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# Teorema de De Moivre: fórmulas, explicación y ejemplos Quien soy
Pau Monfort
@paumonfort Autor y referencias
Teorema de De Moivre: fórmulas, explicación y ejemplos

De Moivre’s Theorem is an essential theorem when working with complex numbers. This theorem can help us easily find the powers and roots of complex numbers in polar form, so we must learn about De Moivre’s theorem.

De Moivre’s Theorem states that the power of a complex number in polar form is equal to raising the modulus to the same power and multiplying the argument by the same power.  This theorem helps us find the power and roots of complex numbers easily.

This pattern was first observed by the French mathematician Abraham De Moivre (1667 – 1754) and was used to find the powers, roots, and even solve equations involving complex numbers.

Before we dive right into De Moivre’s theorem, make sure that we have refreshed our knowledge on complex numbers and polar forms of complex numbers.

• Make sure to review your knowledge of complex numbers and their trigonometric forms.
• It’s also important to review how we convert rectangular forms to polar forms and vice-versa.
• For the proof of De Moivre’s theorem, master your knowledge on adding, multiplying, subtracting, and dividing complex numbers as well.

In this article, we’ll learn about De Moivre’s theorem, learn how we can apply them, and appreciate this theorem for how useful it is in manipulating complex numbers.

We’ll also provide a special section for the proof of the theorem for the curious minds and those eager to learn how the theorem was established.

## What is De Moivre’s Theorem?

De Moivre’s theorem helps us raise power and find the roots of complex numbers in trigonometric form. Let’s say we have \$z = r (cos theta + isin theta)\$, according to De Moivre’s theorem, we can easily raise \$z\$ to the power of \$n\$.

Let’s observe how \$z\$ behaves when we raise it to the second and third power to check for patterns.

Starting at \$z\$ and \$z^2\$, we have the following result shown below.

\$begin{aligned}z&= r(cos theta + i sin theta )z^2&=r^2(cos theta + isin theta)^2&= r^2(cos^2 theta + i2sin theta costheta + i^2 sin^2 theta )&=r^2(cos^2 theta +i 2sin theta cos theta – sin ^2 theta)&= r^2(cos^2 theta – sin^2 theta + i2 sin theta cos theta&= r^2(cos 2theta + i2sin theta cos theta )phantom{xxxxxx}color{green} cos 2theta = cos^2 theta – sin^2 theta &= r^2(cos 2theta + isin 2theta  )phantom{xxxxxxxxxx}color{green} sin 2theta = 2sin theta cos theta end{aligned}\$

We can also use the FOIL method and the sum formulas for sine and cosine to find \$z^3\$.

\$begin{aligned}z^3 &= z cdot z^2&r^3=(cos theta + isin theta)(cos 2theta + isin 2theta  ) &= r^3[(cos theta cos 2theta – sin theta sin 2theta)+ i(cos theta sin 2theta + sin theta cos 2 theta)] &=r^3[cos(theta + 2theta) + isin( theta  +2theta)]&= r^3(cos 3theta + i sin 3theta) end{aligned}\$

Have you noticed any patterns so far? Let’s list down \$z\$, \$z^2\$, and \$z^3\$ first, and maybe you’ll be able to spot a pattern.

\$begin{aligned}z&= r(cos theta + i sin theta)z^2 &=r^2 (cos 2theta + isin 2theta)z^3 &= r^3(cos 3theta + i sin 3theta)end{aligned}\$

Do you have a good guess for \$z^4\$? Yes, \$r^4 (cos 4 theta + i sin 4theta)\$ is actually a good guess! You can apply a similar process from \$z^3\$ to find \$z^4\$, so try verifying the expression yourself also to help you review your knowledge of algebraic and trigonometric techniques.

Notice how tedious it will be if we want to find \$z^8\$? This is why De Moivre’s theorem is extremely helpful when finding complex numbers’ powers and roots.

The formula below states how we can apply the theorem to find \$z^n\$ easily. We can even extend this to finding the \$n\$th roots of \$z\$.

### De Moivre’s theorem formula

When \$n\$ is a rational number and a complex number in polar or trigonometric form, we can raise the complex number by a power of \$n\$ using the formula shown below.

\$ z^n = r^n (cos ntheta + isin ntheta)\$

This means that to raise \$z = r (cos theta + isin theta)\$ to the power of \$n\$, we simply:

• Raise the modulus, \$r\$, by the power of \$n\$.
• Multiply the value of \$theta\$ inside the parenthesis by \$n\$.

Also, we can find the roots of the complex numbers using De Moivre’s theorem.

\$ sqrt[n]{z} = sqrt[n]{r}left( cos dfrac{theta + 2pi k}{n} + isin dfrac{ theta + 2pi k }{n}right) \$.

From the formula, we can see that we can find the \$n\$th root of \$z\$ by:

• Taking the \$n\$th root of the modulus, \$r\$.
• Divide the values of the angle by \$n\$.
• Repeat the process while increasing the angle by \$2pi k\$, where \$k = 1, 2, …n-1\$.
• Make sure you have a total of \$n\$ complex numbers before stopping.

In the next section, you’ll see how helpful it is to know these two formulas when finding the powers, roots, and even solving equations involving the complex system.

## How to use De Moivre’s Theorem?

Now that we know the two essential formulas established from De Moivre’s theorem. Let’s explore the common problems involving complex numbers that we might make use of these identities.

• We can raise any complex number (in either rectangular or polar form) to the \$n\$th power easily using De Moivre’s theorem. When given a complex number in rectangular form, make sure to convert it to polar form first.
• Similarly, we can find the \$n\$th root of complex numbers.
• We can also solve equations that involve complex number roots using De Moivre’s theorem.

This means that if we want to find \$(1 + i)^4\$, we can use De Moivre’s theorem by:

• Converting \$1 + i\$ to polar form.
• Applying the formula \$ z^n = r^n (cos ntheta + isin ntheta)\$.

Let’s find the modulus and argument of \$1 + i\$ first then write it in trigonometric form.

We can now use the formula \$ z^n = r^n (cos ntheta + isin ntheta)\$, to raise \$(1 + i)^4\$.

\$begin{aligned}(1 + i)^4 &= left[sqrt{2}left(cos dfrac{pi}{4} + isin dfrac{pi}{4}right)right]^4&=(sqrt{2})^4 left(cos 4cdot dfrac{pi}{4} + isin 4cdot dfrac{pi}{4}right )&=4(cos pi + i sin pi)end{aligned}\$

If we want to return an answer in rectangular, we simply evaluate \$cos pi\$ and \$sin pi\$ then distribute \$4\$ to each of the resulting values.

\$begin{aligned}4(cos pi + i sin pi) &= 4(-1 + 0i)&=-4end{aligned}\$

Hence, \$(1 + i)^4\$ is equal to \$4(cos pi + isin pi)\$ or \$-4\$.

We can also find the cube root of \$(1 + i) \$ using the polar form of \$1 + i\$.

\$begin{aligned}sqrt{1 + i} &= sqrt{sqrt{2}left(cos dfrac{pi }{4}+ isin dfrac{pi}{4}right)} end{aligned}\$

Since we’re looking for the cube root, we’re using \$k = {0, 1, 2}\$ in the formula, \$ sqrt[n]{z} = sqrt[n]{r}left( cos dfrac{theta + 2pi k}{n} + isin dfrac{ theta + 2pi k }{n}right) \$.

Meaning, we’re expecting three roots for our answer. It helps to keep in mind as well that we can rewrite \$sqrt{sqrt{2}}\$ as a root of \$6\$ as shown below.

\$begin{aligned} sqrt{sqrt{2}}  & = (2^{frac{1}{2}})^{frac{1}{3}} &= 2^{frac{1}{6}} &= sqrt{6}end{aligned}\$

Why don’t we start with \$k = 0\$?

\$begin{aligned}sqrt{sqrt{2}left(cos dfrac{pi }{4}+ isin dfrac{pi}{4}right)}&= sqrt{sqrt{2}}left( cos dfrac{dfrac{pi}{4} + 2pi (0)}{3} + isin dfrac{ dfrac{pi}{4} + 2pi(0) }{3}right) &=sqrt{sqrt{2}} left(cos dfrac{pi}{12} + isin dfrac{pi}{12} right )&=sqrt{2}left(cos dfrac{pi}{12} + isin dfrac{pi}{12} right )end{aligned}\$

We’ll apply a similar when working out the two remaining roots when \$k = 1\$ and \$k = 2\$.

We’ve just shown you how we can apply De Moivre’s theorem to find complex numbers’ power and roots. Don’t worry. We have more examples prepared for you!

Ever wondered how we can confirm the validity of De Moivre’s theorem? Check out the section below to understand how we can prove these formulas. This can also help you master the two formulas when you know how they were established.

If you want to jump right into trying out more problems involving De Moivre’s theorem, you can skip through the section below and start with the four examples we’ve provided.

### De Moivre’s theorem proof

We can prove De Moivre’s theorem using mathematical induction. Let’s recall the process of proving a theorem using mathematical induction first.

If we want to show that \$P(n)\$ is true for all \$n\$ that is greater than or equal to, we have to:

1. Show that \$P(1)\$ exists and is true.
2. If \$P(n)\$ is indeed true, we have to show that \$P(n + 1)\$ is also true.

We’ll have to show these two conditions for De Moivre’s theorem to prove valid.

Starting with the equation, \$(cos theta + i sin theta)^n = cos ntheta + i sin n theta\$.

For this to be true, we have to show that it is true for \$n = 1\$.

\$ begin{aligned}(cos theta + i sin theta)^1 &= cos 1theta + isin 1theta&=cos theta + isin theta&= (cos theta + i sin theta)^1end{aligned}\$

This shows that the theorem is true for \$n = 1\$.

Assuming that \$(cos theta + i sin theta)^n = cos ntheta + i sin n theta\$ is indeed true, we must show that \$(cos theta + i sin theta)^{n + 1} = cos (n + 1) theta + i sin (n + 1) theta\$ is also true.

To do so, let’s express \$(cos theta + i sin theta)^{n + 1}\$ as a product of \$(cos theta + i sin theta)^n\$ and \$cos theta + i sin theta\$.

\$begin{aligned}(cos theta + i sin theta)^{n + 1} &= (cos theta + isin theta)^n(cos theta + isin theta)end{aligned}\$

Replace \$(cos theta + isin theta)^n(cos theta + isin theta)^n\$ with \$cos ntheta + isin ntheta\$.

\$begin{aligned}(cos theta + i sin theta)^{n + 1} &= (cos theta + isin theta)^n(cos theta + isin theta)&= (cos ntheta + isin ntheta)(cos theta + i sin theta)end{aligned}\$

Apply the FOIL method to expand the expression and replace \$i^2\$ with \$-1\$.

\$begin{aligned}(cos theta + i sin theta)^{n + 1} &=cos ntheta cos theta + i cos ntheta sin theta + i sin ntheta cos theta + i^2 sin ntheta sin theta &=cos ntheta cos theta + i cos ntheta sin theta + i sin ntheta cos theta – sin ntheta sin theta&=cos ntheta cos theta – sin ntheta sin theta + i sin n theta cos theta + i cos ntheta sin theta&=(cos ntheta cos theta – sin ntheta sin theta )+ i (sin n theta cos theta + cos ntheta sin theta) end{aligned}\$

Rewrite the grouped terms using the sum formula for cosine and sine.

\$begin{aligned}(cos theta + i sin theta)^{n + 1} &=cos (ntheta + theta) + i sin (ntheta + theta)&= cos (n+1)theta + isin (n + 1)thetaend{aligned}\$

We’ve just shown that \$(cos theta + i sin theta)^{n + 1} = cos (n+1)theta + isin (n + 1)theta\$, meaning De Moivre’s theorem is also true for \$n + 1\$.

By mathematical induction, we’ve just shown that the De Moivre’s theorem, \$[r(cos theta + i sin theta)]^n= r^n(cos ntheta + isin ntheta)\$ is also true.

Since we’ve already established De Moivre’s theorem for raising the power of complex numbers, we can also prove the formula for finding the root.

If we have \$z =r ( cos theta + isin theta)\$, to take the \$n\$th rooth, we want to actually find \$z^{frac{1}{n}}\$.

\$begin{aligned}z^{frac{1}{n}} &= r^{frac{1}{n}}left( dfrac{1}{n}cdot cos theta + dfrac{1}{n}cdot isin theta right)&=r^{frac{1}{n}}left(dfrac{cos theta}{n} + dfrac{sin theta}{n} right )end{aligned}\$

Keep in mind that the cosine and sine values will remain the same for all angles that are conterminal to \$theta\$. This means that we can extend the formula to \$z^{frac{1}{n}} = r^{frac{1}{n}}left(dfrac{cos theta + 2pi k}{n} + dfrac{sin theta + 2pi k}{n} right ) \$, where \$k = 0,1, 2,…n-1\$.

Since \$z^{frac{1}{n}} = sqrt[n]{z}\$ and \$r^{frac{1}{n}} = sqrt[n]{r}\$, we can also rewrite the formula as \$sqrt[n]{z } = sqrt[n]{r }left(dfrac{cos theta + 2pi k}{n} + dfrac{sin theta + 2pi k}{n} right ) \$.

In degrees, we can also write this formula as \$sqrt[n]{z } = sqrt[n]{r }left(dfrac{cos theta + 360^{circ} k}{n} + dfrac{sin theta +360^{circ}k}{n} right ) \$.

ejemplo 1

Find the power of the following complex numbers, then express the answer in rectangular form.

a. \$left(cos dfrac{2pi}{3} + i sin dfrac{2pi}{3}right)^3\$
b. \$left[2left(cos dfrac{pi}{4} + i sin dfrac{5pi}{4}right)right]^5\$
c. \$(1 – sqrt{3}i)^{12}\$

Solución

For the first two items, we use the power formula from De Moivre’s theorem.

\$ z^n = r^n (cos ntheta + isin ntheta)\$.

\$ begin{aligned}left(cos dfrac{2pi}{3} + i sin dfrac{2pi}{3}right)^3 &= (1)^3left[cos left(3 cdotdfrac{2pi}{3}right) + i sin left(3 cdotdfrac{2pi}{3}right)right]&= cos 2pi + i sin 2piend{aligned}\$

We now have the simplified polar form to convert the complex number into a rectangular form.

\$ begin{aligned} cos 2pi + i sin 2pi &= 1 + 0i&=1end{aligned}\$

Hence, \$left(cos dfrac{2pi}{3} + i sin dfrac{2pi}{3}right)^3\$ in rectangular form is actuall equal to \$1\$.

Let’s go ahead and apply a similar process to simplify the second item.

\$ begin{aligned} left[2left(cos dfrac{pi}{4} + i sin dfrac{5pi}{4}right)right]^5 &= 2^5left[cos left(5cdot dfrac{pi}{4} right ) + i sin left(5cdot dfrac{pi}{4} right )right]&=32left(cos dfrac{5pi}{4} + i sin dfrac{5pi}{4} right )&=32 left( – dfrac{sqrt{2}}{2} – idfrac{sqrt{2}}{2}right)&= 32 cdot – dfrac{sqrt{2}}{2} – 32 cdot dfrac{sqrt{2}}{2}&=-16sqrt{2} – 16sqrt{2}end{aligned}\$

Before we can evaluate \$(1 – sqrt{3}i)^12\$, let’s convert \$1 – sqrt{3}i\$ into polar form first.

Let’s go ahead and raise \$2 left(cos dfrac{5pi}{3} + i sin dfrac{5pi}{3}right)\$ to the \$12\$th power.

\$begin{aligned}(1 – sqrt{3}i)^{12}&= left[2 left(cos dfrac{5pi}{3} + i sin dfrac{5pi}{3}right) right ]^{12}&= (2^{12})left[cos left(12 cdot dfrac{5pi}{3} right ) + isin left(12 cdot dfrac{5pi}{3} right ) right ]&= 4096 (cos 30 pi + i sin 30 pi)&=4096(1 + 0i)&= 4096end{aligned}\$

This means that \$(1 – sqrt{3}i)^{12}\$, in rectangular form, is equal to \$4096\$.

ejemplo 2

Find all the complex cube roots of \$27\$.

Solución

We can express \$27\$ as a complex number in rectangular form: \$27  = 27 + 0i\$. We can then convert \$27 +  0i\$ to polar form. It’s expected to lie on the positive part of the real axis (or when \$theta = 0). We can still confirm this by using the traditional approach:

To find the three complex roots of \$sqrt 27\$, we use the formula for the \$n\$th root of \$r(cos theta + isin theta)\$, \$ sqrt[n]{z} = sqrt[n]{r}left( cos dfrac{theta + 2pi k}{n} + isin dfrac{ theta + 2pi k }{n}right) \$.

For \$sqrt 27 = sqrt{27 (cos 0 + i sin 0)} \$, we’ll use \$n = 3\$ and \$k = {0, 1, 2}\$.

In the past, we only know that the cube root of \$27\$ is equal to \$3\$, but with our knowledge of complex numbers and De Moivre’s theorem, we can find the two remaining roots!

This means that the three complex roots of \$27\$ are \$left{3, -dfrac{3}{2} + idfrac{3sqrt{3}}{2}, -dfrac{3}{2} – idfrac{3sqrt{3}}{2}right}\$.

ejemplo 3

Plot all the complex fourth roots of \$64(cos 240^{circ} + isin 240^{circ})\$ in one complex plane.

Solución

In degrees, we have the root formula from De Moivre’s theorem as \$ sqrt[n]{z} = sqrt[n]{r}left( cos dfrac{theta + 360^{circ} k}{n} + isin dfrac{ theta + 360^{circ} k }{n}right) \$. This time, we’ll use \$n = 4\$ and \$k = {0, 1, 2, 3}\$.

Hence, the four fourth roots of \$64(cos 240^{circ} + isin 240^{circ})\$ are \${2 + 2sqrt{3}i , -2sqrt{3} + 2i, -2 -2sqrt{3}i, 2sqrt{3} -2i }\$.

Let’s plot the four roots on one complex plane, as shown below. Notice something? The four roots are each \$90^{circ}\$ away from each other.  The segments are also all equal to \$4\$.

ejemplo 4

Solve the equation \$x^3 – (1 + sqrt{3}i) = 0\$ in the complex system.

Solución

First, let’s isolate \$x^3\$ on the left-hand side of the equation.

\$ begin{aligned}x^3 – (1 + sqrt{3}i) &= 0 x^3 &= 1 + sqrt{3}i end{aligned}\$

This means that to find the solution to a complex system equation, we need to find the cube root of \$1 + sqrt{3}i\$.

For us to do this, we need to convert \$1 + sqrt{3}i\$ to polar form.

Let’s find the cube root using the formula, \$ sqrt[n]{z} = sqrt[n]{r}left( cos dfrac{theta + 2pi k}{n} + isin dfrac{ theta + 2pi k }{n}right) \$, where \$n = 3\$ and \$k = {0, 1, 2}\$.

This means that the equation has three solutions at: \$ x = left{sqrt{2} left(cos dfrac{pi}{9} + i sin dfrac{pi}{9}right), sqrt{2} left(cos dfrac{7pi}{9} + i sin dfrac{7pi}{9}right), sqrt{2} left(cos dfrac{13pi}{9} + i sin dfrac{13pi}{9}right)right}\$.  This actually makes sense since we expect three solutions for a cubic equation.

### Preguntas de práctica

1. Find the power of the following complex numbers then express the answer in rectangular form.
a. \$left(cos dfrac{3pi}{4} + i sin dfrac{3pi}{4}right)^4\$
b. \$left[-4left(cos dfrac{pi}{12} + i sin dfrac{pi}{12}right)right]^6\$
c. \$(1 + sqrt{3}i)^8\$
2. Find all the complex cube roots of \$125\$.
3. Plot all the complex fourth roots of \$16(cos 240^{circ} + isin 240^{circ})\$ in one complex plane.
4. Solve the equation \$x^4 – (4 – 4sqrt{3}i) = 0\$ in the complex system.

### clave de respuestas

1.
a. \$-1 = -1 + 0i\$
b. \$4096left( cos dfrac{pi}{2} + isin dfrac{pi}{2}right) = 4096i\$
c. \$256left( cos dfrac{2pi}{3} + isin dfrac{2pi}{3}right) = -128 +128sqrt{3}i\$
2. \$dfrac{5}{2} + dfrac{5sqrt{3}}{2}i\$, \$ dfrac{5}{2} – dfrac{5sqrt{3}}{2}i\$, and \$-5\$
3. 4.
\$begin{aligned}k&= dfrac{sqrt{2}}{2}left( cos -dfrac{pi}{12} + isin -dfrac{pi}{12}right) &= dfrac{sqrt{2}}{2}left( cos dfrac{5pi}{12} + isin -dfrac{5pi}{12}right) &= dfrac{sqrt{2}}{2}left( cos dfrac{11pi}{12} + isin dfrac{11pi}{12}right) &= dfrac{sqrt{2}}{2}left( cos dfrac{17pi}{12} + isin dfrac{17pi}{12}right)end{aligned}\$

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