical point of $R$ outside of $D_1\cup D_2$. This immediately implies the Gauss--Lucas theorem.

PROOF: Suppose that there is a critical point $z_0$ of $R$ outside of $D_1\cup D_2$. We may as well assume that $z_0=0$, and that $D_1$ is contained in the half-plane $\{z:Re(z)<-1\}$ and that $D_2$ is contained in the half-plane $\{z:Re(z)>1\}$.

Define $\epsilon=|R(0)|$, and consider the level set $\mathcal{L}=\{z:|R(z)|=\epsilon\}$. Since $0$ is a critical point of $R$, $\mathcal{L}$ has a branching at $0$. In fact, if $0$ is a critical point of $R$ of multiplicity $k$, then $\mathcal{L}$ has a $(2k+2)$--fold branching at $0$. Thus there is some horizontal line segment $H_m=\{x+im:-1\leq x\leq1\}$ which intersects $\mathcal{L}$ in at least two points distinct, say $z_1$ and $z_2$, with $Re(z_1)<Re(z_2)$. However since $z_1$ is closer to every zero and further from every pole of $R$ than $z_2$, it must be that $|R(z_1)|<|R(z_2)|$, contradicting the assumption that $z_1$ and $z_2$ are in the same level set of $R$.

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  • $\begingroup$ Thank you so much for your very interesting answer and your elegant result on generalization of Gauss Lucas theorem. My apology that I did not pay attention to the details, however I have given my +1, with my most pleasure. $\endgroup$ Sep 16, 2018 at 19:05
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I think (but am not sure), that the first and most enlightening proof is due to Gauss himself: The derivative of a polynomial is given by

$$ f^\prime(z) = f(z) \sum \frac1{z-a_i}, $$

where the $a_i$ are the zeros. So, thinking of point charges at $a_i,$ with Coulomb repulsion, the critical points are where the Coulomb forces balance. This can obviously only happen in the convex hull of the zeros.

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Charles R. Johnson and I gave a proof of the Gauss–Lucas theorem that utilizes the field of values (or numerical range) of a matrix, which can be viewed as geometric in nature (the field itself is an uncountably infinite intersection of half-planes). The proof can be found in the article Matricial Proofs of Some Classical Results about Critical Point Location, which recently appeared in The American Mathematical Monthly.

Our methods also yield a proof of Marden's theorem and a weaker version of Siebeck's theorem.

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