In this explainer, we will learn how to use de Moivreβs theorem to find the roots of a complex number and explore their properties.
We are interested in finding the complex-valued solutions to equations of the form , where is a positive integer and is a given complex number. Solutions to equations in this form are known as the roots of . In particular, when , we recall that the equation has the distinct complex-valued solutions , which are called the roots of unity. In this explainer, we want to replace the right-hand side of this equation with a general complex number 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 , determine the square roots of 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 by for real-valued variables and . Since this complex number is a square root of , we can write
We can distribute the square on the left-hand side of this equation as
Hence,
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
While these two equations are sufficient to finding the two unknowns and , these simultaneous equations are messy. Instead, we can use another equation involving and . We recall that the property of the modulus tells us, for any complex number , . Since we have , this tells us
We recall that the modulus of a complex number is given by . Hence,
Now, we can use the simultaneous equations
Adding the two equations leads to , which is the same as . Hence, . Subtracting these two equations leads to , which means ; hence, .
At first sight, it appears that we have four solutions, since and . However, we should remember that and should still satisfy our earlier equation . In particular, the product of and should be positive. This restricts the pairs of our solutions to
We can verify that both of these pairs satisfy the equation . This gives us the roots and .
Thus, the square roots of are
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 Moivreβs theorem for roots.
Theorem: De Moivreβs Theorem for Roots
For a complex number in polar form , the roots are
Equivalently, for a complex number in exponential form , its roots can be written as
In the next example, we will apply de Moivreβs theorem to find the square roots of a given complex number.
Example 2: Finding the Square Roots of Complex Numbers Using De Moivreβs Theorem
Use de Moivreβs theorem to find the two square roots of .
Answer
In this example, we need to compute a square root of a complex number given in polar form. We recall de Moivreβs theorem for roots, which states that for a complex number , the roots of are given by
We can use de Moivreβs theorem for computing the roots, but first, we need to make sure that we are starting with the correct form , which is the polar form of a complex number. The provided form 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
Hence, we can rewrite the given form of the complex number as
Now that we have the polar form of our complex number, we can use the values and . Also, since we are looking for the square root, we can use , which means . Substituting these values into de Moivreβs theorem above, the square roots are given by
Let us convert these square roots to the Cartesian form. We know that . Also, using the unit circle, we can find the trigonometric ratios
Substituting these values into the roots, we have
Multiplying through the parentheses, we obtain the square roots and .
Hence, the two square roots of are .
In the previous example, we found the square roots of a given complex number using de Moivreβs theorem for roots. This method can be generalized for roots of higher powers since de Moivreβs 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
- Solve .
- 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 , which is the Cartesian form of the complex number. We recall de Moivreβs theorem for roots, which states that for a complex number in exponential form , the roots of are given by
We can use de Moivreβs 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 and argument is .
We begin by finding the modulus and argument of so we can express it in polar form. Recall that the modulus of a complex number is given by . We can find the modulus of the given complex number by substituting and , which gives
Hence, . Also, we recall that the argument of a complex number lying in the first quadrant of an Argand diagram is given by . Since and 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
Hence, . Then, the exponential form of is
We can now find the solutions of the equation
We can find the quintic roots by applying de Moivreβs theorem for roots with , , , which gives
Hence, considering each value of , the solutions are
We recall that the argument of a complex number, by convention, should lie in the standard range . The last two quintic roots have arguments and , which do not lie in this range. Since these arguments are over the upper bound , we can obtain an equivalent argument by subtracting the full revolution radians from this value:
Using these arguments in the standard range, the last two quintic roots can be written as and . Hence, the quintic roots of are
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 has modulus and argument . 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
Starting with the initial term , the arguments increase by the common difference . This means that beginning with the point on the circle at the angle radians, which is a clockwise angle of radians from the positive real axis of an Argand diagram, we can rotate counterclockwise on the circle by 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 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 roots of unity and arbitrary roots of a complex number.
Example 4: Relation between Arbitrary Roots and Roots of Unity
- Find the solutions to the equation . What are their geometrical properties?
- State the 6th roots of unity.
- What is the relationship between the 6th roots of unity and the solutions to the equation ?
Answer
Part 1
We know that the solutions of equation are the sixth roots of the complex number on the right-hand side of the equation. We recall de Moivreβs theorem for roots, which states that the roots of a complex number are given by
Applying de Moivreβs theorem with , the roots of the equation are given by
We know that , which gives . Substituting in each value of and simplifying so that the arguments of each complex number lies in the standard range , we have
Now, let us plot these numbers on an Argand diagram. We recall that a complex number in exponential form has modulus and argument . Observing the sixth roots above, we can see that the modulus of all roots is equal to . This means that all the sixth roots lie on a circle centered at the origin with radius in an Argand diagram.
We can also see that their arguments form an arithmetic sequence
Starting with the initial term , the arguments increase by the common difference . This means that beginning with the point on the circle at the angle radians, which is a clockwise angle of radians from the positive real axis of an Argand diagram, we can rotate counterclockwise on the circle by 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 .
Part 2
Recall that the roots of unity in exponential form are given by
Substituting and finding equivalent arguments in the range , we obtain the sixth roots of unity:
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 and rotated counterclockwise by radians.
We recall the geometric property for multiplication of a pair of complex numbers: Given a pair of nonzero complex numbers and ,
- ,
- .
Hence, dilating a complex number by a scale factor of and rotating counterclockwise by radians is equivalent to multiplying a complex number by another number that has modulus and argument . This number can be written in exponential form as .
This tells us that we can obtain the sixth root of our number by multiplying the sixth roots of unity by . We can verify this by using the property of multiplication of complex numbers in polar form. We recall that
Thus, multiplying each of the sixth roots of unity by ,
Hence, the solutions to the equation are the 6th roots of unity multiplied by .
In previous examples, we noticed that some roots of a complex number formed the vertices of a regular hexagon centered at the origin. We can generalize this observation to any roots of a complex number. De Moivreβs theorem for roots tells us that the roots of a complex number in exponential form are given by
From this expression, we can see that all these roots have the same modulus, which is . This tells us that all these complex numbers lie on a circle centered at the origin with radius of an Argand diagram. Also, we can see that the arguments of these complex numbers form an arithmetic sequence with terms, with the initial term and the common difference . To plot these roots on an Argand diagram, we can find the first point with argument on the circle centered at the origin with radius . Then, we can rotate counterclockwise times by 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 roots of a complex number with modulus and argument form the vertices of a regular -gon inscribed in a circle of radius , where one of the vertices is the point on the circle with argument .
Using the same reasoning as the previous example, we can use this geometric property to relate arbitrary roots of a complex number to the roots of unity. We know that the roots of unity form a regular -gon inscribed in the unit circle, where one vertex is at the trivial root of unity 1. Hence, the -gon representing arbitrary roots for a complex number with modulus and argument can be obtained by dilating the -gon from the roots of unity with scale factor and rotating counterclockwise by the angle radians. This is the effect given by multiplying each root of unity by the complex number .
This leads to the following statement.
Property: Arbitrary Roots of a Complex Number and πth Roots of Unity
The roots of a complex number with modulus and argument can be obtained by multiplying each root of unity by .
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 . 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 roots of a complex number with modulus and argument form the vertices of a regular -gon inscribed in a circle of radius , where one of the vertices is the point on the circle with argument .
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 correspond to the complex number 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 , which corresponds to the complex number in the Argand diagram. This tells us that 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 and then applying de Moivreβs 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 distinct complex numbers, which are the roots of a given complex number. The moduli of all roots are the same, and the arguments of the roots of a complex number form an arithmetic sequence with common difference . In this example, we know that one of the quintic roots of a complex number is . 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 , we can form an arithmetic sequence with common difference of 5 terms to obtain the arguments of the other four quintic roots.
Let us find the modulus and argument of . Recall that the modulus of a complex number is given by . We can find the modulus of the given complex number by substituting and , which gives
Hence, the modulus of is , 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 lying in the first quadrant of an Argand diagram is given by . Since and 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
Hence, the argument of is radians. We can compute the arguments of the other four quintic roots by forming an arithmetic sequence starting with this argument with common difference . We can write
We recall that the argument of a complex number, by convention, should lie in the standard range . The last three arguments are larger than , so we will subtract the full revolution radians from these arguments to write the equivalent arguments , , and respectively.
Finally, we recall that a complex number with modulus and argument can be expressed in polar form
Then, the quintic roots are
The first root can be simplified by using . When we substitute these values into the first root, we obtain 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
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 with Cartesian coordinates .
Hence, the coordinates of the regular pentagon centered at the origin with one vertex at are
So far, we have discussed properties of arbitrary roots of a complex number. From previous examples, it is apparent that, for any complex number , the expression 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 πth Roots
Find the possible real values of .
Answer
From the given expression, we note the terms and , which are the third roots of and the reciprocals. We know that, given a complex number , there are distinct values, which are its roots. This means that the expression has three possible values and also that its reciprocal has three possible values. Let us first identify all possible values of these expressions.
To find the third root of , we recall de Moivreβs theorem for roots, which states that for a complex number in exponential form , the roots of are given by
To apply this formula, we first need to write in the exponential form. Recall that the exponential form of a complex number with modulus and argument is . We know that the complex number has modulus 1. We also know that it lies on the positive imaginary axis in an Argand diagram, which means that its argument is . Hence,
Then, we can apply de Moivreβs theorem with to find the possible values of :
This gives
Next, let us find all possible values of the expression . We can write hence, we can find all possible values of this expression by raising each value for to the power . We recall de Moivreβs theorem for integer powers of a complex number, which states that the power of a complex number is given by
Applying this theorem with for each possible outcome of , we obtain
Hence, there are 3 possible values for and 3 possible values for . 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 to Cartesian form by computing
Hence,
We can simplify the expressions for by recalling the identities
We can write
Using the unit circle, we can find the trigonometric ratios
Substituting these values, we obtain
We can write out the 9 different combinations to find the possible values of . We should remember to multiply the result by in the end. Then, we can compute the sum by constructing a table where the first row has possible values of and the first column has possible values of :
| 0 | |||
| 0 | |||
| 0 |
This gives us seven distinct values for the sum, which are
Multiplying each number by , all possible values for the given multivalued expression are
The question asks about the real values, so the solution will be .
Let us now summarize the points we have learned in this explainer.
Key Points
- Any nonzero complex number has distinct values for the expression , which are called the roots of . We can find the roots of a complex number by applying de Moivreβs theorem for roots.
- In an Argand diagram, the roots of a complex number with modulus and argument form the vertices of a regular -gon inscribed in a circle of radius , where one of the vertices is the point on the circle with argument .
- The roots of a complex number with modulus and argument can be obtained by multiplying each root of unity by .
