In this explainer, we will learn how to find powers and roots of complex numbers and how to use De Moivreβs theorem to simplify calculations of powers and roots.

By considering the identities for multiplying complex numbers in polar or exponential form with , we might guess that we can generalize this rule to any positive power. Furthermore, if we consider the reciprocal relationships too, we might also predict that we can further generalize these into a relationship for all integer powers. We can indeed do this, and the result is known as de Moivreβs theorem.

### De Moivreβs Theorem

For any integer ,

Using induction, we can prove this for positive powers. We begin by showing that this is true in the case where . With , the left-hand side is which is the right-hand side. Hence, de Moivreβs theorem is true for .

We now assume it is true for some positive integer :

Now we need to show that this implies that de Moivreβs theorem is true for . Hence, we write

Using the assumption that this is true for , we can rewrite this as

Expanding the brackets, we have

Using and gathering the real and imaginary terms, we get

Using the addition and subtraction trigonometric identities, we can rewrite this as

Hence, since de Moivreβs theorem is true for , for , and for , then by mathematical induction, it is true for all positive integers . To prove de Moivreβs theorem for negative integers, we can use the result we have just proved and the reciprocal identities. We let be a positive integer. Then,

Using de Moivreβs theorem for positive integers, we have

We can now apply the reciprocal relationship to arrive at

Hence, we have shown this is the case for negative integers. The case when is trivial to prove. Hence, we have shown that de Moivreβs theorem holds for all .

For a more concise proof, we can use Eulerβs formula as follows:

Since is an integer, we can rewrite this as

Using Eulerβs formula again, we get

We will now look at a number of examples where using this theorem significantly simplifies our calculations.

### Example 1: Using de Moivreβs Theorem

Simplify .

### Answer

Applying de Moivreβs theorem to each complex number, we have

Using the rule for multiplying complex numbers in polar form, we can rewrite this as

Simplifying, we have

The last example demonstrates that using de Moivreβs theorem significantly simplifies calculations. With this in mind, if we need to solve a problem involving high powers of complex numbers, it is preferable to start by expressing them in polar or exponential form. The next example will demonstrate this process.

### Example 2: Division of Complex Numbers with Large Powers

Simplify .

### Answer

We begin by converting the complex numbers in the numerator and denominator to polar form. Starting with the numerator, its modulus is given by . Since its real part is positive and its imaginary part is negative, it lies in the fourth quadrant, so we can find its argument by evaluating the inverse tangent function as follows:

Therefore, we can express this in polar form as

Similarly, for the denominator, its modulus is . Since both its real and imaginary parts are positive, it lies in the first quadrant and we can find its argument by evaluating

Hence, the denominator can be expressed as in polar form. Now we can rewrite the whole fraction as

Applying de Moivreβs theorem to the complex numbers in the numerator and the denominator, we can rewrite this as

Using the quotient rules for a complex number in polar form, if and , we can rewrite this as

One of the implications of de Moivreβs theorem is that we can generalize the properties of the modulus and argument to arbitrary integer powers. This gives us the following identities, which are true for any complex number and :

Sometimes, using these identities is more useful for solving some problems than directly using de Moivreβs theorem as the following example will demonstrate.

### Example 3: Solving Problems with Powers of Complex Numbers

Given that , determine the principal amplitude of .

### Answer

Substituting the value of into gives

Using the properties of the modulus, we can rewrite this as

Now ; hence,

Therefore, . Hence,

Taking the argument of both sides gives

Using the properties of the argument, we have

Now, we can find the argument of . Since its real part is positive and its imaginary part is negative, it lies in the fourth quadrant and, consequently, we can find its argument by evaluating

Hence,

### Example 4: The Difference of Complex Powers

What is ?

### Answer

In this example, we could convert each number to polar form and apply de Moivreβs theorem. However, it is often worth noting first that this is of the form . Given this fact, it is worth considering whether we might be able to apply some of the properties of complex conjugates to simplify our calculation. First, let us consider a general complex number in polar form and its conjugate . Therefore, applying de Moivreβs theorem, we can rewrite

Using the property of complex conjugation, we have

Hence,

Now, we can find the modulus and argument of . Firstly, its modulus . Since its real part is negative and its imaginary part is positive, it lies in the second quadrant and we can find its argument by evaluating

Using de Moivreβs theorem, we can write

Hence,

Notice that in the previous example, using de Moivreβs theorem, we showed that for any complex number ,

We will now consider how we can use de Moivreβs theorem to find the roots of complex numbers.

### Example 5: Using de Moivreβs Theorem to Find Roots

Consider the equation .

- Express in polar form using the general form of the argument.
- By applying de Moivreβs theorem to the left-hand side, rewrite the equation in polar form.
- By equating the moduli and arguments and considering different values of the general argument, find the 3 cube roots of , expressing them in exponential form.

### Answer

**Part 1**

First, we calculate the modulus of as follows:

Second, we calculate the argument. Since both its real and imaginary parts are positive, we have a complex number in the first quadrant and we can calculate its principal argument by evaluating

We get the general form of the argument from the principal argument by adding integer multiples of . Hence, we can write its general argument as , where . Therefore, we can express in polar form using the general form of the argument as follows: for .

**Part 2**

We can express in polar form as follows

Hence, we can rewrite the equation as

By applying de Moivreβs theorem, we get

**Part 3**

Equating the moduli gives us and, hence, . Equating the arguments gives us

Hence,

We now consider three consecutive values of to find the three distinct roots. Starting with , we have . Next, we consider , which gives . Finally, considering , we get . Since this is not in the range , we can subtract to get the principal argument: . Therefore, the three distinct roots of are

Abstracting out the method used in the previous question, we arrive at de Moivreβs theorem for roots.

### De Moivreβs Theorem for Roots

For a complex number , the th roots are given by for .

To finish this explainer, we will look at one more example where we apply de Moivreβs theorem to find roots.

### Example 6: Complex Roots

Find the fourth roots of , giving your answers in trigonometric form.

### Answer

We begin by expressing in polar form. Clearly, its modulus is 1 and its argument is . Therefore, applying de Moivreβs theorem for roots, its 4 fourth roots are given by for , and 3. We consider each value of in turn. Starting with , we have

For , we have

For , we have

However, since this argument is not in the range of the principal argument, we can subtract to get

Finally, for , we have

Once again, this argument is not in the range of the principal argument, so we can subtract to get

Putting all of these values together, we have that the fourth roots of are , , , and .

### Key Points

- De Moivreβs theorem, enables us to significantly simplify calculations involving large integer powers of complex numbers.
- De Moivreβs theorem extends to finding the distinct th roots of complex numbers, by evaluating for .
- Although we can use de Moivreβs theorem to evaluate th roots and integer powers, we cannot assume that it applies to all real or complex exponents.