Starshapeness of polynomial tracts with respect to the (entire collection of) critical points contained in the tract

I recently found out (Piranian, "The Shape of Level Curves") that a polynomial tract (ie a connected component of a set of the form $G=\{z:|p(z)|<\epsilon\}$ for some $\epsilon>0$) need not be starshaped with respect to the zeros of $p$ contained in $G$. This disappointed me bitterly, as that starshapeness was a pivotal step in a proposed "proof" I had of Smale's mean value conjecture.

The places where $G$ is not starshaped with respect to the zeros of $p$ in $G$ are near critical points of $p$ in $G$ or in $\partial G$, so I still hold out a tiny bit of hope for the starshapeness of $G$ with respect to the critical points of $p$ contained in $G$:

Conjecture: If $G$ is a tract of $p$ with smooth boundary containing more than one distinct zero of $p$, then $G$ is starshaped with respect to the critical points of $p$ contained in $G$.

Intuitions/proofs/disproofs/references are all welcome.

EDIT: Note that when I say that $G$ should be "starshaped with respect to the critical points", I mean that each point in $G$ can be seen by some one of the critical points of $p$ in $G$, not of course that some single critical point can see all points in $G$.

Note also that I added the assumption that $G$ contains more than one distinct zero of $p$ (since otherwise $G$ will not contain any critical points of $p$.

One reason I think this is plausible: If we consider the lemniscate of Bernoulli, and let $G$ be the interior of a level curve of $p$ which is a bit bigger, the critical point of $p$ is right in the center, so should be able to "see" both lobes. In the counter-example of Piranian to my desired conjecture (that tracts are star-shaped with respect to the zeros they contain), the points that killed the starshapeness were close to the boundary of $G$, so perhaps if we assume $\partial G$ is smooth, $G$ will contain enough critical points to see into all "corners".

• Do you assume that $G$ is connected? – Malik Younsi Dec 4 '14 at 20:06
• Yes, a tract is a component of the set $\{z:|p(z)|<\epsilon\}$. I will add that to the definition. – Trevor J Richards Dec 5 '14 at 0:23

There is no hope: according to a theorem of Hilbert, every analytic Jordan curve $J$ can be approximated by a lemniscate $\{z:|P(z)|=\epsilon\}$. So the set does not have to be starlike with respect to any point. For this theorem of Hilbert, see, for example J. L. Walsh, Interpolation and Approximation by Rational Functions in the Complex Plane,'' 5th ed., Amer. Math. Society, Providence, RI, 1969.
• Moreover, it seems that if the Jordan Curve $J$ that the lemniscate $\Lambda=\{z:|p(z)|=\epsilon\}$ approximates is very complicated, then the bounded face of $\Lambda$ may need to contain many zeros of $p$ for $\Lambda$ to approximate $J$ well. – Trevor J Richards Dec 4 '14 at 15:41
• (continued) and thus of course the bounded face of $A$ must contain many zeros of $p'$ if it contains many zeros of $p$. – Trevor J Richards Dec 6 '14 at 22:50