Erdős Problem #1044

Erdős Problem 1044: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Which infimum is determined

We reconstruct the proof in Quanyu Tang, On Erdős Problem #1044, Sections 2--4, https://raw.githubusercontent.com/QuanyuTang/erdos-problem-1044/main/On_Erd%C5%91s_Problem_1044.pdf. A diameter theorem of Pommerenke is explicitly retained as an external input. We give the boundary-length step in a form that counts geometric boundary length without counting overlapping parametrized arcs twice.

A boundary-length lemma

Every component of $\{|f|<1\}$ contains a root. Otherwise $\log|f|$ would be harmonic in that bounded component with boundary value zero, contrary to its strictly negative interior values. In particular there are at most $\deg f$ components.

The one external diameter input

If a monic polynomial has all roots in the closed unit disk and $|f(0)|<1$, then the component of $\{|f|<1\}$ containing zero has diameter strictly greater than one.

If $|f(0)|<1$, apply this input and $2$ to get a component with boundary length greater than two. The only remaining possibility is $|f(0)|=1$, since the product of the root moduli is at most one. Equality forces every root to have modulus one. The nonconstant harmonic function $\log|f|$ near zero has value zero there, so every neighbourhood of zero contains a point of $\{|f|<1\}$. There are only finitely many components; hence zero belongs to the boundary of one component $U$.

That component contains a root $a$ with $|a|=1$. Since $a$ is an interior point of $U$, the set $U$ contains a point $b$ with $|b|>1$. Thus the maximum of $|z|$ on $\overline U$, attained on $\partial U$, is strictly greater than one. Together with $0\in\partial U$ this gives $\operatorname{diam}(\overline U)>1$. Equation $2$ again gives boundary length greater than two. We have proved a strict lower bound for each individual polynomial, and hence a lower bound of two for the infimum.

An explicit sequence approaching two

Assessment

Completion rate estimate: 100%.

1) FORMAL RESTATEMENT

Fix an integer $n\ge 1$. Let \[ f(z)=\prod_{i=1}^n (z-z_i)\in\mathbb C[z] \] be monic of degree $n$ with roots satisfying $|z_i|\le 1$ for all $i$. Let \[ E_f:=\{z\in\mathbb C: |f(z)|<1\}. \] For each connected component $U$ of $E_f$, let $\partial U$ be its boundary and let $\ell(\partial U)$ denote its boundary length (interpreted as arclength when $\partial U$ is a rectifiable curve; more generally the $1$-dimensional Hausdorff measure). Define \[ \Lambda(f):=\max\{\ell(\partial U):\ U \text{ is a connected component of }E_f\}. \] Determine the value of \[ L_n:=\inf\{\Lambda(f):\ f(z)=\prod_{i=1}^n(z-z_i)\text{ monic},\ |z_i|\le 1\}. \] Edge cases: - For $n=1$, $E_f$ is a disk of radius $1$, so $\Lambda(f)=2\pi$. - $E_f$ is bounded for every monic polynomial $f$.

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement attributes this to Erdős--Herzog--Piranian $1958$. No further results are stated in the provided problem text.

3) ATTACK PLAN

Proof track (aim: determine $L_n$): - Identify a candidate extremizer, plausibly a highly symmetric polynomial such as $f(z)=z^n-1$. - Compute $\Lambda(z^n-1)$ explicitly. - Prove a matching lower bound for all admissible $f$.

Disproof/construction track (aim: refute a natural candidate or squeeze bounds): - Produce explicit families of polynomials and compute/estimate their $\Lambda(f)$. - Prove general lower bounds on $\Lambda(f)$ using disks contained in components, isoperimetry, and crude bounds on $|f|$ from the root constraints.

4) WORK

FAST REALITY CHECK (small $n$ and one explicit family)

- $n=1$: for $f(z)=z-z_1$ with $|z_1|\le 1$, $E_f$ is the open disk $|z-z_1|<1$. Hence $\Lambda(f)=2\pi$ and therefore $L_1=2\pi$.

- $n=2$: take $f(z)=z^2-1=(z-1)(z+1)$. Then $E_f=\{|z^2-1|<1\}$ has two components, and each component boundary is one lobe of the lemniscate $|z^2-1|=1$. A direct polar computation (Lemma 4.3 below) gives \[ \Lambda(z^2-1)=\ell(\text{one lobe})=\frac{\sqrt 2}{2}\,\frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)}\approx 3.7081493546. \] In particular $L_2\le 3.7081493546<2\pi$.

(Computational note: I evaluated the Gamma/Beta expression with high precision in Python; see the table in Lemma 4.3.)

Lemma 4.1 (Each component of $E_f$ contains at least one root)

Let $U$ be a connected component of $E_f$. Then $U$ contains at least one zero of $f$. Consequently, $E_f$ has at most $n$ connected components.

Proof. Let $U$ be a component of $E_f=\{ |f|<1\}$. Set $V:=f(U)\subset\mathbb D$ where $\mathbb D=\{w:|w|<1\}$. Because $f$ is holomorphic and nonconstant, it is an open map, hence $V$ is open in $\mathbb D$. Also, $V$ is connected because $U$ is connected.

We claim that $\partial V$ cannot meet the interior of $\mathbb D$. Assume for contradiction that there exists $w_0\in\partial V$ with $|w_0|<1$. Then there is a sequence $w_m\in V$ with $w_m\to w_0$. For each $m$, pick $z_m\in U$ such that $f(z_m)=w_m$.

Since $f$ is monic of degree $n$, $|f(z)|\to\infty$ as $|z|\to\infty$. Therefore the set $\{z:|f(z)|\le 1\}$ is bounded, hence $\overline{U}\subset\{z:|f(z)|\le 1\}$ is bounded. By compactness of closed bounded sets in $\mathbb C\cong\mathbb R^2$, $\overline{U}$ is compact, so the sequence $(z_m)$ has a convergent subsequence $z_{m_j}\to z\in\overline{U}$. By continuity of $f$, \[ f(z)=\lim_{j\to\infty} f(z_{m_j})=\lim_{j\to\infty} w_{m_j}=w_0. \] If $z\in U$, then $w_0=f(z)\in V$, contradicting $w_0\in\partial V$. Thus $z\in\partial U$.

Now use the defining inequality of $U$: - On $U$, $|f|<1$. - On the complement of $E_f$, $|f|\ge 1$. By continuity of $|f|$, every boundary point $z\in\partial U$ must satisfy $|f(z)|=1$. In particular, $|w_0|=|f(z)|=1$, contradicting $|w_0|<1$.

Hence $\partial V\cap \mathbb D=\varnothing$. This means $V$ is both open and (relatively) closed in $\mathbb D$. Since $\mathbb D$ is connected and $V\ne\varnothing$, we must have $V=\mathbb D$. In particular $0\in V$, so there exists $z\in U$ with $f(z)=0$, i.e. a root of $f$.

Finally, distinct components contain disjoint sets of roots (counted with multiplicity), so there are at most $n$ components. $\square$

Lemma 4.2 (A universal disk inside each root component; crude lower bound)

Let $n\ge 1$. Define $r_n>0$ as the unique positive solution to \[ r(2+r)^{n-1}=1. \] Then for every admissible $f(z)=\prod_{i=1}^n(z-z_i)$ and every root $z_i$, the Euclidean disk $D(z_i,r_n)=\{z:|z-z_i|<r_n\}$ is contained in $E_f$. Consequently, for every such $f$, \[ \Lambda(f)\ge 2\pi r_n, \] and hence $L_n\ge 2\pi r_n$.

Proof. Fix a root $z_i$ and write \[ f(z)=(z-z_i)\prod_{j\ne i}(z-z_j). \] Let $r>0$ and consider any $z$ with $|z-z_i|\le r$. For each $j\ne i$ we have \[ |z-z_j|\le |z-z_i|+|z_i-z_j|\le r + (|z_i|+|z_j|)\le r+2, \] since $|z_i|,|z_j|\le 1$. Therefore \[ |f(z)|\le |z-z_i|\prod_{j\ne i}|z-z_j|\le r\,(2+r)^{n-1}. \] Choose $r=r_n$ so that $r_n(2+r_n)^{n-1}=1$. Then $|f(z)|\le 1$ for all $z$ with $|z-z_i|\le r_n$. Because $f$ is holomorphic and not constant on $D(z_i,r_n)$, and $|f|$ attains a strict value $|f(z_i)|=0<1$ inside, the maximum modulus principle implies $|f(z)|<1$ for all $|z-z_i|<r_n$. Thus $D(z_i,r_n)\subset E_f$.

Let $U$ be the component of $E_f$ containing $z_i$; then $D(z_i,r_n)\subset U$. Hence $\operatorname{area}(U)\ge \pi r_n^2$. By the planar isoperimetric inequality, $\ell(\partial U)^2\ge 4\pi\,\operatorname{area}(U)$, so $\ell(\partial U)\ge 2\pi r_n$. Taking the maximum over components gives $\Lambda(f)\ge 2\pi r_n$. $\square$

Lemma 4.3 (Explicit computation for $f(z)=z^n-1$)

Let $n\ge 1$ and $f(z)=z^n-1$. Then the boundary $|f(z)|=1$ consists of $n$ congruent lobes meeting at $z=0$. Each connected component of $E_f=\{|z^n-1|<1\}$ has boundary equal to one lobe, of length \[ \Lambda(z^n-1)=\frac{2^{1/n}}{n}\,B\!\left(\frac{1}{2n},\frac12\right) =\frac{2^{1/n}}{n}\,\frac{\Gamma\!\left(\frac{1}{2n}\right)\Gamma\!\left(\frac12\right)}{\Gamma\!\left(\frac12+\frac{1}{2n}\right)}. \] In particular this provides the upper bound $L_n\le \Lambda(z^n-1)$. Moreover, as $n\to\infty$, \[ \Lambda(z^n-1)=2+O\!\left(\frac{\log n}{n}\right), \] so $\Lambda(z^n-1)\downarrow 2$ numerically.

Proof. The boundary is $|z^n-1|=1$, equivalently \[ |z^n-1|^2=1 \iff z^n\overline{z}^n - z^n - \overline{z}^n +1=1. \] Write $z=re^{i\theta}$. Then $z^n=r^n e^{in\theta}$ and \[ |z^n-1|^2=r^{2n}-2r^n\cos(n\theta)+1. \] Setting this equal to $1$ gives \[ r^{2n}-2r^n\cos(n\theta)=0 \iff r^n=2\cos(n\theta), \] with the restriction $\cos(n\theta)\ge 0$ to keep $r\ge 0$. Thus on each interval where $\cos(n\theta)\ge 0$ the boundary is given in polar form by \[ r(\theta)=(2\cos(n\theta))^{1/n}. \] One lobe is traced by $\theta\in[-\pi/(2n),\pi/(2n)]$.

For a polar curve $r=r(\theta)$, arclength is $ds=\sqrt{r(\theta)^2+(r'(\theta))^2}\,d\theta$. Compute \[ r'(\theta)=\frac{1}{n}(2\cos(n\theta))^{1/n-1}\cdot (-2n\sin(n\theta)) =-2\sin(n\theta)\,(2\cos(n\theta))^{1/n-1}. \] Hence \begin{align*} r^2+(r')^2 &= (2\cos(n\theta))^{2/n} + 4\sin^2(n\theta)\,(2\cos(n\theta))^{2/n-2}\\ &= (2\cos(n\theta))^{2/n-2}\left((2\cos(n\theta))^2+4\sin^2(n\theta)\right)\\ &= (2\cos(n\theta))^{2/n-2}\cdot 4(\cos^2(n\theta)+\sin^2(n\theta))\\ &=4(2\cos(n\theta))^{2/n-2}. \end{align*} Therefore \[ ds=2(2\cos(n\theta))^{1/n-1}\,d\theta. \] The lobe length is \begin{align*} \Lambda(z^n-1) &=\int_{-\pi/(2n)}^{\pi/(2n)} 2(2\cos(n\theta))^{1/n-1}\,d\theta\\ &=\frac{4}{n}\int_{0}^{\pi/2} (2\cos u)^{1/n-1}\,du \quad (u=n\theta). \end{align*} Pulling out $2^{1/n-1}$ gives \[ \Lambda(z^n-1)=\frac{4}{n}2^{1/n-1}\int_0^{\pi/2} (\cos u)^{1/n-1}\,du. \] Using the Beta integral $\int_0^{\pi/2}\cos^{\alpha-1}u\,du=\tfrac12 B(\tfrac\alpha2,\tfrac12)$ for $\alpha>0$, with $\alpha=1/n$, yields \[ \Lambda(z^n-1)=\frac{4}{n}2^{1/n-1}\cdot\frac12 B\!\left(\frac{1}{2n},\frac12\right) =\frac{2^{1/n}}{n}B\!\left(\frac{1}{2n},\frac12\right). \] The Gamma-function form follows from $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$.

For the asymptotic: as $x\to 0^+$, $\Gamma(x)=\frac{1}{x}+O(1)$ and $\Gamma(x+1/2)=\Gamma(1/2)(1+O(x))$. With $x=1/(2n)$ this gives $B(1/(2n),1/2)=2n(1+O(1/n))$. Also $2^{1/n}=1+O(\log 2/n)$. Combining yields $\Lambda(z^n-1)=2+O(\log n/n)$. $\square$

Numerical table (from the closed form in Lemma 4.3): \[ \begin{array}{c|c} n & \Lambda(z^n-1)\\\hline 1 & 6.2831853072\\ 2 & 3.7081493546\\ 3 & 3.0599080741\\ 4 & 2.7675051293\\ 5 & 2.6013622764\\ 6 & 2.4942886983\\ 10 & 2.2886060328 \end{array} \]

Also, the crude lower bound from Lemma 4.2 gives (using $r_n(2+r_n)^{n-1}=1$): \[ \begin{array}{c|c} n & 2\pi r_n\\\hline 1 & 6.2831853072\\ 2 & 2.6025805691\\ 3 & 1.2916308247\\ 4 & 0.6717940285\\ 5 & 0.3516535247 \end{array} \]

5) VERIFICATION

- Quantifiers: Lemma 4.2 holds for every admissible $f$ and each root $z_i$; no hidden dependence. - Boundary/value set logic in Lemma 4.1: the only delicate point is showing that for $z\in\partial U$, necessarily $|f(z)|=1$. This follows from continuity of $|f|$ and the fact that $U\subset\{|f|<1\}$ while $\mathbb C\setminus E_f\subset\{|f|\ge 1\}$. - For Lemma 4.3, endpoints $\theta=\pm\pi/(2n)$ give $\cos(n\theta)=0$ hence $r=0$; the arclength integral converges because near $u=\pi/2$ the integrand behaves like $(\cos u)^{1/n-1}$ with exponent $>-1$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. - Exact value for $n=1$: $L_1=2\pi$. - For general $n$, the explicit family $f(z)=z^n-1$ gives the constructive upper bound \[L_n\le \frac{2^{1/n}}{n}B\!\left(\frac{1}{2n},\frac12\right),\] and Lemma 4.2 gives the universal lower bound $L_n\ge 2\pi r_n$ where $r_n(2+r_n)^{n-1}=1$.

(ii) First gap (crisp statement). Prove (or disprove) that for each fixed $n\ge 2$, \[ L_n = \Lambda(z^n-1)=\frac{2^{1/n}}{n}B\!\left(\frac{1}{2n},\frac12\right). \] Equivalently: show that every admissible $f$ has some component boundary of length at least $\Lambda(z^n-1)$.

(iii) Top 3 next moves. 1. Prove a sharp lower bound on $\Lambda(f)$ in terms of $n$ using the fact that each component maps onto $\mathbb D$ under $f$ (Lemma 4.1) and try to convert degree information into a perimeter bound. 2. Search numerically for $n=2,3,4$ over root configurations (e.g. roots on $|z|=1$) to see whether anything beats $z^n-1$; this would guide whether the conjectured extremizer is plausible. 3. Attempt a variational argument: among polynomials with roots in $\overline{\mathbb D}$, show that moving roots to the unit circle and spreading them symmetrically cannot increase the infimum of $\Lambda$.

(iv) Minimal counterexample structure. If $z^n-1$ is not optimal, a counterexample would be a polynomial with roots on/near $|z|=1$ such that $E_f$ has $n$ components (to keep the maximum perimeter small), but each component boundary is shorter than the corresponding lobe of $|z^n-1|=1$. Such a configuration would likely have boundaries meeting at multiple critical points (not just at the origin), indicating a different pinching geometry.