Geometric Structure of the nth Roots of Unity

Hello! Last lesson, you found every solution of $z^n=w$ and saw that the solutions share a radius and are equally spaced. Now we will examine the case $w=1$. These roots of unity do more than make a regular polygon: their positions let you recognise powers, sums and distances without calculating every root from scratch.

The goal is to move fluently between the diagram and the algebra. That is particularly useful when an exam question gives you a root $\omega$ but does not tell you to draw a diagram.

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The polygon determined by $z^n=1$

The previous lesson’s formula applies with modulus $1$ and argument $0$. For a positive integer $n$, the $n$th roots of unity are

$$z_k=\mathrm{cis}\left(\frac{2k\pi }{n}\right),k=0,1,\dots ,n-1.$$

They lie on the unit circle, starting at $z_0=1$, with an angle of $2\pi /n$ between consecutive points. For $n\ge 3$, they are the vertices of a regular $n$-gon centred at the origin. For $n=2$, they are the opposite points $1$ and $-1$.

Watch the roots-of-unity portion of Roots of Complex Numbers | HSC Extension 2 Mathematics from mcgrathematics. It begins with the fourth roots, then connects their diagram to factorising $z^n-1$.

The image below depicts eight equally spaced points, so its particular polygon has angular spacing $\pi /4$. The formula $2\pi /n$ gives the spacing for other values of $n$.

The picture includes $i$ and $-i$ because eight is divisible by four; they are not present among the roots for every $n$. Likewise, $-1$ is a root exactly when $n$ is even: it sits halfway through the list of vertices, at angle $\pi$.

For a concise written derivation of the general result, read the end of “Roots of Unity” and Theorem A.3, “$n$th Roots of Unity,” in 12.A: Complex Numbers from Mathematics LibreTexts. The text uses $e^{i\theta }$ where we have been writing $\mathrm{cis}\theta$; here they denote the same point on the unit circle.

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One rotation generates every vertex

For $n\ge 2$, give the first root after $1$ a name:

$$\omega =\mathrm{cis}\left(\frac{2\pi }{n}\right).$$

Multiplying by $\omega$ rotates a point anticlockwise by $2\pi /n$ without changing its distance from the origin. Thus the complete set of roots is

$$1,\omega ,{\omega }^2,\dots ,{\omega }^{n-1}.$$

After $n$ such rotations, ${\omega }^n=1$. Powers therefore repeat every $n$ steps; for example, ${\omega }^{n+2}={\omega }^2$. On the diagram, the exponent tells you how many vertex positions to move from $1$, counting anticlockwise.

There is an important qualification in exam wording. Not every root of unity other than $1$ generates all $n$ roots. The choice $\omega =\mathrm{cis}(2\pi /n)$ does, because its first $n$ powers occupy distinct vertices. By contrast, $-1$ is a sixth root of unity, but its powers alternate between only $1$ and $-1$. If a question says merely that ${\omega }^n=1$, do not assume its powers list all $n$ roots unless it also identifies $\omega$ as a primitive $n$th root, or gives equivalent information.

The polygon also makes conjugates easy to locate. Reflection in the real axis takes the vertex at angle $2k\pi /n$ to the vertex at angle $-2k\pi /n$. Hence, for $1\le k<n$,

$$\overline{{\omega }^k}={\omega }^{n-k}.$$

Because every vertex has modulus $1$, its conjugate is also its reciprocal:

$$\overline{{\omega }^k}=\frac{1}{{\omega }^k}.$$

These identities are useful checks. If your list of roots of $z^n=1$ contains a non-real point but omits its reflection, the list is incomplete.

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Symmetry becomes an algebraic shortcut

Picture the $n$ position vectors from the origin to all the vertices. When $n\ge 2$, their symmetry suggests that their total is zero. We can prove it without relying on a picture, using the chosen $\omega =\mathrm{cis}(2\pi /n)$.

Let

$$S=1+\omega +{\omega }^2+\cdots +{\omega }^{n-1}.$$

Multiplying by $\omega$ shifts every term one place around the polygon. Subtracting the two expressions gives

$$(1-\omega )S=1-{\omega }^n=0.$$

Since $n\ge 2$, our chosen $\omega \ne 1$, so

$$\boxed{1+\omega +{\omega }^2+\cdots +{\omega }^{n-1}=0.}$$

This is both a geometric result and the finite geometric-series identity. It also explains why you must check the condition $\omega \ne 1$ before dividing by $1-\omega$.

For the sixth roots, the full sum is zero, but there is more structure. The vertices with even exponents,

$$1,{\omega }^2,{\omega }^4,$$

are $120^{\circ }$ apart: they form the cube roots of unity. Consequently,

$$1+{\omega }^2+{\omega }^4=0.$$

The three vertices with odd exponents are the same triangle rotated by $60^{\circ }$. Their sum is therefore also zero:

$$\omega +{\omega }^3+{\omega }^5=\omega (1+{\omega }^2+{\omega }^4)=0.$$

This is a general problem-solving move: before expanding a sum of powers, ask whether the selected vertices themselves form a smaller regular polygon.

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Using the structure in an unfamiliar question

Suppose $\omega =\mathrm{cis}(\pi /3)$, so its powers are the six sixth roots of unity. Which of them solve $z^3=-1$?

First, any solution of $z^3=-1$ satisfies $z^6=1$, so it must appear somewhere in our six-vertex list. Also,

$$({\omega }^k{)}^3=({\omega }^3{)}^k=(-1{)}^k.$$

We therefore want the odd exponents. The complete answer is

$$\boxed{z=\omega ,{\omega }^3,{\omega }^5.}$$

On the Argand diagram these are alternate vertices of the hexagon. In Cartesian form they are

$$\frac{1}{2}+\frac{\sqrt{3}}{2}i,-1,\frac{1}{2}-\frac{\sqrt{3}}{2}i.$$

Notice what made the argument complete: showing that every candidate is a sixth root of unity, then selecting exactly the vertices whose cubes equal $-1$. Merely spotting three plausible points on a diagram would not establish that none were missed.

The same diagram can answer a distance question. What is the distance from $1$ to ${\omega }^2$? The points have arguments $0$ and $2\pi /3$, so their position vectors form an isosceles triangle with two sides of length $1$. Equivalently, use their coordinates:

$${\omega }^2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i,|1-{\omega }^2|=\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{3}.$$

The expression $|1-{\omega }^2|$ is a chord length on the unit circle, not the length of an arc. For any two roots, the modulus of their difference measures the straight-line distance between their vertices. This gives you a geometric interpretation before you choose whether coordinates or trigonometry will be quicker.

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Key takeaways

The $n$th roots of unity are the points $\mathrm{cis}(2k\pi /n)$ for $k=0,\dots ,n-1$: equally spaced points on the unit circle, beginning at $1$. Choosing $\omega =\mathrm{cis}(2\pi /n)$ lets its powers name the vertices in order. The diagram then reveals conjugate pairs, smaller polygons within the full set, and distances; identities such as ${\omega }^n=1$ and the zero sum turn those observations into concise proofs.

Next, you will use moduli of complex expressions to describe geometric loci. The chord distance $|1-{\omega }^2|$ is a first glimpse of that idea: a modulus can specify a distance between two points on an Argand diagram.