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Statement

Quoted (pp. 111--112): "Let $f(z)=\prod_{k=1}^m(z-z_k)^{p_k}$ ($z_k$ distinct, $p_k$ positive integers), and let $E=\{|f(z)|\le1\}$ have the maximal number $m$ of components. H. Grunsky (see [2, Problem 16]) raised the question whether all components must be convex. I shall give a counter-example."

Theorem 14 (p. 112). "Let $f(z)=z^p(z-a)$. If $a-(1+p^{-1})\cdot p^{1/(p+1)}$ is positive and sufficiently small, and $p$ is sufficiently large, then the set $E=\{|f(z)|\le1\}$ has two components, one of which is not convex."

Here $m=2$ ($z_1=0$ with $p_1=p$, $z_2=a$ with $p_2=1$), the level is $c=1$ for the closed sublevel set, and the nonconvex component is the one containing $0$. The degree $p+1$ is large and the root $0$ has high multiplicity; a quartic with four simple roots was given later by Goodman (1966), whose paper (p. 358) cites this counterexample and remarks on its high degree and its zero of high multiplicity.

Source. Ch. Pommerenke, On metric properties of complex polynomials, Michigan Math. J. 8 (1961), no. 2, 97--115; the question and Theorem 14 with its proof on printed pp. 111--112 (PDF pp. 15--16 of the publisher's scan), read on the page images (the scan has no text layer). The copy read is identified in the source digest.

Read depth. Claims checked: the question as the paper recalls it and the statement were read clause by clause on the page images; the proof (one paragraph) was read in full and its steps followed, the two limits $f_p(z^*)\to0$ and $\xi\to1$ being checked mentally and the constant $1.1$ taken as printed. Nothing here is independently reviewed.

Proof pointer

Page 112. The proof first takes the borderline value $a=(1+p^{-1})\xi$, $\xi=p^{1/(p+1)}$, and writes $f_p(z)=z^p(z-(1+p^{-1})\xi)$. A direct computation gives $f_p(\xi)=-1$, and the logarithmic derivative $f_p'/f_p=p/z+1/(z-(1+p^{-1})\xi)$ vanishes at $\xi$, so $f_p'(\xi)=0$: the level curve $\{|f_p|=1\}$ crosses itself at $\xi$ with branch tangents $y=\pm(x-\xi)$, and $E_p=\{|f_p|\le1\}$ is made of two pieces meeting only at $\xi$. Close to $\xi$, $E_p$ therefore stays inside the double sector $S=\{x+iy:|y|\le1.1\,|\xi-x|\}$. As $p\to\infty$, $\xi\to1$ and $f_p(z^*)\to0$ at $z^*=0.5+0.6i$ (where $|z^*|<1$), so for large $p$ the point $z^*$ lies in $E_p$ but outside $S$; the chord from $z^*$ to $\xi$ then leaves $E_p$, so the piece through $0$ fails to be convex. Raising $a$ slightly above $(1+p^{-1})\xi$ pulls the two pieces apart, so $E=\{|f(z)|\le1\}$ has two components and the one through $0$ is still not convex.

Dependencies

None outside elementary calculus.

Bears on

• Problem 1047: the negative answer to the problem's question, in the problem's own terms: a monic polynomial with $m=2$ distinct roots and a level $c=1$ at which $\{|f|\le c\}$ has two components, one of which is not convex. The status "DISPROVED (LEAN)" of the site was not traced to a formal proof here; the closed sublevel set matches the problem's inequality.