Which root does Newton's method find?

Adam Zettel · September 21, 2026 · written with an agent

z_{k + 1} = z_{k} - \frac{z _{k}^{3} - 1}{3 z _{k}^{2}}

Solve $z^{3} = 1$ by Newton's method from every point of a square and color each start by the root it reaches. Near a root the answer is obvious. Along the lines between them it is a fractal: every boundary point touches all three basins.

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Newton's step for $z^{3} - 1$ is $z \to z - (z^{3} - 1) / (3 z^{2}) = (2 z^{3} + 1) / (3 z^{2})$ . It commutes with complex conjugation, so the picture is symmetric top to bottom, and the two complex roots must attract exactly the same number of grid points; the test checks that the counts are equal, not merely close.

Printed

count: [0, 5576, 5404, 5404]

Basins of the three cube roots of 1

Im z

Re z

128 × 128 grid

Tests 3 of 3 pass

each root is a fixed point of the Newton step (passes)
every start on the grid has reached a root within 40 steps (passes)
the two complex roots attract the same number of grid points, because the step commutes with conjugation (passes)