Cube Root of Unity - Properties, Values, Examples

What Are the Cube Roots of Unity?

A cube root of unity is any number $x$ that satisfies $x^3 = 1$. Because $x^3 = 1$ is a degree-3 equation, it has exactly three solutions: one real and two complex. Written out, they are

$$1, \quad \omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2},$$

where $i = \sqrt{-1}$ is the imaginary unit and the Greek letter ω (omega) is the standard name for the first complex root. The bare cube root of 1 that a calculator returns is only the real one; the two complex partners are what make the topic worth a chapter.

The Number 1 Has Three Cube Roots, Not One

Ask most students for the cube root of 1 and they answer "1" - which is true, and also only a third of the story. The equation $x^3 = 1$ is a cubic, and a cubic has three roots. Two of them are complex, they are called the cube roots of unity, and they run quietly underneath everything from the Fast Fourier Transform to the way a JPEG is compressed.

The three roots sit evenly spaced around the unit circle, 120° apart - the corners of an equilateral triangle. That even spacing is the whole idea, and every property below falls out of it.

What Are the Values of ω and ω²?

The two complex roots are not independent - ω² is literally the square of ω, which is why we write them as $\omega$ and $\omega^2$ rather than inventing a second letter. Both are complex conjugates of each other, sitting symmetrically above and below the real axis:

$$\omega = \frac{-1 + i\sqrt{3}}{2}, \qquad \omega^2 = \frac{-1 - i\sqrt{3}}{2}.$$

Squaring the first gives the second, and cubing either one returns to 1. Each also has modulus 1, meaning $|\omega| = |\omega^2| = 1$ - they live exactly on the unit circle in the Argand plane, which is why the modulus of a complex number here is the clean value 1.

How Do You Derive the Cube Roots of Unity?

Solving $x^3 = 1$ from scratch shows exactly where ω comes from — this is the derivation, not a rule to memorise. Start by moving everything to one side and factoring the difference of cubes:

$$x^3 - 1 = 0$$ $$(x - 1)(x^2 + x + 1) = 0$$

The first factor gives the obvious root:

$$x - 1 = 0 \implies x = 1.$$

The second factor is a quadratic, and its two roots are the complex ones. Applying the quadratic formula to $x^2 + x + 1 = 0$ with $a = 1$, $b = 1$, $c = 1$:

$$x = \frac{-1 \pm \sqrt{1 - 4}}{2}$$ $$x = \frac{-1 \pm \sqrt{-3}}{2}$$ $$x = \frac{-1 \pm i\sqrt{3}}{2}.$$

Those two values are ω and ω². The quadratic $x^2 + x + 1$ is worth remembering on its own, because "$1 + \omega + \omega^2 = 0$" is just the statement that ω and ω² are its roots of a quadratic equation whose coefficient of $x$ is 1.

What Are the Properties of Cube Roots of Unity?

Every property below is a consequence of the two facts $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$.

• Their sum is zero. $1 + \omega + \omega^2 = 0$. This is the single most-used identity - it lets you replace $1 + \omega$ with $-\omega^2$, or $\omega + \omega^2$ with $-1$, on sight.

• Their product is one. $1 \times \omega \times \omega^2 = \omega^3 = 1$.

• Cubing any root gives 1. $1^3 = 1$, $\omega^3 = 1$, and $(\omega^2)^3 = \omega^6 = 1$.

• Powers cycle with period 3. $\omega^3 = 1$, so any power reduces by taking the exponent modulo 3: $\omega^4 = \omega$, $\omega^5 = \omega^2$, $\omega^6 = 1$, and so on.

• The complex roots are reciprocals and conjugates. $\omega^2 = \dfrac{1}{\omega} = \overline{\omega}$ - squaring one gives the other, and so does taking its reciprocal or its complex conjugate.

Examples Of Cube Root Of Unity

The set runs from confirming ω is genuinely a cube root, through the most common power-reduction mistake, to products, factoring, and finding all cube roots of a non-unit number.

Example 1

Show that $\omega = \dfrac{-1 + i\sqrt{3}}{2}$ satisfies $\omega^2 + \omega + 1 = 0$.

Square ω first:

$$\omega^2 = \left(\frac{-1 + i\sqrt{3}}{2}\right)^2 = \frac{1 - 2i\sqrt{3} + (i\sqrt{3})^2}{4} = \frac{1 - 2i\sqrt{3} - 3}{4} = \frac{-2 - 2i\sqrt{3}}{4} = \frac{-1 - i\sqrt{3}}{2}.$$

Now add ω and 1:

$$\omega^2 + \omega + 1 = \frac{-1 - i\sqrt{3}}{2} + \frac{-1 + i\sqrt{3}}{2} + 1 = \frac{-2}{2} + 1 = -1 + 1 = 0.$$

Final answer: the identity holds, and notice $\omega^2$ came out equal to the second listed root — squaring ω is ω².

Example 2

Evaluate $\omega^{10} + \omega^{5} + 1$.

Wrong attempt. A student substitutes the surd value of ω and tries to raise it to the 10th power directly - expanding $\left(\frac{-1 + i\sqrt{3}}{2}\right)^{10}$ term by term. Several lines of messy binomial expansion later, the arithmetic collapses and the answer is anyone's guess.

The trouble is ignoring the one property that makes ω pleasant: $\omega^3 = 1$. High powers should be reduced, not expanded.

Correct. Reduce each exponent modulo 3:

$$\omega^{10} = \omega^{9}\cdot\omega = (\omega^3)^3\cdot\omega = 1\cdot\omega = \omega$$ $$\omega^{5} = \omega^{3}\cdot\omega^{2} = 1\cdot\omega^2 = \omega^2$$

So the expression becomes

$$\omega^{10} + \omega^{5} + 1 = \omega + \omega^2 + 1 = 0.$$

Final answer: $\omega^{10} + \omega^{5} + 1 = 0$. Reducing the powers turned a page of algebra into one line.

Example 3

Find the value of $(1 + \omega)(1 + \omega^2)$.

The first-instinct error is to expand as $1 + \omega^2 + \omega + \omega^3$ and stop — forgetting that $\omega^3$ collapses to 1. Push through instead. Since $1 + \omega + \omega^2 = 0$, we have $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$. So

$$(1 + \omega)(1 + \omega^2) = (-\omega^2)(-\omega) = \omega^3 = 1.$$

Final answer: $(1 + \omega)(1 + \omega^2) = 1$.

Example 4

Verify that 1, ω, and ω² are exactly the roots of $x^3 - 1 = 0$.

Factor the cubic:

$$x^3 - 1 = (x - 1)(x^2 + x + 1).$$

The factor $x - 1$ gives $x = 1$. The factor $x^2 + x + 1$ has roots ω and ω² (from the derivation above). A cubic has three roots and we have named three distinct values, so the list is complete.

Final answer: the three roots are $1, \omega, \omega^2$ - no more, no fewer.

Example 5

Evaluate $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$.

Use $1 + \omega + \omega^2 = 0$ to simplify each bracket. From $1 + \omega^2 = -\omega$, the first bracket is $1 + \omega^2 - \omega = -\omega - \omega = -2\omega$. From $1 + \omega = -\omega^2$, the second bracket is $1 + \omega - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$. Multiply:

$$(-2\omega)(-2\omega^2) = 4,\omega^3 = 4.$$

Final answer: the product is 4.

Example 6

Find all three cube roots of 8.

Every number has three cube roots, spaced 120° apart. One real cube root of 8 is 2. The other two are that real root multiplied by ω and ω²:

$$\sqrt[3]{8} = 2, \quad 2\omega, \quad 2\omega^2.$$

Writing the complex ones out:

$$2\omega = 2\cdot\frac{-1 + i\sqrt{3}}{2} = -1 + i\sqrt{3}, \qquad 2\omega^2 = -1 - i\sqrt{3}.$$

Final answer: the three cube roots of 8 are $2$, $-1 + i\sqrt{3}$, and $-1 - i\sqrt{3}$. The same trick finds all cube roots of any number: take one root, then multiply by ω and ω².

Why Do the Cube Roots of Unity Matter?

"Why should a number other than 1 cube back to 1?"

The cube roots of unity are the smallest case of a much bigger object - the nth roots of unity, the $n$ evenly spaced points on the unit circle whose $n$th power is 1. Those points are given directly by De Moivre's theorem, and they turn out to be the engine of one of the most-used algorithms ever written.

Where roots of unity do real work:

• Signal processing. The Discrete Fourier Transform and its fast cousin, the FFT, run entirely on roots of unity - the algorithm behind digital audio, Wi-Fi, and image compression. When a phone plays an MP3 or a browser decodes a JPEG, roots of unity are doing the arithmetic.

• Solving higher equations. Any pure equation $x^n = a$ is solved by taking one root and multiplying by the $n$th roots of unity, exactly as in Example 6.

• Symmetry and geometry. The roots always form a regular polygon inscribed in the unit circle - three roots give an equilateral triangle, four give a square, and so on.

A student who first sees ω as "the number that turns $x^3 - 1$ into three tidy pieces" stops treating the complex roots as exotic. They are just the other two corners of a triangle that was always there.

Common Mistakes

Mistake 1: Not reducing powers of ω modulo 3

Where it slips in: Any expression with a power of ω above 2 - $\omega^7$, $\omega^{100}$, $\omega^{2026}$.

Don't do this: Try to compute the high power directly from the surd value of ω.

The correct way: Use $\omega^3 = 1$ and reduce the exponent modulo 3. The exact first-instinct error is reaching for the messy complex value instead of the period-3 rule - once a student writes $\omega^{2026} = \omega^{2026 \bmod 3} = \omega^{1} = \omega$, the fear of large exponents disappears.

Mistake 2: Swapping which root is ω and which is ω²

Where it slips in: Writing out the two complex values from memory.

Don't do this: Attach the $+i\sqrt{3}$ and $-i\sqrt{3}$ signs at random.

The correct way: By convention $\omega = \dfrac{-1 + i\sqrt{3}}{2}$ (the one above the real axis), and its conjugate $\omega^2 = \dfrac{-1 - i\sqrt{3}}{2}$ sits below. The second-guesser who redoes the assignment mid-problem often flips a sign and breaks their own working; fix ω once at the start and leave it.

Mistake 3: Thinking 1 is the only cube root of 1

Where it slips in: The very first line - reading "cube root of unity" as a single number.

Don't do this: Answer "1" and stop.

The correct way: $x^3 = 1$ is a cubic, so it has three roots. Two are complex. Missing them is the same error as saying $x^2 = 4$ has only the root $x = 2$ and forgetting $x = -2$.

Conclusion

• The cube roots of unity are $1$, ω, and ω² - the three solutions of $x^3 = 1$, one real and two complex.

• The two identities $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$ generate every other property.

• High powers of ω reduce by taking the exponent modulo 3 - this is the single most useful working habit.

• The three roots are the vertices of an equilateral triangle on the unit circle, and the smallest case of the nth roots of unity behind the FFT.

• The most common mistake is treating 1 as the only cube root of unity - a cubic has three roots.

To go further with a teacher, explore Bhanzu's algebra tutor or a high school math tutor, or join live math classes online built around this reasoning-first approach.

Practice These To Solidify Your Understanding

1. Evaluate $\omega^{22} + \omega^{11} + 1$.

2. Show that $(1 + \omega - \omega^2)^3 = -8$.

3. Find all three cube roots of 27.

Answer to Question 1: $\omega^{22} = \omega^{1} = \omega$ and $\omega^{11} = \omega^{2}$, so the sum is $\omega + \omega^2 + 1 = 0$. Answer to Question 2: $1 + \omega = -\omega^2$, so $1 + \omega - \omega^2 = -2\omega^2$, and $(-2\omega^2)^3 = -8,\omega^6 = -8$. Answer to Question 3: $3, 3\omega, 3\omega^2$, i.e. $3$, $\frac{-3 + 3i\sqrt{3}}{2}$, $\frac{-3 - 3i\sqrt{3}}{2}$. If Question 1 gave anything but 0, revisit Mistake 1.

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Read More

• Cube root of 1 — the real cube root and how it extends to the complex roots.

• Cube root of 27 — a worked perfect-cube example.

• Cube root of 64 — another perfect cube, computed step by step.

• Quadratic equations — the $x^2 + x + 1 = 0$ factor that produces ω and ω².

• Cubic polynomials — why a degree-3 equation always has three roots.