Lesson Explainer: The 𝑛th Roots of Unity | Nagwa Lesson Explainer: The 𝑛th Roots of Unity | Nagwa

Lesson Explainer: The 𝑛th Roots of Unity Mathematics • Third Year of Secondary School

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

In complex numbers, the 𝑛th roots of unity are complex numbers 𝑧 satisfying 𝑧=1,𝑛.forapositiveinteger

We know that there is only one real-valued solution, 𝑧=1, to this equation if 𝑛 is an odd integer, since we can solve the equation by applying the 𝑛th root on both sides of the equation. If 𝑛 is an even integer, there are two real-valued solutions, 𝑧=1 and 𝑧=βˆ’1, to this equation.

On the other hand, for 𝑛>2, there are other solutions of this equation which are not real numbers. In particular, for 𝑛=3, we know that the solutions of these equation are π‘§βˆˆο―1,βˆ’12+√32𝑖,βˆ’12βˆ’βˆš32𝑖.

We can begin to notice a pattern here. Clearly, there is only one solution to the equation 𝑧=1, which is 𝑧=1. There are two real-valued solutions 𝑧=1 and 𝑧=βˆ’1 for the equation 𝑧=1. Moreover, we observed that there are three distinct solutions to 𝑧=1. At this point, we can conjecture that there are 𝑛 distinct solutions to the equation 𝑧=1.

We can prove this conjecture by applying de Moivre’s theorem for roots.

Theorem: De Moivre’s Theorem for Roots

For a complex number 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin, the 𝑛th roots of 𝑧 are given by ο‘ƒβˆšπ‘Ÿο€½ο€½πœƒ+2πœ‹π‘˜π‘›ο‰+π‘–ο€½πœƒ+2πœ‹π‘˜π‘›ο‰ο‰,cossin for π‘˜=0,1,…,π‘›βˆ’1.

In the first example, we will apply de Moivre’s theorem to compute the 𝑛th roots of unity in polar form.

Example 1: The 𝑛th Roots of Unity

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

Answer

Recall that, by de Moivre’s theorem for roots, the 𝑛th roots of a complex number 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin are given by ο‘ƒβˆšπ‘Ÿο€½ο€½πœƒ+2πœ‹π‘˜π‘›ο‰+π‘–ο€½πœƒ+2πœ‹π‘˜π‘›ο‰ο‰,π‘˜=0,1,…,π‘›βˆ’1.cossinfor

Hence, to take the 𝑛th root of the right-hand side of the equation, which is 1, we can begin by expressing 1 in polar form. We know that the modulus of 1 is equal to 1, and since 1 is on the positive real axis of an Argand diagram, we also know that the argument of 1 is equal to 0 radians. We can write the polar form of 1 by using π‘Ÿ=1 and πœƒ=0: 1=1(0+𝑖0).cossin

Applying de Moivre’s theorem for roots, the 𝑛th roots of unity are given by ο‘ƒβˆš1ο€½ο€½2πœ‹π‘˜π‘›ο‰+𝑖2πœ‹π‘˜π‘›ο‰ο‰=ο€½2πœ‹π‘˜π‘›ο‰+𝑖2πœ‹π‘˜π‘›ο‰,cossincossin for π‘˜=0,1,…,π‘›βˆ’1. Hence, the roots of 𝑧=1 in polar form are cossinο€½2πœ‹π‘˜π‘›ο‰+𝑖2πœ‹π‘˜π‘›ο‰.

In the previous example, we found the polar form of the 𝑛th roots of unity by applying de Moivre’s theorem. In particular, we have proved our conjecture that states that there are 𝑛 distinct complex solutions to the equation 𝑧=1. In other words, there are 𝑛 distinct 𝑛th roots of unity.

Using the polar form of the 𝑛th roots of unity obtained in the previous example, we can also write the exponential form of the roots of unity. Recall that the exponential form of a complex number with modulus π‘Ÿ and argument πœƒ is π‘Ÿπ‘’οƒοΌ, which means the polar and exponential forms of the complex number are related by the identity π‘Ÿ(πœƒ+π‘–πœƒ)=π‘Ÿπ‘’.cossin

We summarize both the polar and exponential forms of the 𝑛th roots of unity below.

Definition: 𝑛th Roots of Unity in Polar and Exponential Form

The 𝑛th roots of unity in polar form are π‘§βˆˆο­1,ο€Ό2πœ‹π‘›οˆ+𝑖2πœ‹π‘›οˆ,…,ο€½2πœ‹(π‘›βˆ’1)𝑛+𝑖2πœ‹(π‘›βˆ’1)𝑛.cossincossin

The 𝑛th roots of unity in exponential form are π‘§βˆˆο­1,𝑒,…,𝑒.οŽ‘ο‘½ο‘ƒοŽ‘ο‘½(οŽͺ)

In particular, the root 1 is called the trivial 𝑛th root of unity.

The expressions for the 𝑛th roots of unity above give the moduli and arguments of the complex numbers. We can see that the moduli of all 𝑛th roots of unity are equal to 1, which means that they all lie on the unit circle in an Argand diagram. The trivial root of unity 1 lies at the intersection of the unit circle and the positive real line in an Argand diagram. The arguments of 𝑛th roots of unity increase in an arithmetic sequence increasing by 2πœ‹π‘› radians. In an Argand diagram, this means that we can plot the 𝑛th roots of unity by starting with 1 and rotating counterclockwise on the unit circle by 2πœ‹π‘› consecutively. If we connect consecutive 𝑛th roots of unity with line segments, we will obtain a regular polygon inscribed in the unit circle. Let us observe this pattern in the diagrams below for several different values of 𝑛.

As expected, the 𝑛th roots of unity for 𝑛β‰₯3 form vertices of a regular 𝑛-gon inscribed in the unit circle in an Argand diagram, with a vertex at the trivial root 1.

We note that the arguments of the 𝑛th roots of unity do not all lie in the standard range, which is ]βˆ’πœ‹,πœ‹] radians. In particular, we note that the cube roots of unity are labeled in the Argand diagram above by 1, π‘’οŽ‘ο‘½οŽ’οƒ, and π‘’οŽ£ο‘½οŽ’οƒ. The last cube root of unity has argument 4πœ‹3, which is outside this range. Since this argument is over the upper bound πœ‹, we can obtain an equivalent argument by subtracting the full revolution of 2πœ‹ radians from this value: 4πœ‹3βˆ’2πœ‹=4πœ‹3βˆ’6πœ‹3=βˆ’2πœ‹3.

We note that this equivalent argument βˆ’2πœ‹3 lies in the standard range ]βˆ’πœ‹,πœ‹], so we can use this argument to write the third root of unity as 𝑒οŽͺοŽ‘ο‘½οŽ’οƒ.

In the next example, we will find the quintic (fifth) roots of unity so that the arguments lie in the standard range and compute their sum.

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

We recall that the 𝑛th roots of unity are given in polar form as π‘’οŽ‘ο‘€ο‘½ο‘ƒοƒ for π‘˜=0,1,…,π‘›βˆ’1. In this example, we want to find the quintic roots of unity, which are the same as the fifth roots of unity. Hence, we can write the quintic roots of unity by substituting 𝑛=5 and π‘˜=1,2,…,4 into this expression: π‘˜=0βˆΆπ‘’=1,π‘˜=1βˆΆπ‘’,π‘˜=2βˆΆπ‘’=𝑒,π‘˜=3βˆΆπ‘’=𝑒,π‘˜=4βˆΆπ‘’=𝑒.οŠ¦οƒοƒοƒοƒοƒοƒοƒοƒοŽ‘ο‘½οŽ€οŽ‘ο‘½Γ—οŽ‘οŽ€οŽ£ο‘½οŽ€οŽ‘ο‘½Γ—οŽ’οŽ€οŽ₯ο‘½οŽ€οŽ‘ο‘½Γ—οŽ£οŽ€οŽ§ο‘½οŽ€

We recall that the argument of a complex number, by convention, should lie in the standard range ]βˆ’πœ‹,πœ‹]. The last two quintic roots of unity have arguments 6πœ‹5 and 8πœ‹5, which do not lie in this range. Since these arguments are over the upper bound πœ‹, we can obtain equivalent arguments by subtracting the full revolution of 2πœ‹ radians from this value: 6πœ‹5βˆ’2πœ‹=βˆ’4πœ‹58πœ‹5βˆ’2πœ‹=βˆ’2πœ‹5.

Using these arguments in the standard range, the last two quintic roots of unity can be written as 𝑒οŽͺοŽ£ο‘½οŽ€ and 𝑒οŽͺοŽ‘ο‘½οŽ€. Hence, the quintic roots of unity are 1,𝑒,𝑒,𝑒,𝑒.οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽͺοŽ£ο‘½οŽ€οŽͺοŽ‘ο‘½οŽ€οƒοƒοƒοƒ

Part 2

In this part, we need to find the sum of the roots. We will show two different methods for this computation. The first method will require a calculator to evaluate the cosine ratios at nonspecial angles, and the second method will use an algebra trick and will not require a calculator. The second method is much simpler once we know the algebra trick.

Method 1

We need to compute 1+𝑒+𝑒+𝑒+𝑒.οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽͺοŽ£ο‘½οŽ€οŽͺοŽ‘ο‘½οŽ€οƒοƒοƒοƒ

We can rearrange the sum to write 1+𝑒+𝑒+𝑒+𝑒.οŽ‘ο‘½οŽ€οŽͺοŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽͺοŽ£ο‘½οŽ€οƒοƒοƒοƒ

Recall the property of the exponential form of a complex number with respect to the complex conjugate: 𝑒=π‘’οƒοΌοŠ±οƒοΌ, for any real number πœƒ. This tells us that the pair of complex numbers inside each of the parentheses above are complex conjugates. We know that, for any complex number 𝑧, 𝑧+𝑧=2𝑧Re, where Re𝑧 is the real part of 𝑧. Hence, the sum of the roots can be written as 1+2𝑒+2𝑒.ReReοŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οƒοƒ

It now remains to find the real parts of the complex numbers π‘’οŽ‘ο‘½οŽ€ and π‘’οŽ£ο‘½οŽ€. We recall that a complex number in the exponential form can be expressed in the polar form by writing π‘Ÿπ‘’=π‘Ÿ(πœƒ+π‘–πœƒ),cossin where π‘Ÿ is the modulus and πœƒ is the argument of the complex number. For π‘’οŽ‘ο‘½οŽ€οƒ and π‘’οŽ£ο‘½οŽ€οƒ, the moduli of both complex numbers are 1, which means π‘Ÿ=1 for both cases. Also, we can see that πœƒ=2πœ‹5 for π‘’οŽ‘ο‘½οŽ€οƒ, and πœƒ=4πœ‹5 for π‘’οŽ£ο‘½οŽ€οƒ. Hence, 𝑒=2πœ‹5+𝑖2πœ‹5𝑒=4πœ‹5+𝑖4πœ‹5.οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οƒοƒcossincossin

This tells us that RecosRecos𝑒=2πœ‹5,𝑒=4πœ‹5.οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οƒοƒ

Substituting these values, the sum of the roots is written as 1+22πœ‹5+24πœ‹5.coscos

Since neither 2πœ‹5 nor 4πœ‹5 is an angle in the unit circle, we cannot evaluate the cosine function at these angles without using a calculator. Using a calculator, we can compute coscos2πœ‹5=0.309016β‹―,4πœ‹5=βˆ’0.809016β‹―.

Substituting these values into the sum, we obtain 1+2Γ—0.309016β‹―+2Γ—(βˆ’0.809016β‹―)=0.

Method 2

Next, we will use an algebra trick to compute the sum. We begin by setting the sum equal to an unknown constant 𝑠:

𝑠=1+𝑒+𝑒+𝑒+𝑒.οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽͺοŽ£ο‘½οŽ€οŽͺοŽ‘ο‘½οŽ€οƒοƒοƒοƒ(1)

We multiply both sides of the equation above by π‘’οŽ‘ο‘½οŽ€οƒ. Then, using the property of the exponential form 𝑒×𝑒=𝑒()οƒοŽ οŽ‘οŽ οŽ‘, we can write 𝑒𝑠=𝑒1+𝑒+𝑒+𝑒+𝑒=𝑒+𝑒+𝑒+𝑒+1.οŽ‘ο‘½οŽ€οŽ‘ο‘½οŽ€οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽͺοŽ£ο‘½οŽ€οŽͺοŽ‘ο‘½οŽ€οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽ₯ο‘½οŽ€οŽͺοŽ‘ο‘½οŽ€οƒοƒοƒοƒοƒοƒοƒοƒοƒοƒ

We know that 𝑒=𝑒οŽ₯ο‘½οŽ€οŽͺοŽ£ο‘½οŽ€οƒοƒ by subtracting 2πœ‹ from the argument of the first complex number. This means that 𝑒𝑠=𝑒+𝑒+𝑒+𝑒+1.οŽ‘ο‘½οŽ€οŽ‘ο‘½οŽ€οŽ£ο‘½οŽ€οŽͺοŽ£ο‘½οŽ€οŽͺοŽ‘ο‘½οŽ€οƒοƒοƒοƒοƒ

We can see that the right-hand side of the equation above is the same as the right side of equation (1). Hence, equating the left sides of these equations leads to 𝑒𝑠=𝑠.οŽ‘ο‘½οŽ€οƒ

Rearranging this equation, we obtain π‘’π‘ βˆ’π‘ =0,π‘ ο€½π‘’βˆ’1=0.οŽ‘ο‘½οŽ€οŽ‘ο‘½οŽ€οƒοƒ

Since the complex number inside the parentheses on the left-hand side of the equation above is not equal to zero, we can divide both sides of this equation by this number to obtain 𝑠=0.

Hence, the sum of the quintic roots of unity is 0.

In the previous example, we found that the sum of the quintic roots of unity is equal to zero. In fact, this is a general property of the 𝑛th roots of unity, as we will see later.

In the next example, we will consider the reciprocal of an 𝑛th root of unity.

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”)
    4. πœ”=(πœ”)
  2. Express πœ”οŠ±οŠ§ in terms of positive powers of πœ”.

Answer

Part 1

In this example, we find πœ”οŠ±οŠ§ where πœ” is an 𝑛th root of unity. Recall that de Moivre’s theorem for integer powers tells us that, given a complex number 𝑧=π‘Ÿπ‘’οƒοΌ, 𝑧=π‘Ÿπ‘’,𝑛.οŠοŠοƒοŠοΌforanyinteger

Here, the integer is 𝑛=βˆ’1. Also, we know that the modulus of an 𝑛th root of unity is equal to 1. In other words, πœ”=𝑒 for some argument πœƒ. Hence, applying de Moivre’s theorem for integer powers, πœ”=𝑒.οŠ±οŠ§οŠ±οƒοΌ

This is the complex number with modulus 1 and argument βˆ’πœƒ. Since the argument of this complex number has the opposite sign to the argument of πœ”, this means that they are located on opposite sides of the real axis in an Argand diagram.

As we can observe in the Argand diagram above, this means that the real parts of πœ” and πœ”οŠ±οŠ§ are equal, while the imaginary parts of these two complex numbers have the opposite signs. In other words, they are complex conjugates of each other. Hence, πœ”=(πœ”).

This is option D.

Part 2

In this part, we want to express πœ”=πœ”οŠ±οŠ§ο‡ for some positive integer π‘˜. Since πœ” is an 𝑛th root of unity, we know that πœ”=1.

We can multiply both sides of this equation by πœ”οŠ±οŠ§ and use the rule of exponents to write πœ”πœ”=πœ”πœ”=πœ”.

Since π‘›βˆ’1 is a positive integer, we have πœ”=πœ”.

In the previous example, we noted that the reciprocal of an 𝑛th root of unity is equal to its complex conjugate. This means that the product of a root of unity with its conjugate is equal to 1.

Property: Conjugate of Roots of Unity

Let πœ” be an 𝑛th root of unity. Then πœ”β‹…πœ”=πœ”β‹…πœ”=1.

So far, we have considered properties of the 𝑛th roots of unity. Some 𝑛th roots of unity are shared for different values of 𝑛. These are called common roots of unity. In the next example, we will discuss the 𝑛 roots of unity which are shared between 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 between the cube roots of unity and the 6th roots of unity?

Answer

Part 1

Recall that the 𝑛th roots of unity in polar form are given as cossinforο€½2πœ‹π‘˜π‘›ο‰+𝑖2πœ‹π‘˜π‘›ο‰,π‘˜=0,1,…,π‘›βˆ’1.

We can find the cubit roots of unity by substituting 𝑛=3, which means that we need to substitute π‘˜=0,1,2. This leads to 1,ο€Ό2πœ‹3+𝑖2πœ‹3,ο€Ό4πœ‹3+𝑖4πœ‹3.cossinandcossin

Finding an equivalent expression for the last cube root of unity so that its argument lies in the standard range ]βˆ’πœ‹,πœ‹], we find that the three cube 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 substituting 𝑛=6 and π‘˜=0,1,…,5, which leads to 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 cube roots of unity are also 6th roots of unity.

In the previous example, we found that all cube roots of unity are also sixth roots of unity. We can also observe this fact from a different perspective. A cube root of unity is a complex number satisfying 𝑧=1, while a sixth root of unity is a complex solution 𝑧 to 𝑧=1. If a complex number satisfies 𝑧=1, then we can square both sides of the equation to write 𝑧=1𝑧=1.

Hence, if a complex number satisfies 𝑧=1, then it also satisfies 𝑧=1. In fact, we can make this statement more general. Say that we have two positive integers π‘š and 𝑛 such that π‘š divides 𝑛; that is, 𝑛=π‘šπ‘˜ for some positive integer π‘˜. If 𝑧 is an π‘šth root of unity, it must satisfy 𝑧=1. Applying the π‘˜th power to both sides of this equation, (𝑧)=1𝑧=1𝑧=1.ο‰ο‡ο‡ο‰ο‡οŠ

This means that 𝑧 is also an 𝑛th root of unity. This tells us that, if π‘š and 𝑛 are positive integers such that π‘š divides 𝑛, any π‘šth root of unity is automatically an 𝑛th root of unity.

We can apply this statement to prove an even more general theorem. This time, say that π‘š does not divide 𝑛, but gcd(π‘š,𝑛)=π‘˜. This tells us that π‘˜ divides both π‘š and 𝑛. Hence, all of π‘˜th roots of unity are also π‘šth and 𝑛th roots of unity. In fact, these are precisely the common roots of unity.

Theorem: Common Roots of Unity When gcd(π‘š,𝑛) β‰  1

Say that π‘š and 𝑛 are positive integers such that gcd(π‘š,𝑛)=π‘˜. Then, the common roots of unity shared by the 𝑛th and π‘šth roots of unity are precisely the π‘˜th roots of unity.

Let us consider an example where we apply this theorem to solve a geometric problem.

Example 5: Number of Common Vertices in Inscribed Polygons

Two regular polygons are inscribed in the same circle where the first has 1β€Žβ€‰β€Ž731 sides and the second has 4β€Žβ€‰β€Ž039. 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 in an Argand diagram lie at the vertices of a regular 𝑛-gon inscribed in a unit circle, where one of the vertices is at 1. We can place a coordinate system on this plane so that the given circle is the unit circle and the common vertex is the point (1,0). Then, we can think of the vertices as complex numbers in an Argand diagram.

In this setting, the vertices of the polygon with 1β€Žβ€‰β€Ž731 vertices are the 1731st roots of unity, while the vertices with 4β€Žβ€‰β€Ž039 sides are the 4039th roots of unity.

We recall that, for any positive integers π‘š and 𝑛, the common roots of unity shared by the 𝑛th and π‘šth roots of unity are precisely the π‘˜th roots of unity where gcd(π‘š,𝑛)=π‘˜. In this example, π‘š=1731 and 𝑛=4039, so we need to find gcd(1731,4039). We note that the first number is divisible by 3 and the second number is divisible by 7. Using these prime factors, we can write the two numbers as 1731=3Γ—577,4039=7Γ—577.

We see here that 577 is a common factor between these numbers, and it is the greatest factor since the other prime factors 3 and 7 are not shared. This gives us π‘˜=(1731,4039)=577.gcd

Hence, the common vertices of these two polygons are represented in the Argand diagram as the 577th roots of unity. We know that there are exactly 577 complex numbers that are the 577th roots of unity.

Therefore, if the two polygons inscribed in the same circle have at least one vertex in common, they will have a total of 577 vertices in common.

In previous examples, we identified common roots of unity. We now turn our attention to another important definition for the 𝑛th roots of unity. Previously, we have listed the 𝑛th roots of unity for 𝑛=2,3,…,7, which are labeled in Argand diagrams. Let us list them here, so that the arguments of the roots of unity lie in the standard range ]βˆ’πœ‹,πœ‹].

𝑛The 𝑛th roots of unity
11
21, βˆ’1
31, π‘’οŽ‘ο‘½οŽ’οƒ, 𝑒οŽͺοŽ‘ο‘½οŽ’οƒ
41, 𝑖, βˆ’1, βˆ’π‘–
51, π‘’οŽ‘ο‘½οŽ€οƒ, π‘’οŽ£ο‘½οŽ€οƒ, 𝑒οŽͺοŽ£ο‘½οŽ€οƒ, 𝑒οŽͺοŽ‘ο‘½οŽ€οƒ
61, π‘’ο‘½οŽ’οƒ, π‘’οŽ‘ο‘½οŽ’οƒ, βˆ’1, 𝑒οŽͺοŽ‘ο‘½οŽ’οƒ, 𝑒οŽͺο‘½οŽ’οƒ
71, π‘’οŽ‘ο‘½οŽ¦οƒ, π‘’οŽ£ο‘½οŽ¦οƒ, 𝑒οŽ₯ο‘½οŽ¦οƒ, 𝑒οŽͺοŽ₯ο‘½οŽ¦οƒ, 𝑒οŽͺοŽ£ο‘½οŽ¦οƒ, 𝑒οŽͺοŽ‘ο‘½οŽ¦οƒ

The red numbers in the table above are the 𝑛th roots of unity that have appeared as an π‘šth root of unity for some π‘š<𝑛. For instance, the root π‘’οŽ‘ο‘½οŽ’οƒ for 𝑛=6 is red because this root has appeared as a root of unity when π‘š=3 and 3<6. This means that the green numbers are the roots of unity that appear for the first time in the list. Such roots of unity are called the primitive roots of unity.

Definition: Primitive 𝑛th Roots of Unity

A primitive 𝑛th root of unity is a complex number πœ” for which π‘˜=𝑛 is the smallest positive integer satisfying πœ”=1. In other words, a primitive 𝑛th root of unity is an 𝑛th root of unity that is also not an π‘šth root of unity for any π‘š<𝑛.

From the table above, we have identified all the primitive 𝑛th roots of unity for 𝑛=1,2,…,7:

𝑛Primitive 𝑛th roots of unity
11
2βˆ’1
3π‘’οŽ‘ο‘½οŽ’οƒ, 𝑒οŽͺοŽ‘ο‘½οŽ’οƒ
4𝑖, βˆ’π‘–
5π‘’οŽ‘ο‘½οŽ€οƒ, π‘’οŽ£ο‘½οŽ€οƒ, βˆ’π‘’οŽ£ο‘½οŽ€οƒ, 𝑒οŽͺοŽ‘ο‘½οŽ€οƒ
6π‘’ο‘½οŽ’οƒ, 𝑒οŽͺο‘½οŽ’οƒ
7π‘’οŽ‘ο‘½οŽ¦οƒ, π‘’οŽ£ο‘½οŽ¦οƒ, 𝑒οŽ₯ο‘½οŽ¦οƒ, 𝑒οŽͺοŽ₯ο‘½οŽ¦οƒ, 𝑒οŽͺοŽ£ο‘½οŽ¦οƒ, 𝑒οŽͺοŽ‘ο‘½οŽ¦οƒ

We can observe that the first nontrivial 𝑛th root of unity for 𝑛β‰₯2 is always a primitive root of unity.

Property: Primitive 𝑛th Roots of Unity

Any complex number of the form πœ”=π‘’οŽ‘ο‘½ο‘ƒοƒ for positive integer 𝑛 is a primitive 𝑛th root of unity.

This property also tells us that there is always a complex number corresponding to a primitive 𝑛th root of unity. However, we should keep in mind that a primitive 𝑛th root of unity for 𝑛>2 is not unique, and this property provides a way to obtain one particular primitive 𝑛th root of unity rather than all of them.

Let us prove this property by showing that there is no positive integer π‘˜, less than 𝑛, such that πœ”=1. Thus, consider an arbitrary integer π‘˜ satisfying 0<π‘˜<𝑛. Then, using de Moivre’s theorem for integer powers, we can write πœ”=𝑒.ο‡οƒοŽ‘ο‘½ο‘€ο‘ƒ

We also note that the condition 0<π‘˜<𝑛 implies 0<π‘˜π‘›<1 when we divide each part of the inequality by 𝑛. Then, the argument of the complex number above satisfies 0<2πœ‹π‘˜π‘›=2πœ‹π‘˜π‘›<2πœ‹.

Since the argument of 1 is equal to 0 or an integer multiple of 2πœ‹, this tells us that πœ”β‰ 1 for any integer π‘˜ satisfying 0<π‘˜<𝑛. In other words, π‘˜=𝑛 is the smallest positive integer for which πœ”=1, which proves the property.

One important property of a primitive 𝑛th root of unity is that it generates all 𝑛th roots of unity by taking consecutive powers.

Property: Primitive 𝑛th Roots of Unity

If πœ” is a primitive root of unity, then all 𝑛th roots of unity are given by 1,πœ”,πœ”,…,πœ”.

We can prove this property by showing the following two facts. Let πœ” be a primitive 𝑛th root of unity. Then,

  • any integer power of πœ” is also an 𝑛th root of unity,
  • the complex numbers 1,πœ”,πœ”,…,πœ”οŠ¨οŠοŠ±οŠ§ are distinct.

If we know these two facts, then we know that 1,πœ”,πœ”,…,πœ”οŠ¨οŠοŠ±οŠ§ are 𝑛 distinct 𝑛th roots of unity, which means that they are all 𝑛th roots of unity.

To show the first statement, let π‘˜ be an arbitrary integer. Then, using the rules of exponents, ο€Ήπœ”ο…=πœ”=(πœ”).ο‡οŠο‡οŠοŠο‡

We know that πœ”=1 since πœ” is an 𝑛th root of unity. Hence, the right-hand side of the equation above is equal to 1. Since ο€Ήπœ”ο…=1ο‡οŠ, πœ”ο‡ is an 𝑛th root of unity. This proves the first statement needed for our proof.

To prove the second statement, we can begin by assuming that this statement is false. Then there are two distinct nonnegative integers π‘š and π‘˜ with π‘š<𝑛 and π‘˜<𝑛 satisfying πœ”=πœ”.

Let us assume π‘š<π‘˜, since the order of these integers is arbitrary. Then, dividing both sides of the equation by πœ”ο‡ and using the rule of exponents, we can write πœ”πœ”=1πœ”=1.ο‰ο‡ο‰οŠ±ο‡

This tells us that 𝑧=πœ” satisfies the equation 𝑧=1ο‰οŠ±ο‡, where π‘šβˆ’π‘˜ is a positive integer by the assumption π‘š>π‘˜. Hence, πœ” is the (π‘šβˆ’π‘˜)th root of unity where π‘šβˆ’π‘˜<𝑛. However, this contradicts the fact that πœ” is a primitive 𝑛th root of unity. This means that it is not possible to have πœ”=πœ”, for any nonnegative integers π‘š and π‘˜ smaller than 𝑛. In other words, the numbers 1,πœ”,πœ”,…,πœ”οŠ¨οŠοŠ±οŠ§ are 𝑛 distinct numbers. This proves the desired property.

Let us now consider a few properties of primitive 𝑛th roots of unity.

Property: Sum of Powers of Primitive 𝑛th Roots of Unity

Let πœ” be a primitive 𝑛th root of unity for 𝑛β‰₯2. Then 1+πœ”+πœ”+β‹―+πœ”=0.

Since the terms in the left-hand side of the equation above are the 𝑛th roots of unity, this statement tells us that the sum of the 𝑛th roots of unity is equal to zero for any 𝑛β‰₯2. Remember that we showed that the sum of the quintic roots of unity is equal to zero in example 2. A similar algebraic trick can be used here to prove this statement.

To prove this property, let us begin by setting the left-hand side of the equation equal to an unknown constant 𝑠: 𝑠=1+πœ”+πœ”+β‹―+πœ”.

Multiplying both sides of this equation by πœ”, πœ”π‘ =πœ”+πœ”+β‹―+πœ”+πœ”.

Since πœ” is an 𝑛th root of unity, we know that πœ”=1. This gives us πœ”π‘ =πœ”+πœ”+β‹―+πœ”+1,𝑠=1+πœ”+πœ”+β‹―+πœ”.

We can see that the right-hand sides of both equations are the same, which tells us πœ”π‘ =π‘ πœ”π‘ βˆ’π‘ =0𝑠(πœ”βˆ’1)=0.

Since πœ” is a primitive 𝑛th root of unity, so we know that πœ”β‰ 1. This means that we can divide both sides of the equation above by (πœ”βˆ’1) to obtain 𝑠=0.

This proves the desired statement, which leads directly to the following result.

Corollary: Sum of 𝑛th Roots of Unity

The sum of the 𝑛th roots of unity for any 𝑛β‰₯2 is equal to zero.

Since the sum of complex numbers is geometrically equivalent to the addition of vectors, the property above gives us another aspect of the symmetry of the 𝑛th roots of unity. To make a physical analogy, if we think of each 𝑛th root of unity as a vector representing the magnitude and the direction of a force acting at the origin of an Argand diagram, this property tells us that the combined effect of all the forces at the origin is zero. In other words, the forces represented by the 𝑛th roots of unity form a perfect equilibrium.

Let us consider an example where we need to apply this property of a primitive 𝑛th root of unity.

Example 6: Summing Power of Primitive Roots of Unity

If πœ” is a primitive 6th root of unity, which of the following expressions is equivalent to πœ”+πœ”+πœ”οŠ¨οŠ©?

  1. πœ”+πœ”+πœ”οŠͺ
  2. 12ο€Ήπœ”+πœ”+πœ”ο…οŠ¨οŠͺ
  3. 1βˆ’πœ”βˆ’πœ”οŠͺ
  4. βˆ’ο€Ή1+πœ”+πœ”ο…οŠͺ
  5. 1

Answer

We recall that a primitive 𝑛th root of unity satisfies 1+πœ”+πœ”+β‹―+πœ”=0.

Since we are given that πœ” is a 6th primitive root of unity, 𝑛=6. Hence, 1+πœ”+πœ”+πœ”+πœ”+πœ”=0.οŠͺ

We can rearrange the equation to find an equivalent expression to the given one: πœ”+πœ”+πœ”=βˆ’ο€Ή1+πœ”+πœ”ο….οŠͺ

This is option D.

In our final example, we will apply properties of an 𝑛th root of unity to solve a problem.

Example 7: 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¨οŠ¦οŠ¨οŠ¦ hold?

Answer

Recall the properties of the modulus of a complex number, which states |𝑧|=|𝑧|,|𝑧|=||𝑧||.

If we denote 𝑧=π‘Ž+𝑏𝑖, we have 𝑧=π‘Žβˆ’π‘π‘–; hence the given equation is 𝑧=π‘§οŠ¨οŠ¦οŠ¨οŠ¦. Taking the modulus of this equation, we have ||𝑧||=||𝑧||.

Then, the properties of the modulus stated above give |𝑧|=|𝑧|.

Hence, subtracting |𝑧| from both sides of the equation, we have |𝑧|βˆ’|𝑧|=0,|𝑧|ο€Ί|𝑧|βˆ’1=0.

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 that 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.

Recall that, for any complex number 𝑧, 𝑧𝑧=|𝑧|. Since we know that |𝑧|=1, we can rewrite the equation above as 𝑧=1.

We know that the solutions of these equations are called the 2021st roots of unity, and we also know there are exactly 2β€Žβ€‰β€Ž021 unique roots of unity satisfying this equation. Therefore, in the case that |𝑧|=1, there are 2β€Žβ€‰β€Ž021 pairs of (π‘Ž,𝑏) which satisfy the equation.

In total, there are 2β€Žβ€‰β€Ž022 pairs of real numbers (π‘Ž,𝑏) for which the given relationship holds.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • For any positive integer 𝑛, there are exactly 𝑛 distinct complex numbers 𝑧 satisfying the equation 𝑧=1. The complex solutions of this equation are called the 𝑛th roots of unity.
  • The 𝑛th roots of unity in exponential form are given by π‘§βˆˆο­1,𝑒,…,𝑒.οŽ‘ο‘½ο‘ƒοŽ‘ο‘½(οŽͺ)
  • The 𝑛th roots of unity for 𝑛β‰₯3 form vertices of a regular 𝑛-gon inscribed in the unit circle in an Argand diagram, with a vertex at the trivial root 1.
  • Say that π‘š and 𝑛 are positive integers such that gcd(π‘š,𝑛)=π‘˜. Then the common roots of unity shared by the 𝑛th and π‘šth roots of unity are precisely the π‘˜th roots of unity.
  • A primitive 𝑛th root of unity is a complex number πœ” for which π‘˜=𝑛 is the smallest positive integer satisfying πœ”=1. In other words, a primitive 𝑛th root of unity is an 𝑛th root of unity which is also not an π‘šth root of unity for any π‘š<𝑛. In particular, any complex number of the form πœ”=π‘’οŽ‘ο‘½ο‘ƒοƒ is a primitive 𝑛th root of unity.
  • If πœ” is a primitive root of unity, then all 𝑛th roots of unity are given by 1,πœ”,πœ”,…,πœ”. They satisfy 1+πœ”+πœ”+β‹―+πœ”=0.

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