Explainer: De Moivre’s Theorem

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.

By considering the identities for multiplying complex numbers in polar or exponential form with 𝑧=𝑧=π‘§οŠ§οŠ¨, (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(2πœƒ+𝑖2πœƒ),ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘’,cossincossinοŠ¨οŠ¨οƒοΌοŠ¨οŠ¨οŠ¨οƒοΌ we might guess that we can generalize this rule to any positive power. Furthermore, if we consider the reciprocal relationships too, (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ((βˆ’πœƒ)+𝑖(βˆ’πœƒ)),ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘’,cossincossinοŠ±οŠ§οŠ±οŠ§οƒοΌοŠ±οŠ§οŠ±οŠ§οŠ±οƒοΌ 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 Moivre’s theorem.

De Moivre’s Theorem

For any integer 𝑛, (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(π‘›πœƒ+π‘–π‘›πœƒ).cossincossin

Using induction, we can prove this for positive powers. We begin by showing that this is true in the case where 𝑛=1. With 𝑛=1, the left-hand side is (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(πœƒ+π‘–πœƒ)=π‘Ÿ((1Γ—πœƒ)+𝑖(1Γ—πœƒ)),cossincossincossin which is the right-hand side. Hence, de Moivre’s theorem is true for 𝑛=1.

We now assume it is true for some positive integer π‘˜: (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(π‘˜πœƒ+π‘–π‘˜πœƒ).cossincossin

Now we need to show that this implies that de Moivre’s theorem is true for π‘˜+1. Hence, we write (π‘Ÿ(πœƒ+π‘–πœƒ))=(π‘Ÿ(πœƒ+π‘–πœƒ))(π‘Ÿ(πœƒ+π‘–πœƒ)).cossincossincossinο‡οŠ°οŠ§ο‡

Using the assumption that this is true for π‘˜, we can rewrite this as (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(π‘˜πœƒ+π‘–π‘˜πœƒ)(π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(π‘˜πœƒ+π‘–π‘˜πœƒ)(πœƒ+π‘–πœƒ).cossincossincossincossincossinο‡οŠ°οŠ§ο‡ο‡οŠ°οŠ§

Expanding the brackets, we have (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿο€Ήπ‘˜πœƒπœƒ+π‘–π‘˜πœƒπœƒ+π‘–π‘˜πœƒπœƒ+π‘–π‘˜πœƒπœƒο….cossincoscoscossinsincossinsinο‡οŠ°οŠ§ο‡οŠ°οŠ§οŠ¨

Using 𝑖=βˆ’1 and gathering the real and imaginary terms, we get (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(π‘˜πœƒπœƒβˆ’π‘˜πœƒπœƒ+𝑖(π‘˜πœƒπœƒ+π‘˜πœƒπœƒ)).cossincoscossinsincossinsincosο‡οŠ°οŠ§ο‡οŠ°οŠ§

Using the addition and subtraction trigonometric identities, sinsincoscossincoscoscossinsin(𝐴±𝐡)=𝐴𝐡±𝐴𝐡,(𝐴±𝐡)=π΄π΅βˆ“π΄π΅, we can rewrite this as (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ((π‘˜πœƒ+πœƒ)+𝑖(π‘˜πœƒ+πœƒ))=π‘Ÿ(((π‘˜+1)πœƒ)+𝑖((π‘˜+1)πœƒ)).cossincossincossinο‡οŠ°οŠ§ο‡οŠ°οŠ§ο‡οŠ°οŠ§

Hence, since de Moivre’s theorem is true for 𝑛=π‘˜, for 𝑛=π‘˜+1, and for 𝑛=1, then by mathematical induction, it is true for all positive integers 𝑛. To prove de Moivre’s theorem for negative integers, we can use the result we have just proved and the reciprocal identities. We let 𝑛 be a positive integer. Then, (π‘Ÿ(πœƒ+π‘–πœƒ))=ο€Ή(π‘Ÿ(πœƒ+π‘–πœƒ)).cossincossin

Using de Moivre’s theorem for positive integers, we have (π‘Ÿ(πœƒ+π‘–πœƒ))=(π‘Ÿ(π‘›πœƒ+π‘–π‘›πœƒ)).cossincossin

We can now apply the reciprocal relationship to arrive at (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ((βˆ’π‘›πœƒ)+𝑖(βˆ’π‘›πœƒ)).cossincossin

Hence, we have shown this is the case for negative integers. The case when 𝑛=0 is trivial to prove. Hence, we have shown that de Moivre’s theorem holds for all π‘›βˆˆβ„€.

For a more concise proof, we can use Euler’s formula as follows: (π‘Ÿ(πœƒ+π‘–πœƒ))=ο€Ήπ‘Ÿπ‘’ο….cossinοŠοƒοΌοŠ

Since 𝑛 is an integer, we can rewrite this as (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿπ‘’.cossinοŠοŠοƒοŠοΌ

Using Euler’s formula again, we get (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(π‘›πœƒ+π‘–π‘›πœƒ).cossincossin

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

Example 1: Using de Moivre’s Theorem

Simplify ο€Όβˆš5ο€Όο€Ό3πœ‹14+𝑖3πœ‹14οˆοˆοˆο€Όβˆš3ο€Όο€Ό5πœ‹22+𝑖5πœ‹22cossincossin.

Answer

Applying de Moivre’s theorem to each complex number, we have ο€Όβˆš5ο€Όο€Ό3πœ‹14+𝑖3πœ‹14οˆοˆοˆο€Όβˆš3ο€Όο€Ό5πœ‹22+𝑖5πœ‹22=ο€»βˆš57Γ—3πœ‹14+𝑖7Γ—3πœ‹14οˆοˆο€»βˆš311Γ—5πœ‹22+𝑖11Γ—5πœ‹22=ο€»125√53πœ‹2+𝑖3πœ‹2οˆοˆο€»243√35πœ‹2+𝑖5πœ‹2.cossincossincossincossincossincossin

Using the rule for multiplying complex numbers in polar form, 𝑧𝑧=π‘Ÿπ‘Ÿ((πœƒ+πœƒ)+𝑖(πœƒ+πœƒ)),cossin we can rewrite this as ο€Όβˆš5ο€Όο€Ό3πœ‹14+𝑖3πœ‹14οˆοˆοˆο€Όβˆš3ο€Όο€Ό5πœ‹22+𝑖5πœ‹22=ο€»125√5243√33πœ‹2+5πœ‹2+𝑖3πœ‹2+5πœ‹2=30,375√15(4πœ‹+𝑖4πœ‹).cossincossincossincossin

Simplifying, we have ο€Όβˆš5ο€Όο€Ό3πœ‹14+𝑖3πœ‹14οˆοˆοˆο€Όβˆš3ο€Όο€Ό5πœ‹22+𝑖5πœ‹22=30,375√15.cossincossin

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.

Example 2: Division of Complex Numbers with Large Powers

Simplify 18(βˆ’π‘–+1)(𝑖+1)οŠͺ.

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 |βˆ’π‘–+1|=1+(βˆ’1)=√2. 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: argarctan(βˆ’π‘–+1)=ο€Όβˆ’11=βˆ’πœ‹4.

Therefore, we can express this in polar form as βˆ’π‘–+1=√2ο€»ο€»βˆ’πœ‹4+π‘–ο€»βˆ’πœ‹4.cossin

Similarly, for the denominator, its modulus is |𝑖+1|=√1+1=√2. Since both its real and imaginary parts are positive, it lies in the first quadrant and we can find its argument by evaluating argarctan(𝑖+1)=ο€Ό11=πœ‹4.

Hence, the denominator can be expressed as 𝑖+1=√2ο€»ο€»πœ‹4+π‘–ο€»πœ‹4cossin in polar form. Now we can rewrite the whole fraction as 18(βˆ’π‘–+1)(𝑖+1)=18ο€»βˆš2ο€»ο€»βˆ’ο‡+π‘–ο€»βˆ’ο‡ο‡ο‡ο€»βˆš2+𝑖.οŠͺοŠ§οŽ„οŠͺοŽ„οŠͺοŠ©οŠ―οŽ„οŠͺοŽ„οŠͺοŠͺcossincossin

Applying de Moivre’s theorem to the complex numbers in the numerator and the denominator, we can rewrite this as 18(βˆ’π‘–+1)(𝑖+1)=18ο€»βˆš2ο‡ο€»ο€»βˆ’39+π‘–ο€»βˆ’39ο‡ο‡ο€»βˆš241+𝑖41.οŠͺοŠ§οŠ©οŠ―οŽ„οŠͺοŽ„οŠͺοŠͺοŠ§οŽ„οŠͺοŽ„οŠͺcossincossin

Using the quotient rules for a complex number in polar form, if 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin and 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin, 𝑧𝑧=π‘Ÿπ‘Ÿ((πœƒβˆ’πœƒ)+𝑖(πœƒβˆ’πœƒ)),cossin we can rewrite this as 18(βˆ’π‘–+1)(𝑖+1)=18ο€»βˆš2ο‡ο€»ο€»βˆ’39πœ‹4βˆ’41πœ‹4+π‘–ο€»βˆ’39πœ‹4βˆ’41πœ‹4=9((βˆ’20πœ‹)+𝑖(βˆ’20πœ‹))=9.οŠͺcossincossin

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, which are true for any complex number 𝑧 and π‘›βˆˆβ„€: |𝑧|=|𝑧|,(𝑧)=𝑛(𝑧).argarg

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 𝑍=ο€»βˆš3βˆ’π‘–ο‡|𝑍|=32and, determine the principal amplitude of 𝑍.

Answer

Substituting the value of 𝑍=ο€»βˆš3βˆ’π‘–ο‡οŠ into |𝑍|=32 gives ||ο€»βˆš3βˆ’π‘–ο‡||=32.

Using the properties of the modulus, we can rewrite this as ||√3βˆ’π‘–||=32.

Now ||√3βˆ’π‘–||=ο„žο€»βˆš3+(βˆ’1)=√4=2; hence, 2=32.

Therefore, 𝑛=5. Hence, 𝑍=ο€»βˆš3βˆ’π‘–ο‡.

Taking the argument of both sides gives argarg(𝑍)=ο€½ο€»βˆš3βˆ’π‘–ο‡ο‰.

Using the properties of the argument, we have argarg(𝑍)=5ο€»βˆš3βˆ’π‘–ο‡.

Now, we can find the argument of √3βˆ’π‘–. 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 argarctanο€»βˆš3βˆ’π‘–ο‡=ο€Ώβˆ’1√3=βˆ’πœ‹6.

Hence, arg(𝑍)=5Γ—βˆ’πœ‹6=βˆ’5πœ‹6.

Example 4: The Difference of Complex Powers

What is (βˆ’2+2𝑖)βˆ’(βˆ’2βˆ’2𝑖)οŠͺοŠͺ?

Answer

In this example, we could convert each number to polar form and apply de Moivre’s theorem. However, it is often worth noting first that this is of the form π‘§βˆ’(𝑧)οŠβˆ—οŠ. 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 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin and its conjugate 𝑧=π‘Ÿ((βˆ’πœƒ)+𝑖(βˆ’πœƒ))βˆ—cossin. Therefore, applying de Moivre’s theorem, we can rewrite π‘§βˆ’(𝑧)=π‘Ÿ(π‘›πœƒ+π‘–π‘›πœƒ)βˆ’π‘Ÿ((βˆ’π‘›πœƒ)+𝑖(βˆ’π‘›πœƒ))=π‘§βˆ’(𝑧).οŠβˆ—οŠοŠοŠοŠοŠβˆ—cossincossin

Using the property of complex conjugation, π‘§βˆ’π‘§=2𝑖(𝑧),βˆ—Im we have π‘§βˆ’(𝑧)=2𝑖(𝑧).οŠβˆ—οŠοŠIm

Hence, (βˆ’2+2𝑖)βˆ’(βˆ’2βˆ’2𝑖)=2𝑖(βˆ’2+2𝑖).οŠͺοŠͺοŠͺIm

Now, we can find the modulus and argument of (βˆ’2+2𝑖). Firstly, its modulus |βˆ’2+2𝑖|=(βˆ’2)+2=2√2. 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 argarctan(βˆ’2βˆ’2𝑖)=ο€Ό2βˆ’2+πœ‹=βˆ’πœ‹4+πœ‹=3πœ‹4.

Using de Moivre’s theorem, we can write (βˆ’2+2𝑖)=ο€»2√243πœ‹4+𝑖43πœ‹4=64(3πœ‹+𝑖3πœ‹).οŠͺοŠͺcossincossin

Hence, (βˆ’2+2𝑖)βˆ’(βˆ’2βˆ’2𝑖)=2𝑖(βˆ’2+2𝑖)=128𝑖3πœ‹=0.οŠͺοŠͺοŠͺImsin

Notice that in the previous example, using de Moivre’s theorem, we showed that for any complex number 𝑧, (𝑧)=(𝑧).βˆ—οŠοŠβˆ—

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 Roots

Consider the equation 𝑧=2√3+2π‘–οŠ©.

  1. Express 2√3+2𝑖 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 2√3+2𝑖, expressing them in exponential form.

Answer

Part 1

First, we calculate the modulus of 2√3+2𝑖 as follows: ||2√3+2𝑖||=ο„žο€»2√3+2=√12+4=√16=4.

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 argarctanο€»2√3+2𝑖=ο€Ώ22√3=πœ‹6.

We get the general form of the argument from the principal argument by adding integer multiples of 2πœ‹. Hence, we can write its general argument as πœ‹6+2πœ‹π‘˜, where π‘˜βˆˆβ„€. Therefore, we can express 2√3+2𝑖 in polar form using the general form of the argument as follows: 2√3+2𝑖=4ο€»ο€»πœ‹6+2πœ‹π‘˜ο‡+π‘–ο€»πœ‹6+2πœ‹π‘˜ο‡ο‡cossin for π‘˜βˆˆβ„€.

Part 2

We can express 𝑧 in polar form as follows 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ).cossin

Hence, we can rewrite the equation as (π‘Ÿ(πœƒ+π‘–πœƒ))=4ο€»ο€»πœ‹6+2πœ‹π‘˜ο‡+π‘–ο€»πœ‹6+2πœ‹π‘˜ο‡ο‡.cossincossin

By applying de Moivre’s theorem, we get π‘Ÿ(3πœƒ+𝑖3πœƒ)=4ο€»ο€»πœ‹6+2πœ‹π‘˜ο‡+π‘–ο€»πœ‹6+2πœ‹π‘˜ο‡ο‡.cossincossin

Part 3

Equating the moduli gives us π‘Ÿ=4 and, hence, π‘Ÿ=√4. Equating the arguments gives us 3πœƒ=πœ‹6+2πœ‹π‘˜.

Hence, πœƒ=+2πœ‹π‘˜3.οŽ„οŠ¬

We now consider three consecutive values of π‘˜ to find the three distinct roots. Starting with π‘˜=0, we have πœƒ=πœ‹18. Next, we consider π‘˜=1, which gives πœƒ=13πœ‹18. Finally, considering π‘˜=2, we get πœƒ=25πœ‹18. Since this is not in the range (βˆ’πœ‹,πœ‹], we can subtract 2πœ‹ to get the principal argument: πœƒ=βˆ’11πœ‹18. Therefore, the three distinct roots of 2√3+2𝑖 are 𝑧=√4𝑒,√4𝑒,√4𝑒.οŽ’ο‘½οŽ οŽ§οŽ’οŽ οŽ’ο‘½οŽ οŽ§οŽ’οŽ οŽ ο‘½οŽ οŽ§οƒοƒοŠ±οƒand

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

De Moivre’s Theorem for Roots

For a complex number 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin, the 𝑛th roots are given by π‘Ÿο€½ο€½πœƒ+2πœ‹π‘˜π‘›ο‰+π‘–ο€½πœƒ+2πœ‹π‘˜π‘›ο‰ο‰οŽ ο‘ƒcossin for π‘˜=0,1,…,π‘›βˆ’1.

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

Example 6: Complex Roots

Find the fourth roots of βˆ’1, giving your answers in trigonometric form.

Answer

We begin by expressing βˆ’1 in polar form. Clearly, its modulus is 1 and its argument is πœ‹. Therefore, applying de Moivre’s theorem for roots, its 4 fourth roots are given by 1ο€½ο€½πœ‹+2πœ‹π‘˜4+π‘–ο€½πœ‹+2πœ‹π‘˜4=ο€½πœ‹+2πœ‹π‘˜4+π‘–ο€½πœ‹+2πœ‹π‘˜4ο‰οŽ οŽ£cossincossin for π‘˜=0,1,2, and 3. We consider each value of π‘˜ in turn. Starting with π‘˜=0, we have cossinο€»πœ‹4+π‘–ο€»πœ‹4.

For π‘˜=1, we have cossinο€Ό3πœ‹4+𝑖3πœ‹4.

For π‘˜=2, we have cossinο€Ό5πœ‹4+𝑖5πœ‹4.

However, since this argument is not in the range of the principal argument, we can subtract 2πœ‹ to get cossinο€Όβˆ’3πœ‹4+π‘–ο€Όβˆ’3πœ‹4.

Finally, for π‘˜=3, we have cossinο€Ό7πœ‹4+𝑖7πœ‹4.

Once again, this argument is not in the range of the principal argument, so we can subtract 2πœ‹ to get cossinο€»βˆ’πœ‹4+π‘–ο€»βˆ’πœ‹4.

Putting all of these values together, we have that the fourth roots of βˆ’1 are ο€»ο€»πœ‹4+π‘–ο€»πœ‹4cossin, ο€Όο€Ό3πœ‹4+𝑖3πœ‹4cossin, ο€»ο€»βˆ’πœ‹4+π‘–ο€»βˆ’πœ‹4cossin, and ο€Όο€Όβˆ’3πœ‹4+π‘–ο€Όβˆ’3πœ‹4cossin.

Key Points

  1. De Moivre’s theorem, (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(π‘›πœƒ+π‘–π‘›πœƒ),cossincossin enables us to significantly simplify calculations involving large integer powers of complex numbers.
  2. De Moivre’s theorem extends to finding the distinct 𝑛th roots of complex numbers, by evaluating π‘Ÿο€½ο€½πœƒ+2πœ‹π‘˜π‘›ο‰+π‘–ο€½πœƒ+2πœ‹π‘˜π‘›ο‰ο‰οŽ ο‘ƒcossin for π‘˜=0,1,…,π‘›βˆ’1.
  3. Although we can use de Moivre’s theorem to evaluate 𝑛th roots and integer powers, we cannot assume that it applies to all real or complex exponents.

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