GPT 6 Astra Ultra attempt on Math-6102

Erdős Problem 1048: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Statement and counterexample

The proposed conclusion is false for every fixed $1<r<2$. For all sufficiently large $n$, take \[ f_n(z)=z^n-r^n. \] Its roots all have modulus exactly $r$. We prove that its unit lemniscate has exactly $n$ connected components, each of diameter at most \[ \frac{2}{n(r^n-1)^{(n-1)/n}}\longrightarrow0. \tag{1} \] In particular every component eventually has diameter smaller than the fixed positive number $2-r$, contradicting the required existence of a component with diameter greater than $2-r$.

This is Pommerenke's counterexample as recorded at https://www.erdosproblems.com/1048, based on On metric properties of complex polynomials, https://doi.org/10.1307/mmj/1028998561. We prove the component and diameter assertions explicitly.

Separate inverse branches

The disk $D=\{w:|w-r^n|<1\}$ does not contain zero, because $r^n>1$. It lies in the right half-plane, so the principal logarithm is holomorphic on a neighborhood of its closure. Its $n$ root branches are \[ g_j(w)=\exp\!\left(\frac{\log w+2\pi i j}{n}\right), \qquad j=0,\ldots,n-1. \] Every solution of $z^n=w$ is one of these, and their images $U_j=g_j(D)$ are disjoint: equality of two image points first forces the same $w$, then the same $n$th root of unity. Each $g_j$ is injective and holomorphic with nonzero derivative, so $U_j$ is connected and open. Their union is exactly $\{|f_n|<1\}$. A finite disjoint union of these open sets has them as its connected components, since each one is also relatively closed in that union. Each contains its corresponding root $re^{2\pi i j/n}$.

Uniform diameter control

For every $w\in D$, \[ |g_j'(w)|=\frac1{n|w|^{(n-1)/n}} \le\frac1{n(r^n-1)^{(n-1)/n}}. \] The disk is convex. Integrating the derivative along the segment between $w_1,w_2\in D$ gives \[ |g_j(w_1)-g_j(w_2)| \le\frac{|w_1-w_2|}{n(r^n-1)^{(n-1)/n}} \le\frac2{n(r^n-1)^{(n-1)/n}}, \] proving $1$. For example, once $r^n\ge2$ the right side is at most $4/(nr^{n-1})$, which tends to zero. The same argument on the closed disk bounds the closed components too; their branch images remain disjoint there.

Boundary of the result and assessment

The construction uses $r>1$ precisely to keep the root disk away from zero and to make the inverse branches contract exponentially. It does not contradict the affirmative results recorded for $0<r\le1$. At $r=1$, the disk touches zero and the limiting closed petals meet there; estimate $1$ no longer applies. A counterexample in the allowed range $1<r<2$ is sufficient to settle the stated universal question negatively, and that counterexample is fully established here. No stronger classification for the other parameter range is claimed.

Completion rate estimate: 100%.