The Gauss--Lucas Theorem states that all zeros of a degree $n$ complex polynomial $p(z)$ are contained in the convex hull of the zeros of $p$. By iteration, this implies that the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$ are contained in the convex hull of the zeros of $p$.

The Riemann--Hurwitz Theorem (among others) implies that if a tract $D$ of $p$ (namely a component of the set $\{z:|p(z)|<\epsilon\}$ for some $\epsilon>0$) contains all the zeros of $p$ in its bounded face, then all the critical points of $p$ are contained in $D$.

My conjecture is that in fact, if $D$ is a tract of $p$ and contains all the zeros of $p$, then $D$ also contains all the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$.

This certainly does not follow by straight-forward iteration, since in general there need not be a tract of $p'$ containing all the zeros of $p'$ which is contained in $D$. It seems that the tracts and level curves of $p'$ do not interact very nicely with the tracts and level curves of $p$ (even worse for $p'',p''',\ldots$).

I have taken a look at attempting to apply the Cauchy Integral Formula (some sort of integration by parts application perhaps?), but don't seem to be able to make progress there. Any ideas for proof or counter-example?

  • $\begingroup$ The assumption "$D$ is a tract of $p$ and contains all the zeros of $p$" amounts to saying $\{z: |p(z)|<\epsilon\} $ is connected. There is some research on polynomials with connected lemniscates, but I haven't seen anything about higher derivatives. $\endgroup$
    – user147263
    Aug 12, 2014 at 18:32
  • $\begingroup$ Also, a lemniscate is connected iff it contains all of the critical points of the polynomial. Thus, your question is equivalent to the following: let $M=\max \{|p(z)| : p'(z)=0\}$; is it true that $|p|\le M$ at all zeros of all derivatives of $p$? $\endgroup$
    – user147263
    Aug 12, 2014 at 18:50
  • $\begingroup$ @Thursday Yes, I think your second comment is very helpful, and may be the way to go. It may allow more integral methods to be brought to bear. $\endgroup$ Sep 5, 2014 at 15:59
  • $\begingroup$ Now posted to MO, mathoverflow.net/questions/189245/… $\endgroup$ Dec 9, 2014 at 4:00
  • 1
    $\begingroup$ A counter-example has been posted on the M.O. page (mathoverflow.net/questions/189245/…). $\endgroup$ May 4, 2015 at 15:33

1 Answer 1


This is a community wiki answer to remove this question from the unanswered list: Bobby Ocean posted a counter-example at Mathoverflow: Set $p(z) = (z^4+2z^2+2) (z-1)$. Then $D:= \{ z : |p(z)|<1.45\}$ is connected, but some of the zeroes of $p'''$ and $p''''$ are outside $D$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .