Explainer: The 𝑛th Roots of Unity

In this explainer, we will learn how to use De Moivre’s theorem to find the 𝑛th roots of unity and explore their properties.

One of the interesting and useful applications of De Moivre’s theorem is finding the roots of unit, which means that for some 𝑛, we find 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 explore some of the properties of these roots.

We begin with an example to demonstrate how we can apply De Moivre’s theorem to find the 𝑛th roots of unity.

Example 1: The 𝑛th Roots of Unity

Write a general form for the roots 𝑧=1, giving your answer in polar form.

Answer

De Moivre’s theorem for roots 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 𝑛th roots of unity are given by cossinο€½2πœ‹π‘˜π‘›ο‰+𝑖2πœ‹π‘˜π‘›ο‰, for π‘˜=0,1,…,π‘›βˆ’1. We can express these in exponential form as π‘’οŽ‘ο‘½ο‘€ο‘ƒοƒ, where π‘˜=0,1,…,π‘›βˆ’1.

The solutions to the equation 𝑧=1 are called the 𝑛th roots of unity. We call a root primitive if it is not an π‘šth root of unity for some π‘š<𝑛. Given any primitive 𝑛th root of unity, we can generate the other 𝑛th roots by raising it to the π‘˜th power. Hence, if we take the primitive root to be πœ”=π‘’οŽ‘ο‘½ο‘ƒοƒ, we can write the set of all 𝑛th roots of unity in the form 1,πœ”,πœ”,πœ”,…,πœ”.οŠ¨οŠ©ο‡οŠ±οŠ§

One of the first interesting things we note about the 𝑛th roots of unity is that they are all evenly spaced around the unit circle centered at the origin, which means that they form a regular 𝑛-gon with one vertex at 𝑧=1. The figures below show 𝑛th roots of unity plotted on an Argand diagram for different values of 𝑛.

Example 2: The Sum of the 𝑛th Roots of Unity

  1. Find the quintic roots of unity.
  2. What is the value of their sum?

Answer

Part 1

To find the quintic roots of unity we can use the general form for the 𝑛th roots of unity: π‘’οŽ‘ο‘½ο‘€ο‘ƒοƒ where π‘˜=0,1,…,π‘›βˆ’1. Hence, for 𝑛=5, we have the following roots: 1,𝑒,𝑒,𝑒,𝑒.οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽ₯ο‘½οŽ€οŽ§ο‘½οŽ€οƒοƒοƒοƒ

Rewriting last two of these roots so that their arguments are given in the principal range, we have 1,𝑒,𝑒,𝑒,𝑒.οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽ£ο‘½οŽ€οŽ‘ο‘½οŽ€οƒοƒοŠ±οƒοŠ±οƒ

Part 2

To find the sum of the roots, we can use the fact that they satisfy the equation 𝑧=1. Subtracting 1 from both sides, we have 0=π‘§βˆ’1.

Factorizing out π‘§βˆ’1, we have 0=(π‘§βˆ’1)𝑧+𝑧+𝑧+𝑧+1.οŠͺ

Setting 𝑧=π‘’οŽ‘ο‘½οŽ€οƒ, we see that π‘§βˆ’1β‰ 0; hence, 𝑧+𝑧+𝑧+𝑧+1=0οŠͺ. However, since 𝑧 is a primitive root, 𝑧+𝑧+𝑧+𝑧+1οŠͺ is the sum of all the roots. Hence, the sum of the five 5th roots of unity is zero.

The result from the last example generalizes to state that the sum of the 𝑛th roots of unity for any 𝑛>1 is zero. Equivalently, we can state that if πœ” is a primitive 𝑛th root of unity for some integer 𝑛>1, then 0=ο„šπœ”.οŠοŠ±οŠ§ο‡οŠ²οŠ¦ο‡

This result can be proved in a similar way to the method used in example 2. Alternatively, we can appeal to Vieta’s formulas, in particular, that for a general polynomial 𝑃(π‘₯)=π‘Žπ‘₯+π‘Žπ‘₯+β‹―+π‘Žπ‘₯+π‘Ž, where π‘Žβ‰ 0, the sum of its 𝑛 roots equals βˆ’π‘Žπ‘ŽοŠοŠ±οŠ§οŠ. In the case of the 𝑛th roots of unity, we have the polynomial π‘§βˆ’1. Therefore, π‘Ž=0 and consequently the sum of the roots is zero.

Example 3: Reciprocals of the 𝑛th Roots of Unity

Let πœ” be an 𝑛th root of unity.

  1. Which of the following is the correct relationship between πœ”οŠ±οŠ§ and πœ”?
    1. πœ”=πœ”οŠ±οŠ§
    2. πœ”=βˆ’πœ”οŠ±οŠ§
    3. πœ”=βˆ’(πœ”)οŠ±οŠ§βˆ—
    4. πœ”=(πœ”)οŠ±οŠ§βˆ—
  2. Express πœ”οŠ±οŠ§ in terms of positive powers of πœ”.

Answer

Part 1

We begin by expressing πœ” in exponential form as πœ”=𝑒, where πœƒ=2π‘˜πœ‹π‘› for some integer π‘˜. We can now consider πœ”=𝑒.οŠ±οŠ§οŠ±οƒοΌ

Recalling the properties of the complex conjugate, we can rewrite this as πœ”=𝑒=πœ”.οŠ±οŠ§οƒοΌβˆ—βˆ—

Hence, the correct answer is (d).

Part 2

Once again, we consider the exponential form of πœ”=𝑒, where πœƒ=2π‘˜πœ‹π‘› for some integer π‘˜. Using this, we can express πœ”=𝑒.οŠ±οŠ§οŠ±οƒοΌ

Since πœ” is an 𝑛th root of unity, we know that π‘›πœƒ is some multiple of 2πœ‹. Hence, we can add π‘›πœƒ to the argument without changing the value of the complex number. Therefore, πœ”=𝑒=𝑒.οŠ±οŠ§οƒ()()

Using the rules of integer exponent, we can rewrite this as πœ”=𝑒=πœ”.οŠ±οŠ§οƒοΌοŠοŠ±οŠ§οŠοŠ±οŠ§

The previous example demonstrated another one of the key properties of the 𝑛th roots of unity: they are closed under division. Hence, the 𝑛th roots of unity form a group under multiplication and division. Furthermore, an alternative way to represent the properties demonstrated in the previous example is as follows. If we think of the 𝑛th roots of unity as powers of a primitive root πœ”, then 1πœ”=πœ”=πœ”=ο€Ήπœ”ο….ο‡οŠ±ο‡οŠοŠ±ο‡ο‡βˆ—

In the next example, we look at how the 𝑛th roots of unity are related for different values of 𝑛.

Example 4: Relationship between the 𝑛th Roots of Unity for Different Values of 𝑛

  1. Find the cube roots of unity.
  2. Find the solutions to 𝑧=1.
  3. What is the relationship of the cubic roots of unity to the 6th roots of unity?

Answer

Part 1

Using De Moivre’s theorem, we can find the cube roots of unity by considering cossinο€½2πœ‹π‘˜3+𝑖2πœ‹π‘˜3, for π‘˜=0,1, and 3. Therefore, evaluating this expression for the three values of π‘˜ and expressing them with their arguments in the principal range, we find that the three cubic roots of unity are 1,ο€Ό2πœ‹3+𝑖2πœ‹3,ο€Όβˆ’2πœ‹3+π‘–ο€Όβˆ’2πœ‹3.cossincossin

Part 2

In a similar way, we can find the 6th roots of unity by considering π‘˜=0,1,…,5 for cossinο€½2πœ‹π‘˜6+𝑖2πœ‹π‘˜6.

Hence, expressing them with arguments in the principal range, we have 1,ο€Ό2πœ‹6+𝑖2πœ‹6,ο€Ό2πœ‹3+𝑖2πœ‹3,βˆ’1,ο€Όβˆ’2πœ‹3+π‘–ο€Όβˆ’2πœ‹3,ο€Όβˆ’2πœ‹6+π‘–ο€Όβˆ’2πœ‹6.cossincossincossincossin

Part 3

Comparing the cube roots of unity to the 6th roots of unity we find that all the cubic roots of unity are also 6th roots of unity. This should not be surprising because given the equation 𝑧=1, we can square both sides to get 𝑧=1. Hence, we have some solution to the first equation and it will clearly also be a solution to the second equation.

The previous example demonstrated a special case of the following property: if 𝑛=π‘šπ‘, then all of the π‘šth roots of unity are also 𝑛th roots of unity. Similarly, we can state that all of the 𝑝th roots of unity are 𝑛th roots. We can further generalize this result as stated in the box below.

Common Roots of Unity

The common roots of π‘§βˆ’1=0 and π‘§βˆ’1=0 are the roots of π‘§βˆ’1=0 where 𝑑=(π‘š,𝑛)gcd.

This can be proved by considering the constraints this places on the argument of the common roots. We will not present the proof here, although it can be a nice exercise to help a student gain confidence working with the 𝑛th roots of unity. Furthermore, this theorem gives an indication of some of the connections that exist between the 𝑛th root of unity and the number theory.

We will finish by looking at a couple of examples where the solutions require application of the properties of the 𝑛th roots of unity.

Example 5: Applications of the 𝑛th Roots of Unity

For how many pairs of real numbers (π‘Ž,𝑏) does the relation (π‘Ž+𝑏𝑖)=π‘Žβˆ’π‘π‘–οŠ¨οŽ•οŠ¦οŠ¨οŠ¦ hold?

Answer

Letting 𝑧=π‘Ž+𝑏𝑖, we can rewrite the equation in the form 𝑧=π‘§οŠ¨οŽ•οŠ¦οŠ¨οŠ¦βˆ—. Taking the modulus of this equation we have ||𝑧||=|𝑧|.οŠ¨οŽ•οŠ¦οŠ¨οŠ¦βˆ—

Using the properties of the argument that |𝑧|=|𝑧| and that |𝑧|=|𝑧|βˆ—, we have |𝑧|=|𝑧|.οŠ¨οŽ•οŠ¦οŠ¨οŠ¦

Hence, subtracting |𝑧| from both sides of the equation, we have |𝑧|βˆ’|𝑧|=0,|𝑧|ο€Ί|𝑧|βˆ’1=0.οŠ¨οŽ•οŠ¦οŠ¨οŠ¦οŠ¨οŽ•οŠ¦οŠ§οŠ―

Therefore, either |𝑧|=0 or |𝑧|=1. If |𝑧|=0, π‘Ž and 𝑏 are both zero, so we have one pair of real numbers which satisfy the equation in this case. Now we consider the case where |𝑧|=1. Multiplying both sides of our original equation by 𝑧 gives 𝑧=𝑧𝑧.οŠ¨οŽ•οŠ¦οŠ¨οŠ§βˆ—

Since 𝑧𝑧=|𝑧|=1βˆ—οŠ¨, we can rewrite this as 𝑧=1.οŠ¨οŽ•οŠ¦οŠ¨οŠ§

Any 2,021st root of unity satisfies this equation. Furthermore, we know there are 2,021 unique roots. Therefore, in the case that |𝑧|=1, there are 2,021 pairs of numbers which satisfy the equation. Therefore, in total, there are 2,022 pairs of real numbers (π‘Ž,𝑏) for which the relationship holds.

Example 6: Common Vertices Problem

Two regular polygons are inscribed in the same circle: the first has 1,731 sides and the second has 4,039. If the two polygons have at least one vertex in common, how many vertices in total will coincide?

Answer

Recall that the 𝑛th roots of unity lie at the vertices of a regular 𝑛-gon. Hence, we can restate the problem as finding the number of common roots of π‘§βˆ’1=0οŠ§οŽ•οŠ­οŠ©οŠ§ and π‘§βˆ’1=0οŠͺοŽ•οŠ¦οŠ©οŠ―. Using the theorem about the common roots of unity, we know that the common roots will be the roots of π‘§βˆ’1=0 where 𝑑=(1,731,4,039)=577gcd. Therefore, if the two polygons have at least one vertex in common, they will have a total of 577 vertices in common.

Key Points

  1. The 𝑛th roots of unity for any given 𝑛 can be found using De Moivre’s theorem. Expressed in exponential form, they are given by π‘’οŽ‘ο‘½ο‘€οΌο‘ƒ, where π‘˜=0,1,…,π‘›βˆ’1.
  2. The 𝑛th roots of unity have interesting geometric properties. In particular, they lie at the vertices of regular 𝑛-gons centered at the origin with one vertex at 𝑧=1.
  3. The 𝑛th roots of unity form a group which is closed under multiplication and division.
  4. The common roots of π‘§βˆ’1=0 and π‘§βˆ’1=0 are the roots of π‘§βˆ’1=0 where 𝑑=(π‘š,𝑛)gcd.

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