Roots of unity — cube roots, nth roots, geometric interpretation: Visual walkthrough

Nothing below is memorised. Every symbol is drawn before it is used.

Step 1 — What is a complex number, as a picture?

WHAT. A complex number is written $z = x + i y$ . The two everyday numbers $x$ and $y$ are just coordinates. We plot $z$ as a point (or an arrow from the origin to that point) on a flat sheet: $x$ measured rightward, $y$ measured upward. That sheet is the Argand Diagram.

WHY. Because we are about to multiply complex numbers, and multiplication is impossible to "feel" as $x + i y$ . An arrow, on the other hand, has two honest physical properties we can track: how long it is, and which direction it points. Those two are all we need.

PICTURE. Look at the amber arrow. Its length is called the modulus $r$ (how far the tip is from the centre). The angle it makes with the rightward axis, swept anticlockwise, is the argument $θ$ .

$z = how long r (cos which way θ + i sin θ)$

Here $cos θ$ recovers the rightward part of a unit-length arrow, $i sin θ$ recovers the upward part, and multiplying by $r$ stretches the whole arrow to its true length. This is the Polar Form of Complex Numbers.

Step 2 — What does multiplying two arrows DO?

WHAT. Take two arrows, $z_{1}$ with length $r_{1}$ and angle $θ_{1}$ , and $z_{2}$ with length $r_{2}$ and angle $θ_{2}$ . Their product is the arrow whose lengths multiply and whose angles add:

$z_{1} z_{2} = multiply the lengths r_{1} r_{2} (cos (ADD the angles θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})) .$

WHY this rule and not something simpler? Because this is what actually happens when you expand $(x_{1} + i y_{1}) (x_{2} + i y_{2})$ using $i^{2} = - 1$ — the algebra secretly performs "stretch and turn." We take it as our engine: multiplication turns arrows. This is the single most important fact on the page.

PICTURE. The cyan arrow $z_{1}$ turns by the angle of the second arrow and grows by its length. Watch the angle sweep add up.

Step 3 — What does raising to the $n$ th power do?

WHAT. Multiplying $z$ by itself $n$ times means applying the "stretch-and-turn" rule $n$ times. The length gets multiplied by itself $n$ times, and the angle gets added to itself $n$ times:

$z^{n} = length^{n} r^{n} (cos angle \times n (n θ) + i sin (n θ)) .$

This is De Moivre's Theorem: powering scales the length as a power but only multiplies the angle.

WHY this is the whole game. Our target equation is $z^{n} = 1$ . The number $1$ is an arrow of length $1$ pointing straight right, i.e. angle $0$ . So $z^{n} = 1$ is really two demands at once: the powered length must be $1$ , and the powered angle must point at $1$ .

PICTURE. Start with a short arrow at a small angle; each power stretches it and swings it further round. Three copies of the same starting arrow, powered $1$ , $2$ , $3$ times, fan out.

Step 4 — Force the length: every root sits on the unit circle

WHAT. Match the length demand. We need $r^{n} = 1$ where $r$ is a real length, so $r \geq 0$ .

$r^{n} = 1, r \geq 0 ⟹ r = 1.$

WHY only $r = 1$ ? Among non-negative real numbers, $1$ is the only one whose $n$ th power is $1$ . If $r > 1$ the arrow would blow up when powered; if $0 \leq r < 1$ it would shrink to nothing. Neither lands on length $1$ . So every root of unity has length exactly $1$ .

PICTURE. The dashed cyan circle of radius $1$ — the unit circle — is the only place the roots can live. We have just eliminated the entire rest of the plane.

Recall Why does the length condition kill the whole plane except one circle?

Because $r^{n} = 1$ has exactly one non-negative real solution, $r = 1$ . ::: Every root lies on the unit circle.

Step 5 — Force the angle: careful, angles repeat!

WHAT. Now match the direction demand. The powered arrow must point at $1$ , i.e. at angle $0$ . Naïvely you'd write $n θ = 0$ . But an arrow pointing at angle $2 π$ (a full turn) points at exactly the same spot as angle $0$ . And so does $4 π$ , $6 π$ , $\dots$ . So the honest condition is

$n θ = any whole number of full turns 2 π k, k \in Z .$

Dividing by $n$ :

$θ = \frac{2 π k}{n} .$

WHY the $2 π k$ and not just $0$ ? This is the crux the beginner misses. "Pointing the same way" is not the same as "same angle number." A full extra spin looks identical. If we forgot the $2 π k$ , we'd find only one root ( $θ = 0$ ) and miss all the others. The $k$ is what unlocks the crowd.

PICTURE. All these arrows — at $0$ , at $2 π$ , at $4 π$ — land on the same point $1$ after powering, but before powering they were different starting arrows.

Step 6 — Why exactly $n$ roots (and not more)?

WHAT. We now feed $k = 0, 1, 2, \dots$ into $θ = \frac{2 π k}{n}$ and watch the angles:

$k$	angle $θ = 2 π k / n$	where it points
$0$	$0$	start, at $1$
$1$	$2 π / n$	one step round
$2$	$4 π / n$	two steps round
$⋮$	$⋮$	$⋮$
$n - 1$	$2 π (n - 1) / n$	last new spot
$n$	$2 π$	back to $1$ — repeat!

WHY it stops at $n - 1$ . At $k = n$ the angle is $2 π$ , which is the same direction as $k = 0$ . From there every value of $k$ just re-treads points we already have. So there are exactly $n$ distinct roots — no more, no fewer. This matches the Fundamental Theorem of Algebra: the degree- $n$ polynomial $z^{n} - 1 = 0$ must have exactly $n$ roots, and here they are, all different.

PICTURE. The staircase of angles: each step is the same size $\frac{2 π}{n}$ , and after $n$ steps you have walked exactly one full lap.

Step 7 — The picture snaps into a regular polygon

WHAT. Every root sits at length $1$ (Step 4) and at an angle that is a whole multiple of the same step $\frac{2 π}{n}$ (Step 6). Equal radius, equal angular spacing — that is the exact definition of the vertices of a regular polygon inscribed in the unit circle, with one corner pinned at $1$ .

WHY it must be regular. "Same length" forces every vertex onto the one circle; "same angular step" forces the gaps between neighbours to be identical. Identical gaps + identical radius = perfect symmetry. There is no freedom left to make it lopsided.

PICTURE. For $n = 6$ the six roots are the corners of a regular hexagon; the arrow from the centre to each corner is a root, and each is a $\frac{2 π}{6} = 6 0^{\circ}$ turn from the last.

Step 8 — Degenerate and edge cases (never leave a gap)

WHAT & WHY & PICTURE, three tiny cases the general story must cover:

$n = 1$ . The equation $z^{1} = 1$ has the single root $z_{0} = 1$ . The "polygon" is one lonely point. A degree- $1$ polynomial, one root. Consistent.
$n = 2$ . Roots at angles $0$ and $π$ : that is $1$ and $- 1$ . The "polygon" flattens to a line segment through the centre. Sum $1 + (- 1) = 0$ ✓.
The sum, geometrically. For any $n \geq 2$ the roots are placed with perfect rotational symmetry, so the arrows pull equally in all directions and cancel to the origin. Algebraically this is a Geometric Series: with $ω = e^{i 2 π / n}$ , $1 + ω + \dots + ω^{n - 1} = \frac{ω ^{n} - 1}{ω - 1} = \frac{1 - 1}{ω - 1} = 0,$ since $ω^{n} = 1$ (a full turn) but $ω \neq = 1$ (so we may divide).

The one-picture summary

Everything on this page compressed into a single diagram: the length condition draws the circle, the angle condition drops evenly spaced dots, the dots are a polygon, and the balanced arrows sum to zero.

Recall Feynman retelling — the whole walkthrough in plain words

A complex number is an arrow. Multiplying arrows means multiply their lengths and add their angles — so squaring, cubing, or $n$ -th-powering just spins the arrow faster and stretches it. We want arrows that, after $n$ powers, land exactly on the point " $1$ " (length $1$ , pointing right). Two rules pop out. First, length: only a length of $1$ survives being powered, so every answer sits on the circle of radius $1$ . Second, angle: after $n$ powers the arrow must point right — but "pointing right" also allows a whole number of extra full spins, so before powering the angle could be any of $0, \frac{2 π}{n}, \frac{4 π}{n}, \dots$ Feed those in and you get $n$ evenly spaced dots on the circle. Equal radius plus equal spacing means a regular polygon, one corner nailed at $1$ . And because the dots are perfectly symmetric, the arrows to them tug equally in every direction and add up to the exact centre — zero. That is the whole story.

See also: De Moivre's Theorem · Polar Form of Complex Numbers · Argand Diagram · Geometric Series · Fundamental Theorem of Algebra · Regular Polygons.