Explainer: Cube Roots of Unity

In this explainer, we will learn how to identify the cubic roots of unity using de Moivre’s theorem.

One of the interesting and useful applications of de Moivre’s theorem is finding the roots of unity, by which we mean, for some 𝑛, finding all the complex numbers 𝑧 such that 𝑧=1. The roots of unity play an important role in group theory, number theory, and discrete Fourier transforms. In this explainer, we will focus on the cubic roots of unity. We will begin by using de Moivre’s theorem to find the three cubic roots of one.

Example 1: Cubic Roots of Unity

Find all the values of 𝑧 for which 𝑧=1.

Answer

There are, in fact, multiple methods we could use to solve this. We will demonstrate two. The first is by using de Moivre’s theorem for roots which states that, for a complex number 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin, the 𝑛th roots are given by π‘Ÿο€½ο€½πœƒ+2πœ‹π‘˜π‘›ο‰+π‘–ο€½πœƒ+2πœ‹π‘˜π‘›ο‰ο‰οŽ ο‘ƒcossin for π‘˜=0,1,…,π‘›βˆ’1. We begin by expressing 1 in polar form as 1=1(0+𝑖0).cossin

Therefore, the cubic roots of one are given by √1ο€½ο€½2πœ‹π‘˜3+𝑖2πœ‹π‘˜3=ο€½2πœ‹π‘˜3+𝑖2πœ‹π‘˜3cossincossin

for π‘˜=0,1,2and. Starting with π‘˜=0, we get 1(0+𝑖0)=1cossin. For π‘˜=1, we have cossinο€Ό2πœ‹3+𝑖2πœ‹3.

Finally, for π‘˜=2, we have cossinο€Ό4πœ‹3+𝑖4πœ‹3.

Since the argument of this complex number is not in the range (βˆ’πœ‹,πœ‹], we can subtract 2πœ‹ so that we can express it with its principal argument as cossinο€Όβˆ’2πœ‹3+π‘–ο€Όβˆ’2πœ‹3.

Therefore, the cubic roots of one are 1,ο€Ό2πœ‹3+𝑖2πœ‹3,ο€Όβˆ’2πœ‹3+π‘–ο€Όβˆ’2πœ‹3.cossincossin

We can express these in algebraic form as 1,βˆ’12+√32𝑖,βˆ’12βˆ’βˆš32𝑖.

An alternative way to find the cubic roots of one is by solving the equation π‘§βˆ’1=0. By inspection, we can see that 1 is a trivial solution to this equation. We can, therefore, factor out π‘§βˆ’1 from this equation and solve the resulting quadratic. Firstly, to factor π‘§βˆ’1 from the equation, we can write π‘§βˆ’1=(π‘§βˆ’1)ο€Ήπ‘Žπ‘§+𝑏𝑧+𝑐.

Expanding the right-hand side, we have π‘§βˆ’1=π‘Žπ‘§+𝑏𝑧+π‘π‘§βˆ’π‘Žπ‘§βˆ’π‘π‘§βˆ’π‘=π‘Žπ‘§+(π‘βˆ’π‘Ž)𝑧+(π‘βˆ’π‘)π‘§βˆ’π‘.

Equating coefficients, we immediately see that π‘Ž=1 and 𝑐=1; using this, we quickly see that 𝑏=1 too.

Hence, 0=π‘§βˆ’1=(π‘§βˆ’1)𝑧+𝑧+1.

Using the quadratic formula or otherwise, we can solve 𝑧+𝑧+1 to find 𝑧=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž=βˆ’1±√1βˆ’42=βˆ’12±√32𝑖.

Hence, the three cubic roots of one are 𝑧=1,βˆ’12+√32𝑖,βˆ’12βˆ’βˆš32𝑖.

If we compare these with the answers we got using de Moivre’s theorem, we see that they are the same.

We will now consider the product of the cubic roots of unity.

Example 2: Products of the Cubic Roots of Unity

Let 𝑧=π‘’οŠ§οƒοŽ‘ο‘½οŽ’ and 𝑧=π‘’οŠ¨οŠ±οƒοŽ‘ο‘½οŽ’ be the complex cubic roots of unity.

  1. Evaluate π‘§οŠ¨οŠ§. How does this compare with π‘§οŠ¨?
  2. Evaluate π‘§οŠ¨οŠ¨. How does this compare with π‘§οŠ§?

Answer

Part 1

Using the rules of integer exponents, we can write (𝑧)=𝑒=𝑒=𝑒.οŠ§οŠ¨οƒοŠ¨οŠ¨οƒοƒοŽ‘ο‘½οŽ’οŽ‘ο‘½οŽ’οŽ£ο‘½οŽ’

Subtracting 2πœ‹ from the argument, we can express this in exponential form with its principal argument as 𝑧=𝑒.οŠ¨οŠ§οŠ±οƒοŽ‘ο‘½οŽ’

Hence, we find that 𝑧=π‘§οŠ¨οŠ§οŠ¨.

Part 2

Similarly, using the rules of exponents, we can write (𝑧)=𝑒=𝑒=𝑒.οŠ¨οŠ¨οŠ±οƒοŠ¨οŠ±οŠ¨οƒοŠ±οƒοŽ‘ο‘½οŽ’οŽ‘ο‘½οŽ’οŽ£ο‘½οŽ’

Adding 2πœ‹ from the argument, we can express this in exponential form with its principal argument as 𝑧=𝑒.οŠ¨οŠ¨οƒοŽ‘ο‘½οŽ’

Hence, we find that 𝑧=π‘§οŠ¨οŠ¨οŠ§.

The results from the previous example can be extended to higher powers of π‘§οŠ§. For example, if we are given π‘§οŠοŠ§, we can express 𝑛 as 3π‘˜+π‘Ž, where π‘Ž is 0, 1, or 2. Hence, we can write 𝑧=𝑧=π‘§π‘§π‘Ž=𝑧+𝑧.οŠοŠ§οŠ©ο‡οŠ°οŒΊοŠ§οŠ©ο‡οŠ§οŠ§οŠ©οŠ§ο‡οŒΊοŠ§

By definition, 𝑧=1. Hence, π‘§οŠοŠ§ reduces to π‘§οŒΊοŠ§, where π‘Žβ‰‘π‘›(3)mod. Using the result of the previous example, we can summarize this as follows: for positive 𝑛, 𝑧=1𝑛≑0(3),𝑧𝑛≑1(3),𝑧𝑛≑2(3).formodformodformod

Definition: Cubic Roots of Unity

There are three cubic roots of unity. We often use the symbol πœ” to represent the primitive root which is the root with the smallest strictly positive argument. We can express πœ” in exponential, polar, and algebraic form as follows: πœ”=𝑒=ο€Ό2πœ‹3+𝑖2πœ‹3=βˆ’12+√32𝑖.οƒοŽ‘ο‘½οŽ’cossin

Sometimes πœ” is somewhat ambiguously referred to as β€œthe” cubic root of unity or β€œthe” complex cubic root of unity. We can represent all three cubic roots of unity as 1, πœ”, and πœ”οŠ¨.

As we have seen, the cubic roots of unity form a cycle under multiplication as represented in the figure below.

In the next question, we will explore the properties of the cubic root of unity when raised to negative powers.

Example 3: Cubic Roots of Unity and Negative Powers

Let πœ” be the primitive cubic root of unity.

  1. Find πœ”οŠ±οŠ§. How is this related to the other cubic roots of unity?
  2. Find πœ”οŠ±οŠ¨. How is this related to the other cubic roots of unity?

Answer

Part 1

Expressing πœ” in exponential form and using the properties of integer exponents, we have πœ”=𝑒=𝑒.οŠ±οŠ§οƒοŠ±οŠ§οŠ±οƒοŽ‘ο‘½οŽ’οŽ‘ο‘½οŽ’

Clearly this is equal to πœ”οŠ¨.

Part 2

Similarly, using the exponential form, we have πœ”=𝑒=𝑒=𝑒.οŠ±οŠ¨οƒοŠ±οŠ¨οŠ±οŠ¨οƒοŠ±οƒοŽ‘ο‘½οŽ’οŽ‘ο‘½οŽ’οŽ£ο‘½οŽ’

Since, in this form, the argument is not between βˆ’πœ‹ and πœ‹, we can add 2πœ‹ to represent it in exponential form with the principal argument as πœ”=𝑒,οŠ±οŠ¨οƒοŽ‘ο‘½οŽ’ which we can see is equal to πœ”.

The previous example demonstrates that the cubic roots of unity also form a cycle under division.

We can summarize the properties of the cubic roots of unity under multiplication and division as πœ”=1𝑛≑0(3),πœ”π‘›β‰‘1(3),πœ”π‘›β‰‘2(3),formodformodformod for any π‘›βˆˆβ„€.

Plotting the cube roots of unity on an Argand diagram highlights their symmetry. Furthermore, we observe that πœ”=πœ”.οŠ¨βˆ—

Noticing this will help us as we consider the properties of the cubic roots of unity under addition and subtraction.

Example 4: Sums and Differences of the Cubic Roots of Unity

Let πœ” be the primitive cubic root of unity.

  1. Find πœ”+πœ”οŠ¨.
  2. Find πœ”βˆ’πœ”οŠ¨.
  3. What is πœ”+1 and how is it related to the other roots of unity?
  4. What is πœ”+1 and how is it related to the other roots of unity?

Answer

Part 1

Given that πœ”=πœ”οŠ¨βˆ—, we can rewrite this expression as πœ”+πœ”=πœ”+πœ”.οŠ¨βˆ—

From the properties of complex conjugates, we have πœ”+πœ”=2(πœ”).Re

Using the algebraic form πœ”=βˆ’12+√32𝑖, we see πœ”+πœ”=βˆ’1.

An alternative way we could have derived this result is by noting that πœ”βˆ’1=0. Hence, 0=(πœ”βˆ’1)ο€Ήπœ”+πœ”+1.

Since πœ”βˆ’1β‰ 0, we know that πœ”+πœ”+1=0. Hence, πœ”+πœ”=βˆ’1.

Part 2

Once again, using the fact that πœ”=πœ”οŠ¨βˆ—, we can rewrite the expression as πœ”βˆ’πœ”=πœ”βˆ’πœ”.οŠ¨βˆ—

Using the properties of conjugation, we have πœ”βˆ’πœ”=2𝑖(πœ”).Im

Considering the algebraic form πœ”=βˆ’12+√32𝑖, we see πœ”βˆ’πœ”=π‘–βˆš3.

Part 3

Using the algebraic form of πœ”, we can write πœ”+1=βˆ’12+√32𝑖+1=12+√32𝑖.

To see how this is related to the cubic roots of unity, we first notice that both πœ” and πœ”οŠ¨ have negative real parts. Hence, if we rewrite this as πœ”+1=βˆ’ο€Ώβˆ’12βˆ’βˆš32𝑖,=12+√32𝑖, we can see that πœ”+1=βˆ’πœ”οŠ¨. We could have also derived this by rearranging the formula πœ”+πœ”+1=0; this is the approach we will take to answer the next part of the question.

Part 4

Given the fact that πœ”+πœ”+1=0, by subtracting πœ” from both sides, we get πœ”+1=βˆ’πœ”.

Hence, πœ”+1=12βˆ’βˆš32𝑖.

The previous example demonstrated a number of properties of the cubic roots of unity which we summarize below.

Properties of the Cubic Roots of Unity

The cubic roots of unity have the following properties:

  1. 1+πœ”+πœ”=0,
  2. πœ”=πœ”οŠ¨βˆ—,
  3. πœ”βˆ’πœ”=π‘–βˆš3.

We will now consider some examples which will demonstrate how we can use the properties of the cubic roots of unity to solve problems.

Example 5: Using the Properties of the Cubic Roots of Unity

Evaluate (πœ”βˆ’πœ”)οŠͺ where πœ” is a complex cube root of unity.

Answer

Using the properties of the cubic roots of unity, we know that πœ”βˆ’πœ”=π‘–βˆš3.

Hence, ο€Ήπœ”βˆ’πœ”ο…=ο€»π‘–βˆš3.οŠͺοŠͺ

Using the properties of integer indices, we have ο€Ήπœ”βˆ’πœ”ο…=ο€½ο€»π‘–βˆš3=(βˆ’3)=βˆ’2,187.οŠͺ

Example 6: Simplifying Expressions Using the Properties of the Cubic Roots of Unity

Evaluate ο€Όβˆ’8πœ”+1οˆο€Ό5+5πœ”+1πœ”οˆ where πœ” is a complex cube root of unity.

Answer

Let us consider some of the properties of the cubic roots of unity that might help us simplify this expression. We notice that in the denominator of the fraction in the first set of parenthesis is the term πœ”+1. From the properties of the cubic roots of unity, we know that πœ”+1=βˆ’πœ”οŠ¨. We also notice that, in the second set of parenthesis, we have 5+5πœ”, which we can also simplify using the same result. Hence, we can rewrite the expression as ο€Όβˆ’8πœ”+1οˆο€Ό5+5πœ”+1πœ”οˆ=ο€Όβˆ’8βˆ’πœ”οˆο€Όβˆ’5πœ”+1πœ”οˆ=ο€Ό8πœ”οˆο€Όβˆ’5πœ”+1πœ”οˆ.

We can now use the properties of the powers of the cubic roots of unity to replace 1πœ” with πœ”οŠ¨ and 1πœ”οŠ¨ with πœ”.

Hence, ο€Όβˆ’8πœ”+1οˆο€Ό5+5πœ”+1πœ”οˆ=(8πœ”)ο€Ήβˆ’5πœ”+πœ”ο…=(8πœ”)ο€Ήβˆ’4πœ”ο…=βˆ’32πœ”.

Since πœ”=1, we have ο€Όβˆ’8πœ”+1οˆο€Ό5+5πœ”+1πœ”οˆ=βˆ’32.

Example 7: Simplifying Expressions Using the Properties of the Cubic Roots of Unity

Evaluate ο€Ή9βˆ’πœ”+9πœ”ο…+ο€Ή6+6πœ”+6πœ”ο…οŠ¨οŠͺοŠͺ.

Answer

We begin by replacing higher powers of πœ” with their equivalent power between 0 and 2. Hence, replacing πœ”οŠͺ with πœ”, we have ο€Ή9βˆ’πœ”+9πœ”ο…+ο€Ή6+6πœ”+6πœ”ο…=ο€Ή9βˆ’πœ”+9πœ”ο…+ο€Ή6+6πœ”+6πœ”ο….οŠͺοŠͺ

Factorizing out the constants in each set of parenthesis gives ο€Ή9βˆ’πœ”+9πœ”ο…+(6+6πœ”+6πœ”ο…=ο€Ή9(1+πœ”)βˆ’πœ”ο…+ο€Ή6(1+πœ”+πœ”).οŠͺοŠͺ

Using the properties of the cubic roots of unity, 1+πœ”=βˆ’πœ”,1+πœ”+πœ”=0, we can rewrite this as ο€Ή9βˆ’πœ”+9πœ”ο…+ο€Ή6+6πœ”+6πœ”ο…=ο€Ή9ο€Ήβˆ’πœ”ο…βˆ’πœ”ο…+(6(0))=100πœ”.οŠͺοŠͺοŠͺ

Since πœ”=πœ”οŠͺ, finally, we can state that ο€Ή9βˆ’πœ”+9πœ”ο…+ο€Ή6+6πœ”+6πœ”ο…=100πœ”.οŠͺοŠͺ

Many of the properties of the cubic roots of unity have their analogues in the more general case of the 𝑛th roots of unity.

Key Points

  1. There are three cubic roots of unity which we denote 1, πœ”, and πœ”οŠ¨, where πœ” is referred to as the primitive cubic root of unity. We can express it in exponential, polar, and algebraic forms as πœ”=𝑒=ο€Ό2πœ‹3+𝑖2πœ‹2=βˆ’12+√32𝑖.οƒοŽ‘ο‘½οŽ’cossin
  2. Both the positive and the negative powers of πœ” form a closed cycle of three elements, which we can summarize as πœ”=1,πœ”=πœ”,πœ”=πœ”, for any π‘›βˆˆβ„€.
  3. The cubic roots of unity also have the following properties:
    1. 1+πœ”+πœ”=0,
    2. πœ”=πœ”οŠ¨βˆ—,
    3. πœ”βˆ’πœ”=π‘–βˆš3.
  4. We can use the properties of the cubic roots of unity to simplify complex looking expressions.

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