# Lesson Explainer: De Moivre’s Theorem Mathematics

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

Recall the identity for multiplying complex numbers in polar form.

For two complex numbers and , their product is

Note that if we set in the above equation, we get

This equation shows that for the square of a complex number, we can apply the square directly to the modulus and multiply the argument by two. We might speculate whether this rule can be generalized to other positive powers of a complex number.

In fact, it is possible to derive a similar formula for a negative power of a complex number as well. Recall the identity for division of complex numbers in polar form, using the same and as above:

Setting and above, we get a relation for the reciprocal equation: which we simplify to get

That is, taking a complex number to the power of is the same as taking the modulus to the power of and multiplying the argument by .

Having seen similar formulas for both positive and negative powers of a complex number, we might predict that we can further generalize these rules into a relationship for all integer powers.

We can indeed do this, and the result is known as de Moivre’s theorem.

### Theorem: De Moivre’s 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 Moivre’s theorem is true for .

We now assume it is true for some positive integer :

Now we need to show that this implies that de Moivre’s 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 Moivre’s theorem is true for , and given that it is true for , it is true for , then by mathematical induction, it is true for all positive integers . To prove de Moivre’s 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 Moivre’s 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 Moivre’s theorem holds for all .

For a more concise proof, we can use Euler’s formula as follows:

Since is an integer, we can rewrite this as

Using Euler’s formula again, we get

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

Simplify .

### Answer

Applying de Moivre’s theorem to each complex number, we have

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

Simplifying, using and , we have

The last example demonstrates that using de Moivre’s 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.

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 Moivre’s theorem to the complex numbers in the numerator and the denominator, we can rewrite this as

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

One of the implications of de Moivre’s theorem is that we can generalize the properties of the modulus and argument to arbitrary integer powers. This gives us the following identities.

### Identity: Powers Applied to the Modulus and Argument

For any complex number and integer ,

Sometimes, using these identities is more useful for solving some problems than directly using de Moivre’s 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,

We can confirm that this is indeed the principal part of the argument, since . Hence, the principal amplitude of is .

Sometimes it can be useful to simplify the expression we are working with, or to make note of key properties it might have, before applying de Moivre’s theorem. In the following example, we will see how powers of complex conjugates can be dealt with using de Moivre’s theorem.

What is ?

### Answer

In this example, we could convert each number to polar form and apply de Moivre’s theorem. However, it is worth noting first that this equation is of the form . Given this fact, we should consider 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 Moivre’s theorem, we can rewrite

 𝑧−𝑧=𝑟(𝑛𝜃+𝑖𝑛𝜃)−𝑟((−𝑛𝜃)+𝑖(−𝑛𝜃))=𝑧−(𝑧).cossincossin (1)

Using the property of complex conjugation, letting , we have

 𝑧−(𝑧)=2𝑖(𝑧).Im (2)

Combining equations (1) and (2), we find

Hence, we find that

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 Moivre’s theorem, we can write

Hence,

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

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

### Example 5: Using de Moivre’s Theorem to Find the Roots of a Complex Number

Consider the equation .

1. Express in polar form using the general form of the argument.
2. By applying de Moivre’s theorem to the left-hand side, rewrite the equation in polar form.
3. 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 Moivre’s 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 Moivre’s theorem for roots.

### Theorem: De Moivre’s Theorem for Roots

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

Note that to find the principal arguments of the roots in the above theorem, it may be necessary to subtract from the resulting arguments.

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

### Example 6: Using de Moivre’s Theorem for Roots to Find the Complex Roots of a Number

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 Moivre’s 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 .

Let us finish by summarizing the key points we have learned in this explainer.

### Key Points

• De Moivre’s theorem tells us that, for any , This enables us to significantly simplify calculations involving large integer powers of complex numbers.
• For any complex number and integer , This identity allows us to apply powers directly to the modulus and the argument of a complex number.
• De Moivre’s theorem extends to finding the distinct roots of complex numbers by evaluating for . The principal roots can be found by subtracting from the arguments if necessary.

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