Chapter 9 · The Circle for Free

The Exact Circle

Phase without limits; roots of unity by hand

We have seen i hiding in the powers of zero, and that going twice around comes home: 0^(2ω) = 1. If those powers really sweep a unit around a circle, there is one honest way to test it. Compute points on that circle exactly — by hand, with nothing but ordinary whole-number arithmetic, no sine tables, no infinite series, no π. It works, and this chapter does it.

The step and its undo

First fix what a power of zero means as a point on the circle. Since 0^ω = −1 is the half-turn, 0^(ωx) is the point a fraction x of the way along it — at angle πx. Take a step of that kind, u = 0^(ωt), together with its undo, the reciprocal v = 0^(−ωt). Because one reverses the other their product is 1, and adding a step to its undo gives a single real number that stands in for the step:

u · v = 1 s = u + v

Everything below is built from s. You never have to know what u "is" as a decimal — only that it and v multiply to one.

One rule generates the whole circle

Now ask for the combined step-sizes further around: aₙ = uⁿ + vⁿ, the sum of the n-th step and its undo. Because u and v are reciprocals, these obey one clean rule — multiply the previous one by s and subtract the one before it:

aₙ = s · aₙ₋₁ − aₙ₋₂ a₀ = 2 , a₁ = s

That is the entire engine. Turn the crank and every term is an exact polynomial in s with whole-number coefficients — no rounding, ever:

a₂ = s² − 2 a₃ = s³ − 3s a₄ = s⁴ − 4s² + 2 a₅ = s⁵ − 5s³ + 5s

If those polynomials look familiar to anyone who has met them, they should: this is the Chebyshev recurrence, the same rule that turns the cosine of an angle into the cosine of its multiples. Stepping it is stepping around the circle by equal angles — exactly, with no trigonometry underneath and no limit anywhere in sight.1

From the real engine to the complex corner

The engine only ever returns a real number — s, and the aₙ built from it. The thing we actually want is a complex point on the circle. Crossing from one to the other takes a single observation and one square root — still no trigonometry.

Because u and v multiply to 1 and lie on the unit circle, the undo is just the mirror image of the step across the real axis: if u = x + i y, then v = x − i y. So their sum collapses to twice the horizontal coordinate, while their product only confirms the point is on the circle:

s = u + v = 2x u · v = x² + y² = 1

That is the whole bridge. The real number the recurrence hands you is exactly twice the x-coordinate of the step; the y-coordinate is whatever keeps the point on the circle:

x = s / 2 y = √(1 − x²)

And "divide the circle evenly" stops being a wish and becomes an equation. To reach 0^(ω/3) — a third of a half-turn — demand that three steps make the half-turn itself, 0^ω = −1. Three steps means u³ = −1 (and v³ = −1 with it), so a₃ = u³ + v³ must equal −2. We already turned the crank for a₃ = s³ − 3s, so the step-size is pinned by a polynomial:

s³ − 3s = −2 → s³ − 3s + 2 = 0 → (s − 1)²(s + 2) = 0 → s = 1

The step is not eyeballed; it is a root of that closing polynomial — the other root, s = −2, is the degenerate half-turn itself — and it lands on an exact value. Send s = 1 across the bridge:

x = s / 2 = 1/2 y = √(1 − 1/4) = √3 / 2

and the corner assembles itself — derived, not announced:

0^(ω/3) = x + i y = 1/2 + (√3 / 2) i

Roots of unity, by hand

The same recipe turns the crank for any even division: close the polynomial for the division you want, then read x = s/2 and y = √(1 − x²). The fifth goes exactly so — demand five steps make the half-turn, a₅ = −2, solve s⁵ − 5s³ + 5s + 2 = 0 for the root s = (1 + √5)/2, and its half is the exact cos 36°. Two corners, in exact surds rather than decimal approximations:

0^(ω/3) = 1/2 + (√3 / 2) i (a third of a half-turn) 0^(ω/5) = cos 36° + i sin 36° , cos 36° = (1 + √5) / 4

No calculator produced those; the recurrence did, in exact fractions and roots. The circle isn't being approximated here — it is being computed, corner by corner, in the same whole-number arithmetic you'd use to balance a checkbook.

Climbing the tower

And you can keep halving the step — 0^(ω/4), 0^(ω/8), and so on — each level doubling the number of points and locking in finer angles, exactly, all the way down. This is the "dyadic tower": a staircase of ever-finer exact divisions of the circle.2

What this buys

This is the concrete engine under the phrase "no limits." The circle — phase, rotation, the whole apparatus that usually arrives dressed in sines and π and infinite series — is here a whole-number recurrence you can run by hand and check to the last digit. It is the most solid, most verifiable thing in the book, and it fell out of one normalization and a matched pair of reciprocals.

  1. Worth being honest about what this does and doesn't show: the recurrence aₙ = s·aₙ₋₁ − aₙ₋₂ holds for any pair whose product is 1 — it is a basic fact about reciprocal pairs, not something special to the invertible zero. That it lands on the classical Chebyshev polynomials is therefore natural structure, not a lucky coincidence and not, by itself, independent proof that the zero-machinery is "right." What it genuinely delivers is the exactness: an algebraic, limit-free model of phase.
  2. The reach is exact for every angle, not only the halving steps shown here: the object 0^(ωx) is an exact stand-in for its angle whatever x is, and specific roots of unity come out in exact surds (the thirds and fifths above, not just powers of two). What the construction does not hand you is a finite decimal for a generic angle's coordinates — writing the cosine of an arbitrary angle as a terminating number is where approximation returns. But that is a limitation of decimal notation, not of the reach: the angle itself is represented exactly.