# Explainer: Arbitrary Roots of Complex Numbers

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

We will be interested in the solutions to equations of the form . Finding solutions to equations in this form is equivalent to finding the th roots of . Recall, when taking the square root of a positive real number , that we have two possible solutions, and . Likewise, when taking the root of a complex number, we have distinct roots. Using de Moivre’s theorem for roots, we can determine all possible distinct solutions.

### De Moivre’s Theorem for Roots

For a complex number , the th roots are given by for . Equivalently, we can express this in exponential form as for .

### Example 1: Roots of a Complex Number

1. Solve .
2. By plotting these solutions on an Argand diagram, or otherwise, describe the geometric properties of the solutions.

Part 1

We begin by finding the modulus and argument of so we can express it in polar form. The modulus is given by . Since the complex number lies in the first quadrant, we can find the argument by evaluating the inverse tangent of the imaginary part of divided by the real part: . Hence, we can rewrite the equation as Applying de Moivre’s theorem, the roots are given by for . Hence, considering each value of , the solutions are

Part 2

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

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

### Example 2: Expressions Involving 𝑛th Roots

Find the possible values of

Letting , we can rewrite this expression as

From the definition of , we can see that , from which we can conclude that . Consequently, we can rewrite the expression as

Using the properties of conjugates, we have

We now calculate the . Expressing in exponential form, we have . Using de Moivre’s theorem, its cube roots are given by for . Substituting in the values of and expressing each argument in the principal range, the three possible cube roots of are given by

We can now substitute each value of in the simplified expression: . For , we have

Similarly, for , we can write

Finally, for , we have

Therefore, the three possible values of the expression are , 0, and 1.

The expression in the previous example is the one that you get when applying Tartaglia’s method for solving cubics to the equation . Trying to understand expressions like this led the mathematician Rafael Bombelli to formalize the rules of arithmetic of complex numbers. In fact, later mathematicians showed that when using general methods to solve cubic equations with real roots, the use of complex numbers is unavoidable.

### Example 3: Relation between Arbitrary Roots and Roots of Unity

1. Find the solutions to the equation . 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 ?

Part 1

Applying de Moivre’s theorem, the roots of the equation are given by for . Substituting in each value of and simplifying, we have

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

In a similar way, we can find the 6th roots of unity by substituting into . Hence, the 6th roots of unity are given by

Part 3

Geometrically, we can transform the 6th roots of unity to the roots of through a dilation of scale factor and a counterclockwise rotation by . This can be represented by multiplying by a complex number with modulus and argument , that is, by . Hence, if we represent the 6th roots of unity by 1, , , , , and , the roots of can be expressed as

Notice that there is nothing special about ; we could have equally used any other of the roots of and got the same result.

The result of the previous question generalizes to the th root of any complex number as stated in the box below.

### Relationship between the 𝑛th Root of an Arbitrary Complex Number and the 𝑛th Roots of Unity

If is one root of the equation , and are the th roots of unity, the roots of are given by

This result can be thought about in multiple ways. The first is the one presented in the previous example, in which we consider the transformation that maps the th roots of unity to the roots of . Alternatively, we can consider it in a purely algebraic way: given an equation , we can think of its roots as being , in which case we need only to find one of the th roots of , and then we can use our knowledge of the th roots of unity to find the others. Another way to think about this result is the fact that multiplication by any complex number with modulus one represents a counterclockwise rotation by the argument of the complex number. Hence, starting with one of the roots of , we successively rotate it by an angle of and hence sweep out a regular -gon where each vertex represents an additional root of .

### Example 4: 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.

If we consider the Cartesian plane to be an Argand plane, then we know that the fifth roots of unity form a regular pentagon centered at the origin with one vertex at . We can transform this pentagon to a regular pentagon centered at the origin with one vertex at by multiplying each of the fifth roots of unity by . This is equivalent to finding the fifth roots of . Using de Moivre’s theorem, we know that the fifth roots of unity are 1, , , , and , where

The point can be represented by the complex number . We can express this in exponential form as . Therefore, the vertices of a pentagon centered at the origin with one vertex at will be , , , , and . We now calculate each of these in turn and express it in polar form to then covert it to coordinates in the Cartesian plane. Starting with , we have

Expressing this in polar form, we have

Therefore, the coordinates of the point are .

Considering , we have

Expressing this in polar form, we have which has coordinates .

Similarly, Expressing this in polar form, we have which has coordinates .

Finally, Expressing this in polar form, we have which has coordinates .

Therefore, the coordinates of the regular pentagon are

Considering the transformations that map the th roots of unity to another regular polygon, we use the same technique to find the coordinates of any regular polygon in the plane.

### Example 5: Coordinates of Regular Polygons

Find the coordinates of the vertices of a regular hexagon centered at with one vertex at the origin. Give your answer as exact Cartesian coordinates.

Considering the Cartesian plane to be an Argand plane, we consider the transformations that map the sixth roots of unity onto the given hexagon. Since the roots of unity are centered at the origin, we first consider the translation which takes the given hexagon to one centered at the origin. Since its center is at , the mapping that takes it to a polygon centered at the origin is represented by subtracting the complex number . This mapping takes the vertex that was at the origin to . We can therefore map the sixth roots of unity to the vertices of this polygon by multiplying each one by . Combining these two mappings, we can map the sixth roots of unity to the given hexagon by applying the map . Notice we are adding now since we are mapping a polygon centered at the origin to one centered at .

The sixth roots of unity are given by 1, , , , , and , where

The table represents the calculations of applying this map to each of the sixth roots of unity.

110
2
3
4
5
6

Therefore, the coordinates of the vertices are

### Example 6: Geometric Applications of Complex Numbers

A small robot is programmed to travel units forward and then turn to the left by an angle of . If it does this times, how far will it be from the starting point? Give you answer in exact form.

Let the position of the robot be given by . Using this convention, moving forward by units at an angle of to the left is modeled by adding . Hence, starting from the origin, after iterations, the robot’s position will be given by

Since for any complex number , , we can rewrite as

Factoring out from the numerator and from the denominator, we can express as

Since , we can express the numerator and denominator in terms of sine as follows:

Finally, simplifying gives

This is the position of the robot after iterations. Given that it started at the origin, the distance from the starting point is just the modulus . Therefore, taking the modulus, we have

Since , the distance of the robot from its starting point after iterations is given by .

### Example 7: Geometric Applications of the 𝑛th Roots of a Complex Number

1. Find the roots of .
2. The complex numbers representing the roots of are each squared to form the vertices of a new shape. What is the area of this shape?

Part 1

We begin by expressing in exponential form. Clearly, its modulus is 16 and since it is a negative real number, its argument is . We can express it in polar form as . Hence, we can rewrite the equation as

Applying de Moivre’s theorem, the roots are given by for . Hence, considering each value of , the solutions are

Plotting these points on an Argand diagram, we have the vertices of a regular octagon.

Part 2

If we square each of the complex numbers representing the vertices of the hexagon, we get the following set of four complex numbers:

Plotting these on an Argand diagram, we find they represent the vertices of a square inscribed in a circle of radius 2.

Using the Pythagorean theorem, we find that the side length of the square is . Hence, its area is 8.

### Key Points

1. Using de Moivre’s theorem, we can find arbitrary roots of complex numbers.
2. If is one of the th roots of a complex number, its other roots are given by where is a primitive th root of unity.
3. Using the geometric properties of the roots of complex numbers, we can find the coordinates of regular polygons plotted in the plane.