Lesson Explainer: Arbitrary Roots of Complex Numbers | Nagwa Lesson Explainer: Arbitrary Roots of Complex Numbers | Nagwa

Lesson Explainer: Arbitrary Roots of Complex Numbers Mathematics

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

We are interested in finding the complex-valued solutions 𝑧 to equations of the form 𝑧=π‘§οŠοŠ¦, where 𝑛 is a positive integer and π‘§οŠ¦ is a given complex number. Solutions to equations in this form are known as the 𝑛th roots of π‘§οŠ¦. In particular, when 𝑧=1, we recall that the equation 𝑧=1 has the 𝑛 distinct complex-valued solutions 𝑧, which are called the 𝑛th roots of unity. In this explainer, we want to replace the right-hand side of this equation with a general complex number π‘§οŠ¦ and find the roots of an arbitrary complex number.

Let us begin with an example where we will compute the square root of a complex number using algebraic methods.

Example 1: Finding the Square Roots of Complex Numbers in Cartesian Form

Given that 𝑧=βˆ’28+96𝑖, determine the square roots of 𝑧 without first converting it to trigonometric form.

Answer

In this example, we need to compute the square roots of a complex number without converting it to trigonometric form. We can begin by denoting a square root of 𝑧 by π‘₯+𝑦𝑖 for real-valued variables π‘₯ and 𝑦. Since this complex number is a square root of 𝑧, we can write (π‘₯+𝑦𝑖)=βˆ’28+96𝑖.

We can distribute the square on the left-hand side of this equation as π‘₯+(𝑦𝑖)+2π‘₯𝑦𝑖=π‘₯βˆ’π‘¦+2π‘₯𝑦𝑖.

Hence, ο€Ήπ‘₯βˆ’π‘¦ο…+2π‘₯𝑦𝑖=βˆ’28+96𝑖.

Recall that two complex numbers are equal to each other if the real and imaginary parts of the complex numbers are equal. Hence, the equation above leads to π‘₯βˆ’π‘¦=βˆ’28,2π‘₯𝑦=96.

While these two equations are sufficient to finding the two unknowns π‘₯ and 𝑦, these simultaneous equations are messy. Instead, we can use another equation involving π‘₯ and 𝑦. We recall that the property of the modulus tells us, for any complex number 𝑀, ||𝑀||=|𝑀|. Since we have (π‘₯+𝑦𝑖)=π‘§οŠ¨, this tells us |π‘₯+𝑦𝑖|=|βˆ’28+96𝑖|.

We recall that the modulus of a complex number π‘Ž+𝑏𝑖 is given by βˆšπ‘Ž+π‘οŠ¨οŠ¨. Hence, ο€»βˆšπ‘₯+𝑦=(βˆ’28)+96π‘₯+𝑦=100.

Now, we can use the simultaneous equations π‘₯+𝑦=100,π‘₯βˆ’π‘¦=βˆ’28.

Adding the two equations leads to 2π‘₯=72, which is the same as π‘₯=36. Hence, π‘₯=Β±6. Subtracting these two equations leads to 2𝑦=128, which means 𝑦=64; hence, 𝑦=Β±8.

At first sight, it appears that we have four solutions, since π‘₯=Β±6 and 𝑦=Β±8. However, we should remember that π‘₯ and 𝑦 should still satisfy our earlier equation 2π‘₯𝑦=96. In particular, the product of π‘₯ and 𝑦 should be positive. This restricts the pairs of our solutions to (π‘₯,𝑦)=(6,8)(βˆ’6,βˆ’8).or

We can verify that both of these pairs satisfy the equation 2π‘₯𝑦=96. This gives us the roots 6+8𝑖 and βˆ’6βˆ’8𝑖.

Thus, the square roots of βˆ’28+96𝑖 are (6+8𝑖),βˆ’(6+8𝑖).

In the previous example, we used the algebraic method to find square roots of a given complex number. While this method works well for finding square roots, it cannot be generalized to find roots of higher powers. For computing roots of higher power, it is better to convert a complex number to the polar or exponential form and apply de Moivre’s theorem for roots.

Theorem: De Moivre’s Theorem for Roots

For a complex number in polar form 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin, the 𝑛th roots are 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

Equivalently, for a complex number in exponential form 𝑧=π‘Ÿπ‘’οƒοΌ, its 𝑛th roots can be written as ο‘ƒο‘΅οŽ©οŽ‘ο‘½ο‘€ο‘ƒβˆšπ‘Ÿπ‘’π‘˜=0,1,…,π‘›βˆ’1.for

In the next example, we will apply de Moivre’s theorem to find the square roots of a given complex number.

Example 2: Finding the Square Roots of Complex Numbers Using De Moivre’s Theorem

Use de Moivre’s theorem to find the two square roots of 16ο€Ό5πœ‹3βˆ’π‘–5πœ‹3cossin.

Answer

In this example, we need to compute a square root of a complex number given in polar form. We recall 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 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‰π‘˜=0,1,…,π‘›βˆ’1.cossinfor

We can use de Moivre’s theorem for computing the roots, but first, we need to make sure that we are starting with the correct form π‘Ÿ(πœƒ+π‘–πœƒ)cossin, which is the polar form of a complex number. The provided form 16ο€Ό5πœ‹3βˆ’π‘–5πœ‹3cossin is similar to this form, but it differs from the polar form due to the negative sign inside the parentheses. To remedy this difference, we recall the even/odd identities for the sine and cosine functions sinsincoscos(βˆ’πœƒ)=βˆ’πœƒ,(βˆ’πœƒ)=πœƒ.

Hence, we can rewrite the given form of the complex number as 16ο€Ό5πœ‹3βˆ’π‘–5πœ‹3=16ο€Ό5πœ‹3+π‘–ο€Όβˆ’5πœ‹3=16ο€Όο€Όβˆ’5πœ‹3+π‘–ο€Όβˆ’5πœ‹3.cossincossincossin

Now that we have the polar form of our complex number, we can use the values π‘Ÿ=16 and πœƒ=βˆ’5πœ‹3. Also, since we are looking for the square root, we can use 𝑛=2, which means π‘˜=0,1. Substituting these values into de Moivre’s theorem above, the square roots are given by √16ο€Όο€Όβˆ’5πœ‹6+π‘–ο€Όβˆ’5πœ‹6,√16ο€»ο€»πœ‹6+π‘–ο€»πœ‹6.cossincossin

Let us convert these square roots to the Cartesian form. We know that √16=4. Also, using the unit circle, we can find the trigonometric ratios cossincossinο€Όβˆ’5πœ‹6=βˆ’βˆš32,ο€Όβˆ’5πœ‹6=βˆ’12,ο€»πœ‹6=√32,ο€»πœ‹6=12.

Substituting these values into the roots, we have 4ο€Ώβˆ’βˆš32βˆ’π‘–12,4ο€Ώβˆš32+𝑖12.

Multiplying through the parentheses, we obtain the square roots βˆ’2√3βˆ’π‘–2 and 2√3+𝑖2.

Hence, the two square roots of 16ο€Ό5πœ‹3βˆ’π‘–5πœ‹3cossin are Β±ο€»2√3+2𝑖.

In the previous example, we found the square roots of a given complex number using de Moivre’s theorem for roots. This method can be generalized for roots of higher powers since de Moivre’s theorem can apply for general roots.

In the next equation, we will find the quintic roots of a complex number and plot the roots on an Argand diagram.

Example 3: Roots of a Complex Number

  1. Solve 𝑧=16√2+16π‘–βˆš2.
  2. By plotting these solutions on an Argand diagram, or otherwise, describe the geometric properties of the solutions.

Answer

Part 1

In this part, we need to compute a quintic root of the complex number 16√2+16π‘–βˆš2, which is the Cartesian form of the complex number. We recall de Moivre’s theorem for roots, which states that for a complex number in exponential form 𝑧=π‘Ÿπ‘’οƒοΌ, the 𝑛th roots of 𝑧 are given by ο‘ƒο‘΅οŽ©οŽ‘ο‘½ο‘€ο‘ƒβˆšπ‘Ÿπ‘’π‘˜=0,1,…,π‘›βˆ’1.for

We can use de Moivre’s theorem to find the quintic roots, but first, we need to convert the given complex number into the exponential form. Recall that the exponential form of a complex number with modulus π‘Ÿ and argument πœƒ is π‘Ÿπ‘’οƒοΌ.

We begin by finding the modulus and argument of 16√2+16π‘–βˆš2 so we can express it in polar form. Recall that the modulus of a complex number π‘Ž+𝑏𝑖 is given by βˆšπ‘Ž+π‘οŠ¨οŠ¨. We can find the modulus of the given complex number by substituting π‘Ž=16√2 and 𝑏=16√2, which gives ο„žο€»16√2+ο€»16√2=√1024=32.

Hence, π‘Ÿ=32. Also, we recall that the argument of a complex number π‘Ž+𝑏𝑖 lying in the first quadrant of an Argand diagram is given by arctanπ‘π‘Ž. Since π‘Ž=16√2 and 𝑏=16√2 are both positive for our given complex number, we know that our complex number lies in the first quadrant. Then, its argument is given by arctanarctanο€Ώ16√216√2=(1)=πœ‹4.

Hence, πœƒ=πœ‹4. Then, the exponential form of 16√2+16π‘–βˆš2 is 32𝑒.ο‘½οŽ£οƒ

We can now find the solutions of the equation 𝑧=32𝑒.οŠ«οƒο‘½οŽ£

We can find the quintic roots by applying de Moivre’s theorem for roots with 𝑛=5, π‘Ÿ=32, πœƒ=πœ‹4, which gives 32π‘’π‘˜=0,1,…4.οŽ οŽ€ο‘½οŽ£οŽ©οŽ‘ο‘½ο‘€οŽ€οƒfor

Hence, considering each value of π‘˜, the solutions are 2𝑒,2𝑒,2𝑒,2𝑒,2𝑒.ο‘½οŽ‘οŽŸοŽ¨ο‘½οŽ‘οŽŸοŽ οŽ¦ο‘½οŽ‘οŽŸοŽ‘οŽ€ο‘½οŽ‘οŽŸοŽ’οŽ’ο‘½οŽ‘οŽŸοƒοƒοƒοƒοƒ

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 have arguments 25πœ‹20 and 33πœ‹20, which do not lie in this range. Since these arguments are over the upper bound πœ‹, we can obtain an equivalent argument by subtracting the full revolution 2πœ‹ radians from this value: 25πœ‹20βˆ’2πœ‹=βˆ’15πœ‹20=βˆ’3πœ‹4,33πœ‹20βˆ’2πœ‹=βˆ’7πœ‹20.

Using these arguments in the standard range, the last two quintic roots can be written as 𝑒οŽͺοŽ’ο‘½οŽ£ and 𝑒οŽͺοŽ¦ο‘½οŽ‘οŽŸ. Hence, the quintic roots of 16√2+16π‘–βˆš2 are 2𝑒,2𝑒,2𝑒,2𝑒,2𝑒.ο‘½οŽ‘οŽŸοŽ¨ο‘½οŽ‘οŽŸοŽ οŽ¦ο‘½οŽ‘οŽŸοŽͺοŽ¦ο‘½οŽ‘οŽŸοŽͺοŽ’ο‘½οŽ£οƒοƒοƒοƒοƒ

Part 2

In this part, we need to plot the quintic roots obtained in the previous part on an Argand diagram. The quintic roots in the previous part are in exponential form, and we recall that a complex number in exponential form π‘Ÿπ‘’οƒοΌ has modulus π‘Ÿ and argument πœƒ. Observing the quintic roots from the previous parts, we can see that the modulus of all quintic roots is equal to 2. This means that all of the quintic roots lie on a circle centered at the origin with radius 2 in an Argand diagram.

We can also see that their arguments form an arithmetic sequence βˆ’3πœ‹4,βˆ’7πœ‹20,πœ‹20,9πœ‹20,17πœ‹20.

Starting with the initial term βˆ’3πœ‹4, the arguments increase by the common difference 8πœ‹20=2πœ‹5radians. This means that beginning with the point on the circle at the angle βˆ’3πœ‹4 radians, which is a clockwise angle of 3πœ‹4 radians from the positive real axis of an Argand diagram, we can rotate counterclockwise on the circle by 2πœ‹5 four times to obtain all the quintic roots.

Below is a plot of these complex numbers on an Argand diagram.

From the Argand diagram above, we can see that the roots lie at the vertices of a regular pentagon inscribed in a circle of radius 2 at the origin.

In the previous example, we found the quintic roots of a complex number and observed that these roots form vertices of a regular polygon inscribed in the circle in an Argand diagram. We recall that this exact property is also true for the 𝑛th roots of unity, where one vertex is at the trivial root of unity 1.

In the next example, we will demonstrate the relationship between the 𝑛th roots of unity and arbitrary roots of a complex number.

Example 4: Relation between Arbitrary Roots and Roots of Unity

  1. Find the solutions to the equation 𝑧=125π‘’οŠ¬οƒοŽ‘ο‘½οŽ’. What are their geometrical properties?
  2. State the 6th roots of unity.
  3. What is the relationship between the 6th roots of unity and the solutions to the equation 𝑧=125π‘’οŠ¬οƒοŽ‘ο‘½οŽ’?

Answer

Part 1

We know that the solutions of equation 𝑧=125π‘’οŠ¬οƒοŽ‘ο‘½οŽ’ are the sixth roots of the complex number on the right-hand side of the equation. We recall de Moivre’s theorem for roots, which states that the 𝑛th roots of a complex number 𝑧=π‘Ÿπ‘’οƒοΌ are given by ο‘ƒο‘΅οŽ©οŽ‘ο‘½ο‘€ο‘ƒβˆšπ‘Ÿπ‘’π‘˜=0,1,…,π‘›βˆ’1.for

Applying de Moivre’s theorem with 𝑛=6, the roots of the equation are given by οŽ₯οŽ‘ο‘½οŽ’οŽ©οŽ‘ο‘½ο‘€οŽ₯√125π‘’π‘˜=0,1,…,5.for

We know that 125=5, which gives οŽ₯√125=√5. Substituting in each value of π‘˜ and simplifying so that the arguments of each complex number lies in the standard range ]βˆ’πœ‹,πœ‹], we have √5𝑒,√5𝑒,√5𝑒,√5𝑒,√5𝑒,√5𝑒.ο‘½οŽ¨οŽ£ο‘½οŽ¨οŽ¦ο‘½οŽ¨οŽͺοŽ‘ο‘½οŽ¨οŽͺοŽ€ο‘½οŽ¨οŽͺοŽ§ο‘½οŽ¨οƒοƒοƒοƒοƒοƒ

Now, let us plot these numbers on an Argand diagram. We recall that a complex number in exponential form π‘Ÿπ‘’οƒοΌ has modulus π‘Ÿ and argument πœƒ. Observing the sixth roots above, we can see that the modulus of all roots is equal to √5. This means that all the sixth roots lie on a circle centered at the origin with radius √5 in an Argand diagram.

We can also see that their arguments form an arithmetic sequence βˆ’8πœ‹9,βˆ’5πœ‹9,βˆ’2πœ‹9,πœ‹9,4πœ‹9,7πœ‹9.

Starting with the initial term βˆ’8πœ‹9, the arguments increase by the common difference 3πœ‹9=πœ‹3radians. This means that beginning with the point on the circle at the angle βˆ’8πœ‹9 radians, which is a clockwise angle of 8πœ‹9 radians from the positive real axis of an Argand diagram, we can rotate counterclockwise on the circle by πœ‹3 five times to obtain all the sixth roots.

Plotting these roots on an Argand diagram, we see that they lie at the vertices of a regular hexagon centered at the origin, inscribed in a circle of radius √5.

Part 2

Recall that the 𝑛th roots of unity in exponential form are given by 1,𝑒,…,𝑒.οŽ‘ο‘½ο‘ƒοŽ‘ο‘½(οŽͺ)

Substituting 𝑛=6 and finding equivalent arguments in the range ]βˆ’πœ‹,πœ‹], we obtain the sixth roots of unity: 1,𝑒,𝑒,βˆ’1,𝑒,𝑒.ο‘½οŽ’οŽ‘ο‘½οŽ’οŽͺο‘½οŽ’οŽͺοŽ‘ο‘½οŽ’οƒοƒοƒοƒ

Part 3

Geometrically, we recall that the sixth roots of unity form vertices of a regular hexagon inscribed in the unit circle with one vertex at the real number 1. Comparing this property with the figure of the sixth root given in the first part, we can see that the roots are dilated by scale factor √5 and rotated counterclockwise by πœ‹9 radians.

We recall the geometric property for multiplication of a pair of complex numbers: Given a pair of nonzero complex numbers π‘§οŠ§ and π‘§οŠ¨,

  • |𝑧𝑧|=|𝑧||𝑧|,
  • argargarg(𝑧𝑧)=𝑧+π‘§οŠ§οŠ¨οŠ§οŠ¨.

Hence, dilating a complex number by a scale factor of √5 and rotating counterclockwise by πœ‹9 radians is equivalent to multiplying a complex number by another number that has modulus √5 and argument πœ‹9. This number can be written in exponential form as √5π‘’ο‘½οŽ¨οƒ.

This tells us that we can obtain the sixth root of our number by multiplying the sixth roots of unity by √5π‘’ο‘½οŽ¨οƒ. We can verify this by using the property of multiplication of complex numbers in polar form. We recall that π‘Ÿπ‘’Γ—π‘Ÿπ‘’=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

Thus, multiplying each of the sixth roots of unity by √5π‘’ο‘½οŽ¨οƒ, 1Γ—βˆš5𝑒=√5𝑒,π‘’Γ—βˆš5𝑒=√5𝑒,π‘’Γ—βˆš5𝑒=√5𝑒,βˆ’1Γ—βˆš5𝑒=βˆ’βˆš5𝑒=√5𝑒,π‘’Γ—βˆš5𝑒=√5𝑒,π‘’Γ—βˆš5𝑒=√5𝑒.ο‘½οŽ¨ο‘½οŽ¨ο‘½οŽ’ο‘½οŽ¨οŽ£ο‘½οŽ¨οŽ‘ο‘½οŽ’ο‘½οŽ¨οŽ¦ο‘½οŽ¨ο‘½οŽ¨ο‘½οŽ¨οŽͺοŽ§ο‘½οŽ¨οŽͺο‘½οŽ’ο‘½οŽ¨οŽͺοŽ‘ο‘½οŽ¨οŽͺοŽ‘ο‘½οŽ’ο‘½οŽ¨οŽͺοŽ€ο‘½οŽ¨οƒοƒοƒοƒοƒοƒοƒοƒοƒοƒοƒοƒοƒοƒοƒοƒοƒ

Hence, the solutions to the equation are the 6th roots of unity multiplied by √5π‘’ο‘½οŽ¨οƒ.

In previous examples, we noticed that some 𝑛th roots of a complex number formed the vertices of a regular hexagon centered at the origin. We can generalize this observation to any 𝑛th roots of a complex number. De Moivre’s theorem for roots tells us that the 𝑛th roots of a complex number in exponential form 𝑧=π‘Ÿπ‘’οƒοΌ are given by ο‘ƒο‘΅οŽ©οŽ‘ο‘½ο‘€ο‘ƒβˆšπ‘Ÿπ‘’π‘˜=0,1,…,π‘›βˆ’1.for

From this expression, we can see that all these 𝑛th roots have the same modulus, which is ο‘ƒβˆšπ‘Ÿ. This tells us that all these complex numbers lie on a circle centered at the origin with radius ο‘ƒβˆšπ‘Ÿ of an Argand diagram. Also, we can see that the arguments of these complex numbers form an arithmetic sequence with 𝑛 terms, with the initial term πœƒπ‘› and the common difference 2πœ‹π‘›. To plot these roots on an Argand diagram, we can find the first point with argument πœƒπ‘› on the circle centered at the origin with radius ο‘ƒβˆšπ‘Ÿ. Then, we can rotate counterclockwise π‘›βˆ’1 times by 2πœ‹π‘› to plot the remaining roots. We note that this method will always result in a regular polygon inscribed in the circle. We will summarize this fact below.

Property: Arbitrary Roots of a Complex Number in Argand Diagrams

In an Argand diagram, the 𝑛 roots of a complex number with modulus π‘Ÿ and argument πœƒ form the vertices of a regular 𝑛-gon inscribed in a circle of radius ο‘ƒβˆšπ‘Ÿ, where one of the vertices is the point on the circle with argument πœƒπ‘›.

Using the same reasoning as the previous example, we can use this geometric property to relate arbitrary 𝑛th roots of a complex number to the 𝑛th roots of unity. We know that the 𝑛th roots of unity form a regular 𝑛-gon inscribed in the unit circle, where one vertex is at the trivial root of unity 1. Hence, the 𝑛-gon representing arbitrary roots for a complex number with modulus π‘Ÿ and argument πœƒ can be obtained by dilating the 𝑛-gon from the 𝑛th roots of unity with scale factor ο‘ƒβˆšπ‘Ÿ and rotating counterclockwise by the angle πœƒπ‘› radians. This is the effect given by multiplying each 𝑛th root of unity by the complex number ο‘ƒο‘΅ο‘ƒβˆšπ‘Ÿπ‘’οƒ.

This leads to the following statement.

Property: Arbitrary Roots of a Complex Number and 𝑛th Roots of Unity

The 𝑛th roots of a complex number with modulus π‘Ÿ and argument πœƒ can be obtained by multiplying each 𝑛th root of unity by ο‘ƒο‘΅ο‘ƒβˆšπ‘Ÿπ‘’οƒ.

Let us consider a geometric application of the property for arbitrary roots of a complex number.

Example 5: Coordinates of Regular Polygons at the Origin

Find the coordinates of the vertices of a regular pentagon centered at the origin with one vertex at (3,3). Give your answer as exact Cartesian coordinates.

Answer

In this problem, we need to identify the Cartesian coordinates of the vertices of a regular polygon. We can solve this problem by using the geometric property of arbitrary roots of a complex number. Recall that, in an Argand diagram, the 𝑛 roots of a complex number with modulus π‘Ÿ and argument πœƒ form the vertices of a regular 𝑛-gon inscribed in a circle of radius ο‘ƒβˆšπ‘Ÿ, where one of the vertices is the point on the circle with argument πœƒπ‘›.

To solve this problem, let us assume that the regular pentagon is on an Argand plane rather than on a Cartesian plane, where the Cartesian coordinates (π‘₯,𝑦) correspond to the complex number π‘₯+𝑦𝑖 in the Argand diagram. We will first find the complex numbers corresponding to the five vertices of the regular pentagon on the Argand diagram. Then, we can convert the complex numbers to coordinates of the Cartesian plane using this relationship. Since we have a regular pentagon centered at the origin, this pentagon can be inscribed in a circle.

According to the property of arbitrary roots of a complex number, the vertices of this pentagon on an Argand diagram correspond to quintic roots of a complex number. Let us find this complex number.

We are given that one vertex has the Cartesian coordinates (3,3), which corresponds to the complex number 3+3𝑖 in the Argand diagram. This tells us that 3+3𝑖 is a quintic root of our complex number and the other vertices of the pentagon represent the other quintic roots of the same number. We can find the other quintic roots by computing (3+3𝑖) and then applying de Moivre’s theorem for roots to take the quintic root of this number, but it is simpler to use the property of roots of complex numbers.

We know that there are 𝑛 distinct complex numbers, which are the 𝑛th roots of a given complex number. The moduli of all 𝑛th roots are the same, and the arguments of the 𝑛th roots of a complex number form an arithmetic sequence with common difference 2πœ‹π‘›. In this example, we know that one of the quintic roots of a complex number is 3+3𝑖. Thus, the modulus of this complex number is also the modulus of the other four quintic roots of the same complex number. Also, starting with the argument of 3+3𝑖, we can form an arithmetic sequence with common difference 2πœ‹5 of 5 terms to obtain the arguments of the other four quintic roots.

Let us find the modulus and argument of 3+3𝑖. Recall that the modulus of a complex number π‘Ž+𝑏𝑖 is given by βˆšπ‘Ž+π‘οŠ¨οŠ¨. We can find the modulus of the given complex number by substituting π‘Ž=3 and 𝑏=3, which gives √3+3=√18=3√2.

Hence, the modulus of 3+3𝑖 is 3√2, which is also the modulus of the other four quintic roots.

Next, let us compute the argument of this number. We recall that the argument of a complex number π‘Ž+𝑏𝑖 lying in the first quadrant of an Argand diagram is given by arctanπ‘π‘Ž. Since π‘Ž=3 and 𝑏=3 are both positive for our given complex number, we know that our complex number lies in the first quadrant. Then, its argument is given by arctanarctanο€Ό33=(1)=πœ‹4.

Hence, the argument of 3+3𝑖 is πœ‹4 radians. We can compute the arguments of the other four quintic roots by forming an arithmetic sequence starting with this argument with common difference 2πœ‹5. We can write πœ‹4,13πœ‹20,21πœ‹20,29πœ‹20,37πœ‹20.

We recall that the argument of a complex number, by convention, should lie in the standard range ]βˆ’πœ‹,πœ‹]. The last three arguments are larger than πœ‹, so we will subtract the full revolution 2πœ‹ radians from these arguments to write the equivalent arguments βˆ’19πœ‹20, βˆ’11πœ‹20, and βˆ’3πœ‹20 respectively.

Finally, we recall that a complex number with modulus π‘Ÿ and argument πœƒ can be expressed in polar form π‘Ÿ(πœƒ+π‘–πœƒ).cossin

Then, the quintic roots are 3√2ο€»ο€»πœ‹4+π‘–ο€»πœ‹4,3√2ο€Όο€Ό13πœ‹20+𝑖13πœ‹20,3√2ο€Όο€Όβˆ’19πœ‹20+π‘–ο€Όβˆ’19πœ‹20,3√2ο€Όο€Όβˆ’11πœ‹20+π‘–ο€Όβˆ’11πœ‹20,3√2ο€Όο€Όβˆ’3πœ‹20+π‘–ο€Όβˆ’3πœ‹20.cossincossincossincossincossin

The first root can be simplified by using cossinπœ‹4=πœ‹4=√22. When we substitute these values into the first root, we obtain 3√2ο€Ώβˆš22+π‘–βˆš22=3+3𝑖, which is the first root that we began with. The other roots do not simplify since their arguments do not belong to the special angles of the unit circle. We can multiply through the parentheses for each of these roots to write them as 3+3𝑖,3√2ο€Ό13πœ‹20+3√2𝑖13πœ‹20,3√2ο€Όβˆ’19πœ‹20+3√2π‘–ο€Όβˆ’19πœ‹20,3√2ο€Όβˆ’11πœ‹20+3√2π‘–ο€Όβˆ’11πœ‹20,3√2ο€Όβˆ’3πœ‹20+3√2π‘–ο€Όβˆ’3πœ‹20.cossincossincossincossin

These are vertices of the regular pentagon on the Argand diagram. We can find the equivalent Cartesian coordinates of these points by relating a complex number π‘₯+𝑦𝑖 with Cartesian coordinates (π‘₯,𝑦).

Hence, the coordinates of the regular pentagon centered at the origin with one vertex at (3,3) are (3,3),ο€Ό3√2ο€Ό13πœ‹20,3√2ο€Ό13πœ‹20,ο€Ό3√2ο€Όβˆ’19πœ‹20,3√2ο€Όβˆ’19πœ‹20,ο€Ό3√2ο€Όβˆ’11πœ‹20,3√2ο€Όβˆ’11πœ‹20,ο€Ό3√2ο€Όβˆ’3πœ‹20,3√2ο€Όβˆ’3πœ‹20.cossincossincossincossin

So far, we have discussed properties of arbitrary roots of a complex number. From previous examples, it is apparent that, for any complex number 𝑧, the expression π‘§οŽ ο‘ƒ has multiple possible values. In complex numbers, we say that such expressions are multivalued.

In the final example, we will identify all possible values of a multivalued expression.

Example 6: Expressions Involving 𝑛th Roots

Find the possible real values of 1√3ο€½(𝑖)+(𝑖)ο‰οŽ οŽ’οŽͺ.

Answer

From the given expression, we note the terms (𝑖) and (𝑖)οŽͺ, which are the third roots of 𝑖 and the reciprocals. We know that, given a complex number 𝑧, there are 𝑛 distinct values, which are its 𝑛th roots. This means that the expression (𝑖) has three possible values and also that its reciprocal (𝑖)οŽͺ has three possible values. Let us first identify all possible values of these expressions.

To find the third root of 𝑖, we recall de Moivre’s theorem for roots, which states that for a complex number in exponential form 𝑧=π‘Ÿπ‘’οƒοΌ, the 𝑛th roots of 𝑧 are given by ο‘ƒο‘΅οŽ©οŽ‘ο‘½ο‘€ο‘ƒβˆšπ‘Ÿπ‘’π‘˜=0,1,…,π‘›βˆ’1.for

To apply this formula, we first need to write 𝑖 in the exponential form. Recall that the exponential form of a complex number with modulus π‘Ÿ and argument πœƒ is π‘Ÿπ‘’οƒοΌ. We know that the complex number 𝑖 has modulus 1. We also know that it lies on the positive imaginary axis in an Argand diagram, which means that its argument is πœ‹2. Hence, 𝑖=𝑒.ο‘½οŽ‘οƒ

Then, we can apply de Moivre’s theorem with 𝑛=3 to find the possible values of (𝑖): οŽ’ο‘½οŽ‘οŽ©οŽ‘ο‘½ο‘€οŽ’βˆš1π‘’π‘˜=0,1,…,π‘›βˆ’1.for

This gives (𝑖)βˆˆο­π‘’,𝑒,𝑒.οŽ οŽ’ο‘½οŽ₯οŽ€ο‘½οŽ₯οŽ¨ο‘½οŽ₯

Next, let us find all possible values of the expression (𝑖)οŽͺ. We can write (𝑖)=ο€½(𝑖);οŽͺ hence, we can find all possible values of this expression by raising each value for (𝑖) to the power βˆ’1. We recall de Moivre’s theorem for integer powers of a complex number, which states that the 𝑛th power of a complex number 𝑧=π‘Ÿπ‘’οƒοΌ is given by 𝑧=π‘Ÿπ‘’.οŠοŠοƒοŠοΌ

Applying this theorem with 𝑛=βˆ’1 for each possible outcome of (𝑖), we obtain (𝑖)βˆˆο­π‘’,𝑒,𝑒.οŽͺοŽͺο‘½οŽ₯οŽͺοŽ€ο‘½οŽ₯οŽͺοŽ’ο‘½οŽ‘οƒοƒοƒ

Hence, there are 3 possible values for (𝑖) and 3 possible values for (𝑖)οŽͺ. At first sight, it may appear that we will have 9 different possibilities for the sum of these two expressions, but the sums may have the same value.

To add a pair of complex numbers, it is easier to first convert them to Cartesian form. Recall that we can convert a complex number in exponential form π‘Ÿπ‘’οƒοΌ to Cartesian form by computing π‘Ÿπœƒ+π‘–π‘Ÿπœƒ.cossin

Hence, (𝑖)βˆˆο¬πœ‹6+π‘–πœ‹6,5πœ‹6+𝑖5πœ‹6,3πœ‹2+𝑖3πœ‹2,(𝑖)βˆˆο¬ο€»βˆ’πœ‹6+π‘–ο€»βˆ’πœ‹6,ο€Όβˆ’5πœ‹6+π‘–ο€Όβˆ’5πœ‹6,ο€Όβˆ’3πœ‹2+π‘–ο€Όβˆ’3πœ‹2.οŽͺcossincossincossincossincossincossin

We can simplify the expressions for (𝑖)οŽͺ by recalling the identities sinsincoscos(βˆ’πœƒ)=βˆ’πœƒ,(βˆ’πœƒ)=πœƒ.

We can write (𝑖)βˆˆο¬πœ‹6βˆ’π‘–πœ‹6,5πœ‹6βˆ’π‘–5πœ‹6,3πœ‹2βˆ’π‘–3πœ‹2.οŽͺcossincossincossin

Using the unit circle, we can find the trigonometric ratios cossincossincossinπœ‹6=√32,πœ‹6=12,5πœ‹6=βˆ’βˆš32,5πœ‹6=12,3πœ‹2=0,3πœ‹2=βˆ’1.

Substituting these values, we obtain (𝑖)∈√32+𝑖12,βˆ’βˆš32+𝑖12,βˆ’π‘–ο»,(𝑖)∈√32βˆ’π‘–12,βˆ’βˆš32βˆ’π‘–12,𝑖.οŽͺ

We can write out the 9 different combinations to find the possible values of (𝑖)+(𝑖)οŽͺ. We should remember to multiply the result by 1√3 in the end. Then, we can compute the sum by constructing a table where the first row has possible values of (𝑖) and the first column has possible values of (𝑖)οŽͺ:

√32+𝑖12βˆ’βˆš32+𝑖12βˆ’π‘–
√32βˆ’π‘–12√30√32βˆ’π‘–32
βˆ’βˆš32βˆ’π‘–120βˆ’βˆš3βˆ’βˆš32βˆ’π‘–32
π‘–βˆš32+𝑖32βˆ’βˆš32+𝑖320

This gives us seven distinct values for the sum, which are 0,√3,βˆ’βˆš3,√32+𝑖32,√32βˆ’π‘–32,βˆ’βˆš32+𝑖32,βˆ’βˆš32βˆ’π‘–32.

Multiplying each number by 1√3, all possible values for the given multivalued expression are 0,1,βˆ’1,12+π‘–βˆš32,12βˆ’π‘–βˆš32,βˆ’12+π‘–βˆš32,βˆ’12βˆ’π‘–βˆš32.

The question asks about the real values, so the solution will be 0,1,βˆ’1.

Let us now summarize the points we have learned in this explainer.

Key Points

  • Any nonzero complex number 𝑧 has 𝑛 distinct values for the expression π‘§οŽ ο‘ƒ, which are called the 𝑛th roots of 𝑧. We can find the 𝑛th roots of a complex number by applying de Moivre’s theorem for roots.
  • In an Argand diagram, the 𝑛 roots of a complex number with modulus π‘Ÿ and argument πœƒ form the vertices of a regular 𝑛-gon inscribed in a circle of radius ο‘ƒβˆšπ‘Ÿ, where one of the vertices is the point on the circle with argument πœƒπ‘›.
  • The 𝑛th roots of a complex number with modulus π‘Ÿ and argument πœƒ can be obtained by multiplying each 𝑛th root of unity by ο‘ƒο‘΅ο‘ƒβˆšπ‘Ÿπ‘’οƒ.

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