Spider iteration
z(n+1) = z(n)^2 + c / z(n)
The reciprocal term gives a quadratic orbit a sharp tug around the pole at zero.
Start with c = 0. The reciprocal term vanishes, leaving z², the standard quadratic Julia map; initial values inside the unit disk can still have bounded orbits.
Now take c = 1 and z₀ = (1, 0). The first iteration gives z₁ = 1² + 1/1 = 2, already far from the origin; the next gives z₂ = 4 + 1/2 = 4.5, and escape is rapid. Begin near zero instead—say z₀ = (0.01, 0)—and 1/0.01 = 100 dominates, throwing the orbit out at once. The spider filaments live on the boundary between those regimes.
