شارح الدرس: Cube Roots of Unity | نجوى شارح الدرس: Cube Roots of Unity | نجوى

شارح الدرس: Cube Roots of Unity الرياضيات

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

A cube (or cubic) root of unity is a complex-valued solution 𝑧 to the equation 𝑧=1. If we only consider real-valued solutions to this equation, we can apply the cube root to both sides of the equation to obtain 𝑧=1=1, which means that there is only one real-valued solution. However, there are more solutions to this equation that are not real numbers. To solve this equation algebraically, we first rearrange this equation as 𝑧1=0. Recalling the difference of cubes formula 𝑎𝑏=(𝑎𝑏)𝑎+𝑎𝑏+𝑏, we can factor 𝑧1 to write (𝑧1)𝑧+𝑧+1=0.

The first factor 𝑧1 leads to the real root 𝑧=1, while the quadratic factor 𝑧+𝑧+1 will lead to complex-valued roots when we apply the quadratic roots formula. Applying the quadratic roots formula with 𝑎=1, 𝑏=1, and 𝑐=1, we have 𝑧=𝑏±𝑏4𝑎𝑐2𝑎=1±142=12±32𝑖.

This gives us two other complex-valued solutions to the equation 𝑧=1. Hence, we have obtained three cube roots of unity.

Definition: Cube Roots of Unity in Cartesian Form

The cube roots of unity are the complex-valued solutions of the equation 𝑧=1. The three cube roots of unity in Cartesian form are 𝑧1,12+32𝑖,1232𝑖.

The real root 1 is called the trivial cubic root of unity, and the nontrivial roots 12+32𝑖 and 1232𝑖 are called the complex cubic roots of unity.

From the Cartesian form for cubic roots of unity, we can see that the two complex cubic roots of unity are complex conjugates of each other.

While this method can be used to find the cube roots of unity, it cannot be generalized to obtain higher roots of unity, which are the solutions of the equation 𝑧=1 for 𝑛>3. Another method of obtaining the cube roots of unity involves de Moivre’s theorem, and this method can readily be generalized to find higher roots of unity. Let us recall de Moivre’s theorem for cubic roots.

Theorem: De Moivre’s Theorem for Cubic Roots

For a complex number 𝑧=𝑟(𝜃+𝑖𝜃)cossin, the cubic roots of 𝑧 are given by 𝑟𝜃+2𝜋𝑘3+𝑖𝜃+2𝜋𝑘3cossin for 𝑘=0,1, and 2.

In the first example, we will apply de Moivre’s theorem for roots to find the polar form of the cube roots of unity.

Example 1: Cubic Roots of Unity

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

Answer

In this example, we need to find all solutions for the equation 𝑧=1. If we only want to compute the real-valued solutions, we can apply the cube root on both sides of the equation to obtain 𝑧=1.

To take the cube root to find complex-valued solutions of this equation, we will use the polar form of a complex number. Recall that we can express a complex number with modulus 𝑟 and argument 𝜃 in polar form as 𝑟(𝜃+𝑖𝜃).cossin

We know that de Moivre’s theorem allows us to take the roots of a complex number in its polar form. In particular, we recall de Moivre’s theorem for cubic roots: given a complex number in polar form 𝑟(𝜃+𝑖𝜃)cossin, the cubic roots of this complex number are 𝑟𝜃+2𝜋𝑘3+𝑖𝜃+2𝜋𝑘3cossin for 𝑘=0,1, and 2.

Hence, to take the cube 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 𝑟=1 and 𝜃=0: 1=1(0+𝑖0).cossin

Applying de Moivre’s theorem to take the cubic roots of 1 in polar form, we obtain 12𝜋𝑘3+𝑖2𝜋𝑘3=2𝜋𝑘3+𝑖2𝜋𝑘3cossincossin for 𝑘=0,1, and 2. Hence, we have 𝑘=00+𝑖0=1,𝑘=12𝜋3+𝑖2𝜋3,𝑘=24𝜋3+𝑖4𝜋3.cossincossincossin

We recall that the argument of a complex number, by convention, should lie in the standard range ]𝜋,𝜋]. The first two cubic roots have arguments 0 and 2𝜋3, which are in this range, but the argument of the last root is 4𝜋3, which does not lie in this range. Since this argument is over the upper bound 𝜋, we can obtain an equivalent argument by subtracting the full revolution 2𝜋 radians from this value: 4𝜋32𝜋=4𝜋36𝜋3=2𝜋3.

We note that this equivalent argument 2𝜋3 lies in the standard range ]𝜋,𝜋], so we can use this argument to write the third root of unity cossin2𝜋3+𝑖2𝜋3.

Therefore, the cubic roots of unity are 1,2𝜋3+𝑖2𝜋3,2𝜋3+𝑖2𝜋3.cossincossin

In the previous example, we found the polar form of the cubic roots of unity by applying de Moivre’s theorem. Recall that the exponential form of a complex number with modulus 𝑟 and argument 𝜃 is 𝑟𝑒, which means the polar and exponential forms of the complex number are related by the identity 𝑟(𝜃+𝑖𝜃)=𝑟𝑒.cossin

Using the polar form of the cubic roots of unity obtained in the previous example, we can also write the exponential form of the roots of unity.

Definition: Cube Roots of Unity in Polar and Exponential Forms

The three cube roots of unity in polar form are 𝑧1,2𝜋3+𝑖2𝜋3,2𝜋3+𝑖2𝜋3.cossincossin

The three cube roots of unity in exponential form are 𝑧1,𝑒,𝑒.

In particular, the roots 𝑒 and 𝑒 are called the complex cubic roots of unity.

The expressions for the cube roots of unity above give the moduli and arguments of the complex numbers. We can see that the moduli of all cubic roots of unity are equal to 1, which means that they all lie on the unit circle in an Argand diagram. The first root, 1, lies on the positive real line, and the other two roots are symmetric about the real axis, as we can tell by the fact that their arguments have the opposite signs. Let us draw the cubic roots of unity on an Argand diagram.

From the Argand diagram above, we note that the angle between the two lines from the origin to the two complex cubic roots of unity is 𝜃=2𝜋2𝜋32𝜋3=2𝜋3.

In other words, the three cubic roots can be obtained by beginning at the point (1,0) in an Argand diagram and rotating 2𝜋3 radians counterclockwise or clockwise along the unit circle. We can associate the geometric rotation in an Argand diagram with the power of complex numbers using de Moivre’s theorem.

Theorem: De Moivre’s Theorem for Integer Powers

Let 𝑧=𝑟𝑒 be a nonzero complex number in exponential form. Then, for any integer 𝑛, 𝑧=𝑟𝑒.

In particular, this theorem tells us that if we square a complex number with modulus 1, we double the argument of the complex number. Regarding the Argand diagram above, we can see that squaring a complex cube root of unity will lead to the other complex root of unity.

In the next example, we will demonstrate this fact via explicit computations.

Example 2: Products of the Cubic Roots of Unity

Let 𝑧=𝑒 and 𝑧=𝑒 be the complex cubic roots of unity.

  1. Evaluate 𝑧. How does this compare with 𝑧?
  2. Evaluate 𝑧. How does this compare with 𝑧?

Answer

In this example, we need to take the powers of complex numbers in exponential form. Let us begin by recalling de Moivre’s theorem for integer powers: 𝑟𝑒=𝑟𝑒.

We will apply this theorem to take the squares of 𝑧 and 𝑧 in each of the following parts. In particular, this theorem tells us that if we square a complex number with modulus 1, we double the argument of the complex number.

Part 1

Applying de Moivre’s theorem for integer powers, we can write (𝑧)=𝑒=𝑒=𝑒.

We recall that the argument of a complex number, by convention, should lie in the standard range ]𝜋,𝜋], while the argument of 𝑧 above is 4𝜋3. Since this argument is over the upper bound 𝜋, we can obtain an equivalent argument by subtracting the full revolution 2𝜋 radians from this value: 4𝜋32𝜋=4𝜋36𝜋3=2𝜋3.

We note that this equivalent argument 2𝜋3 lies in the standard range ]𝜋,𝜋], so we can write the exponential form of 𝑧, 𝑧=𝑒, which is the same as 𝑧.

We can also understand this identity by observing an Argand diagram. Remember that, by de Moivre’s theorem, squaring a complex number with modulus 1 is equivalent to doubling the argument of the complex number on the unit circle. Hence, when we square 𝑒, we obtain the complex number with modulus 1 and argument 2×2𝜋3=4𝜋3.

We have already observed that this is equivalent to the argument 2𝜋3, 𝑒, which is the argument for 𝑧. This leads to the following diagram.

Hence, 𝑧=𝑒 (i.e., 𝑧=𝑧).

Part 2

Similarly, applying de Moivre’s theorem for integer powers, we can write (𝑧)=𝑒=𝑒=𝑒.

We can see that the argument of 𝑧 above is 4𝜋3, which does not lie in the standard range ]𝜋,𝜋]. Since this argument is below the lower bound 𝜋, we can obtain an equivalent argument by adding the full revolution 2𝜋 radians to this value: 4𝜋3+2𝜋=4𝜋3+6𝜋3=2𝜋3.

This equivalent argument 2𝜋3 lies in the standard range ]𝜋,𝜋]; hence, the exponential form of 𝑧 is written as 𝑧=𝑒, which is the same as 𝑧. Similarly to the previous part, we can visualize this relationship using an Argand diagram. If we double the argument of 𝑧, we obtain 2𝜋3×2=4𝜋3, which is equivalent to the argument of 𝑧. This leads to the Argand diagram below.

Hence, 𝑧=𝑒 (i.e., 𝑧=𝑧).

In the previous example, we obtained that the square of a complex cubic root of unity leads to the other complex root, as we expected based on the Argand diagram. We also know that the two complex roots of unity are complex conjugates of each other. This leads to the following property.

Property: Square of Complex Cubic Roots of Unity

Let 𝜔 be a complex cubic root of unity. Then, 𝜔=𝜔.

Let us consider an example where we can apply this property to simplify an expression involving 𝜔.

Example 3: Using the Properties of the Cubic Roots of Unity

Evaluate 𝜔𝜔 where 𝜔 is a complex cube root of unity.

Answer

Recall that a complex cube root of unity satisfies 𝜔=𝜔.

Hence, the expression inside the parentheses 𝜔𝜔 can be written as 𝜔𝜔.

We also recall that any complex number 𝑧 satisfies 𝑧𝑧=2𝑖𝑧.Im

Hence, 𝜔𝜔=2𝑖𝑧.Im

The complex cube roots of unity are 12+32𝑖 and 1232𝑖, which means Im𝜔=±32.

Substituting this value above, we obtain 𝜔𝜔=2𝑖×±32=±𝑖3.

Hence, 𝜔𝜔=±𝑖3.

Since we are taking an even power of the number ±𝑖3, the sign of this number can be ignored. Distributing the power, we have ±𝑖3=𝑖3.

By using the rules of exponents, we can write 𝑖3=𝑖3=(1)3=2187.

Hence, if 𝜔 is a complex cube root of unity, 𝜔𝜔=2187.

In the previous example, we used the property that if we square a complex cubic root of unity, we obtain the other complex cubic root of unity, which is the conjugate of the original root. We already know by definition that we obtain 1 when we cube a complex cubic root of unity. Hence, we can observe that 3 is the smallest positive integer power of a complex cubic root of unity that leads to the answer 1. On the other hand, the same cannot be said about the cubic root of unity 1, since 1=1 and also 1=1. This leads to an important definition that we now state.

Definition: Primitive Cubic Roots of Unity

A primitive cubic root of unity is a cubic root of unity 𝜔 for which 𝑘=3 is the smallest positive integer for which 𝜔=1.

As we have observed, both complex cubic roots of unity are also primitive cubic roots of unity, while the real-valued cubic root of unity is not a primitive cubic root of unity. It is not always the case that all complex roots of unity are primitive roots of unity, but this is true for the cubic roots of unity.

Let us consider other powers of a primitive cubic root of unity. For a general integer power 𝑛, we can write 𝑛=3𝑘+𝑎 for some integers 𝑘 and 𝑎=0,1,2. This relationship also implies that 𝑎𝑛(3)mod. Using the rules of exponents, we can write 𝜔=𝜔=𝜔𝜔=𝜔𝜔=(1)𝜔=𝜔.

This leads to the following property for primitive cubic roots of unity.

Property: Integer Powers of Primitive Cubic Roots of Unity

Let 𝜔 be a primitive cubic root of unity. For any integer 𝑛, we have 𝜔=𝜔,𝑎𝑛(3).wheremod

Let us consider a few examples where we can apply this property to compute different powers of a primitive cubic root of unity.

Example 4: Evaluating Powers of the Cube Root of Unity

Write 𝜔 in its simplest form, where 𝜔 is a primitive cube root of unity.

Answer

We recall that for any primitive cubic root of unity 𝜔, any integer power satisfies the property 𝜔=𝜔,𝑎𝑛(3).wheremod

In this example, we need to take the 11th power of a primitive cube root of unity; hence, 𝑛=11. Since 11=3×3+2, we have 211(3).mod

This tells us 𝑎=2. Substituting this value into the property above, we have 𝜔=𝜔.

We can also apply this property for negative integer powers, as we will see in the next example.

Example 5: Simplifying Expressions Involving Cube Roots of Unity

Write 𝜔 in its simplest form, where 𝜔 is a primitive cube root of unity.

Answer

We recall that for any primitive cubic root of unity 𝜔, any integer power satisfies the property 𝜔=𝜔,𝑎𝑛(3).wheremod

In this example, we need to take the 149th power of a primitive cube root of unity; hence, 𝑛=149. We note 149=150+1=3×(50)+1, which leads to 1491(3).mod

This tells us 𝑎=1. Substituting this value into the property above, we have 𝜔=𝜔=𝜔.

In the previous two examples, we computed integer powers of the primitive cubic roots of unity. Let us consider the other property of primitive cubic roots of unity. If 𝜔 is a primitive cubic root of unity, we know that 𝜔=1, which means 𝜔1=0.

Using the difference of cubes formula mentioned earlier, we can write this equation as (𝜔1)𝜔+𝜔+1=0.

We know that 1 is not a primitive cubic root of unity; hence, we must have 𝜔+𝜔+1=0.

This leads to a useful property that can be used to simplify a polynomial expression in 𝜔.

Property: Primitive Cubic Roots of Unity

A primitive cubic root of unity 𝜔 must satisfy 𝜔+𝜔+1=0.

Let us consider a few examples where we will apply this property to simplify polynomial expressions in 𝜔.

Example 6: Evaluating the Negative Powers of an Expression Involving Cube Roots of Unity

Evaluate 1+𝜔, where 𝜔 is a primitive cube root of unity.

Answer

We recall that a primitive cubic root of unity 𝜔 satisfies 𝜔+𝜔+1=0.

The left-hand side of this equation resembles the expression given inside the parentheses. We can rearrange this equation to write 1+𝜔=𝜔.

Substituting this expression, 1+𝜔=(𝜔).

Since 133 is an odd power, we can factor out the negative sign to write (𝜔)=𝜔.

Now, we need to compute the power 𝜔. We recall that for any primitive cubic root of unity 𝜔, any integer power satisfies the property 𝜔=𝜔,𝑎𝑛(3).wheremod

We need to take the 133rd power of a primitive cube root of unity; hence, 𝑛=133. We note 133=135+2=3×(45)+2, which leads to 1332(3).mod

This tells us 𝑎=2. Substituting this value into the property above, we have 𝜔=𝜔.

Substituting this expression above, we have 1+𝜔=𝜔.

In the previous example, we applied the identity 𝜔+𝜔+1=0, which holds for any primitive (or complex) cubic root of unity. The key to reducing expressions in 𝜔 is to identify parts of the given expression that resemble the left-hand side of this equation. Sometimes, the resemblance is not immediately obvious from the given expression and we need to manipulate the given expression before we can see how to apply this identity.

In the next example, we will manipulate the given expression before applying this identity to reduce a given expression in 𝜔.

Example 7: Evaluating Expressions Involving Cubic Roots of Unity

Evaluate 8𝜔+15+5𝜔+1𝜔 where 𝜔 is a complex cube root of unity.

Answer

We begin by recalling that a complex cubic root of unity 𝜔 satisfies the identity 𝜔+𝜔+1=0.

In this example, we will apply this identity to evaluate the given expression.

First, we notice that in the denominator of the fraction in the first set of parentheses is the term 𝜔+1. We can rearrange the identity for complex roots of unity to write 𝜔+1=𝜔.

This means that the expression in the first set of parentheses can be written as 8𝜔+1=8𝜔=8𝜔.

Let us now consider the expression in the second set of parentheses. Here, the first two terms are 5+5𝜔, which can be written as 5(1+𝜔). Using the same identity, we can replace 1+𝜔 with 𝜔 to write this expression as 5𝜔. Then, 5+5𝜔+1𝜔=5𝜔+1𝜔.

We can add these two numbers by making the common denominator 𝜔. This gives 5𝜔+1𝜔=5𝜔𝜔+1𝜔=5𝜔+1𝜔.

Since 𝜔 is a cube root of unity, we know that 𝜔=1. This reduces the expression in the second set of parentheses to 5+5𝜔+1𝜔=4𝜔.

Multiplying these expressions, we obtain 8𝜔+15+5𝜔+1𝜔=8𝜔×4𝜔=32𝜔.

Finally, since 𝜔=1, the given expression is equal to 32.

In our final example, we will simplify a polynomial expression in 𝜔 using this identity.

Example 8: Evaluating Expressions Involving Cubic Roots of Unity

Evaluate 9𝜔+9𝜔+6+6𝜔+6𝜔, where 𝜔 is a nontrivial cubic root of unity.

Answer

We recall that there are three cubic roots of unity. Among the three cubic roots of unity, 1 is called the trivial root of unity and the other two roots are called the nontrivial, or complex, roots of unity. For the cubic roots of unity, we also know that a complex root of unity is also a primitive root of unity. We recall the properties of primitive cubic roots of unity, which we will use to simplify the given expression. Any primitive cubic root of unity 𝜔 satisfies 𝜔=𝜔,𝑎𝑛(3)𝑛,𝜔+𝜔+1=0.wheremodforanyinteger

We begin by replacing higher powers of 𝜔 with their equivalent power between 0 and 2 using the first property for integer powers. In the given expression, we have 𝜔, which means 𝑛=4. Since we know 4=1×3+1, we obtain 41(3)mod, leading to 𝑎=1. This gives 𝜔=𝜔=𝜔.

Then, the given expression can be written as 9𝜔+9𝜔+6+6𝜔+6𝜔.

We can rearrange the terms in the first set of parentheses to write 9𝜔+9𝜔=𝜔+9(𝜔+1).

From the second property of primitive cubic roots of unity, we can rearrange the equation to write 𝜔+1=𝜔. This means 𝜔+9(𝜔+1)=𝜔+9𝜔=10𝜔.

Hence, 9𝜔+9𝜔=10𝜔=100𝜔=100𝜔, where we used the identity 𝜔=𝜔 for the last equality. This simplifies the first term in the given expression.

Let us now consider the second term, which is 6+6𝜔+6𝜔. Since 6 is a common factor inside the parentheses, we can write this term as 61+𝜔+𝜔.

By the property of primitive cubic roots of unity, we know that 𝜔+𝜔+1=0. This means that 61+𝜔+𝜔=0.

Hence, 9𝜔+9𝜔+6+6𝜔+6𝜔=100𝜔.

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

Key Points

  • The cube roots of unity are the complex-valued solutions of the equation 𝑧=1. The three cube roots of unity are
    • in Cartesian form, 𝑧1,12+32𝑖,1232𝑖,
    • in polar form, 𝑧1,2𝜋3+𝑖2𝜋3,2𝜋3+𝑖2𝜋3cossincossin,
    • in exponential form, 𝑧1,𝑒,𝑒.
  • The real root 1 is called the trivial cubic root of unity, and the other roots are called the nontrivial or complex cubic roots of unity.
  • A complex cubic root of unity satisfies 𝜔=𝜔.
  • A primitive cubic root of unity is a cubic root of unity 𝜔 for which 𝑘=3 is the smallest positive integer for which 𝜔=1. In particular, all complex cubic roots of unity are also primitive cubic roots of unity.
  • Let 𝜔 be a primitive cubic root of unity. For any integer 𝑛, we have 𝜔=𝜔,𝑎𝑛(3).wheremod
  • A primitive cubic root of unity 𝜔 must satisfy 𝜔+𝜔+1=0.

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