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Consider a generic complex polynomial $f$ (i.e. one with pairwise distinct zeros [EDITed:] and with pairwise distinct zeros of $f'$) and define for every pair of zeros an open disc which has as diameter the line segment between the two zeros. (Remark: I’m especially interested in the case of zeros that are close to each other, e.g. pairs of zeros defined as vertices of the edges of the minimal weigth spanning tree of the zeros where weight is square of Euclidean distance of zeros.)

Conjecture: In each of these discs there is at least one critical point of $f$ (i.e. a zero of $f’$).

What is known: If $f$ has only real coefficients and (therefore) only real or complex conjugate zeros, then the critical points of $f$ are either real or are located within the open discs which have as diameter the line segment between complex conjugate zeros (Polya, Szeg\“o, Aufgaben und Lehrs\“atze aus der Analysis I, page 90).

Additional conjecture: One of these zeros of $f’$ located in the disc is the one which comes from the center of the disc if one considers in a first step only the quadratic polynomial with the two zeros at the end of the line segment (trivially the zero of $f’$ is then at the center of the line segment) and in a second step one “switches” the other zeros on, e.g. by multiplying with a factor $(\alpha (z-z_i) + 1- \alpha)$ for each additional zero $z_i$ and letting $\alpha$ go from $\alpha=0$ to $\alpha=1$ (which makes the othere zeros move in from $\infty$ to their position $z_i$.

Every hint towards a proof or a refutation is appreciated.

Motivation:

Every complex polynomial with $n$ distinct zeros has $n-1$ critial points (zeros of $f’$). For the location of the zeros there is huge literature (for a starting point see e.g. Q.I. Rahman und G. Schmeisser, Analytic Theory of Polynomials, Oxford University Press, 2002).

I would like to have a deeper understanding on the location of the zeros, also in context with different spanning trees pls. cf. also the following related question: (https://mathoverflow.net/users/6415/andreas-r%c3%bcdinger), Do the bounded isophase lines of a complex polynomial $f$ through the zeros of $f’$ define a spanning tree?, URL (version: 2015-07-12): Do the bounded isophase lines of a complex polynomial $f$ through the zeroes of $f’$ define a spanning tree?

In this context I have proven a slight strengthening of the theorem of Gauss-Lucas for polynomials of degree four (http://arxiv.org/abs/1405.0689).

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No. Take $f(z)=z^n-1$ to get an easy counterexample.

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    $\begingroup$ Thank you, you are right. However, I wanted $f$ also to be generic in the sense that $f'$ has pairwise distinct zeros (which I didn't say in the question; I will edit it accordingly). Perhaps a slight modification of your answer will still give a counterexample. $\endgroup$
    – Andreas Rüdinger
    Commented Feb 21, 2016 at 17:28
  • $\begingroup$ @AndreasRüdinger Yes, this is clear, as the roots and critical points depend continuously on the value of a complex polynomial. $\endgroup$
    – Will Sawin
    Commented Feb 21, 2016 at 17:37
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For seeing immediately that such conjectures cannot be true, it is useful to keep in mind Runge's theorem: every analytic function on a compact set which does not divide the plane is uniform limit of polynomials. Application of Runge always gives sufficiently generic examples: an open set.

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  • $\begingroup$ you're thinking of applying this to some analytic function like $e^z-1$? $\endgroup$
    – Will Sawin
    Commented Feb 22, 2016 at 2:52

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