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# De Moivre’s Theorem: Powers and Roots of Complex Numbers

Created by: CK-12

## Learning Objectives

A student will be able to:

• Use De Moivre’s Theorem to find the powers of complex numbers in polar form.

## Introduction

The basic operations of addition, subtraction, multiplication and division of complex numbers have all been explored in this chapter. The addition and subtraction of complex numbers lent themselves best to those in standard form. However multiplication and division were easily performed when the complex numbers were in polar form. Another operation that is performed using the polar form of complex numbers is the process of raising a complex number to a power.

The polar form of a complex number is $r(\cos \theta + i \sin \theta)$. If we allow $z$ to equal the polar form of a complex number, it is very easy to see the development of a pattern when raising a complex number in polar form to a power. To discover this pattern, it is necessary to perform some basic multiplication of complex numbers in polar form.

If $z = r(\cos \theta + i \sin \theta)$ and $z^2 = z \cdot z$ then:

$z^2 & = r(\cos \theta + i \sin \theta) \cdot r(\cos \theta + i \sin \theta) \\z^2 & = r^2 [\cos (\theta + \theta) + i \sin (\theta + \theta)] \\z^2 & = r^2 (\cos 2\theta + i \sin 2\theta)$

Likewise, if $z = r(\cos \theta + i \sin \theta)$ and $z^3 = z^2 \cdot z$ then

$z^3 & = r^2(\cos 2\theta + i \sin 2\theta) \cdot r(\cos \theta + i \sin \theta) \\z^3 & = r^3 [\cos (2\theta + \theta) + i \sin (2\theta + \theta)] \\z^3 & = r^3 (\cos 3\theta + i \sin 3\theta)$

Again, if $z = r(\cos \theta + i \sin \theta)$ and $z^4 = z^3 \cdot z$ then

$z^4 = r^4(\cos 4\theta + i \sin 4\theta)$

De Moivre’s Theorem

These examples suggest a general rule valid for all $n$. We offer this rule and assume its validity for all $n$ without formal proof, leaving the proof for later studies. The general rule for raising a complex number in polar form to a power is called De Moivre’s Theorem, and has important applications in engineering, particularly circuit analysis. The rule is as follows:

$z^n = [r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)$

Let $z = r(\cos \theta + i \sin \theta)$ and let $n$ be a positive integer.

Notice what this rule looks like geometrically. A complex number taken to the $n$th power has two motions: First, its distance from the origin is taken to the nth power; second, its angle is multiplied by $n$. Conversely, the roots of a number have angles that are evenly spaced about the origin.

Example 1: Find. $[2(\cos 120^\circ + i \sin 120^\circ)]^5$

Solution:

$\theta = 120^\circ = \frac{2 \pi} {3}\ \text{rad}$

Using De Moivre’s Theorem:

$z^n = [r (\cos \theta + i\ \sin \theta)]^n & = r^n (\cos n\theta + i\ \sin n\theta) \\[2(\cos 120^\circ + i\ \sin 120^\circ)]^5 & = 2^5 \left [\cos 5\left (\frac{2\pi} {3}\right ) + i\ \sin 5 \left (\frac{2\pi} {3}\right ) \right ]\\[2(\cos 120^\circ + i\ \sin 120^\circ)]^5 & = 32 \left (\cos \frac{10\pi} {3} + i\ \sin \frac{10\pi} {3}\right ) && \frac{10\pi} {3}\ \text{is in the third quadrant so both are negative.}\\[2(\cos 120^\circ + i\ \sin 120^\circ)]^5 & =32 \left (- \frac{1} {2} + - \frac{\sqrt{3}} {2}\right ) \\[2(\cos 120^\circ + i\ \sin 120^\circ)]^5 & = -16 - 16i \sqrt{3}$

Example 2: Find $\left (-\frac{1} {2} + \frac{\sqrt{3}} {2}\right )^{10}$

Solution:

$r & = \sqrt{x^2 + y^2} && \theta = \tan^{-1} \left (\frac{\sqrt{3}} {2} \cdot - \frac{2} {1}\right ) = - \frac{\pi} {3} \\r & = \sqrt{\left (\frac{-1} {2}\right )^2 + \left (\frac{\sqrt{3}} {2}\right )^2} && \text{The polar form of} \left (-\frac{1} {2} + \frac{\sqrt{3}} {2}\right ) \text{is}\ 1 \left (\cos - \frac{\pi} {3} + i\ \sin - \frac{\pi} {3}\right ) \\r & = \sqrt{\frac{1} {4} + \frac{3} {4}} \\r & = \sqrt{1} = 1$

Now use De Moivre’s Theorem:

$z^n = [r (\cos \theta + i\ \sin \theta)]^n & = r^n (\cos n\theta + i\ \sin n\theta)\\\left (-\frac{1} {2} + \frac{\sqrt{3}} {2}\right )^{10} & = 1^{10} \left [\cos 10\left (-\frac{\pi} {3}\right ) + i\ \sin 10 \left (- \frac{\pi} {3}\right ) \right ] \\\left (-\frac{1} {2} + \frac{\sqrt{3}} {2}\right )^{10} & = 1 \left (\cos - \frac{10\pi} {3} + i\ \sin - \frac{10\pi} {3}\right ) \ \text{Write the result in standard form}.\\\left (-\frac{1} {2} + \frac{\sqrt{3}} {2}\right )^{10} & = 1 \left (-\frac{1} {2} + i \frac{\sqrt{3}} {2}\right )$

## Lesson Summary

In this lesson you discovered the pattern for raising complex numbers in polar form to a power. This pattern was then transferred into a general rule. This general rule is called De Moivre’s Theorem.

## Points to Consider

• If a complex number in polar formed can be raised to a power, can the roots of a complex number be determined?
• If the roots can be determined, will some form of De Moivre’s Theorem be used?
• What do powers and roots of complex numbers look like on the complex plane.

## Review Questions

1. Show that $z^3 = 1$, if $z = -\frac{1} {2} + i \frac{\sqrt{3}} {2}$
2. Rewrite the following in rectangular form: $[2(\cos 315^\circ + i \sin 315^\circ)]^3$

1. Express $z$ in polar form: $r & = \sqrt{x^2 + y^2} && \\r & = \sqrt{\left (-\frac{1} {2}\right )^2 + \left (\frac{\sqrt{3}} {2}\right )^2} &&\\r & = \sqrt{\frac{1} {4} + \frac{3} {4}} = 1 &&\\\theta & = \tan^{-1} \left (-\frac{\sqrt{3}} {1}\right ) = 120^\circ && \text{The polar form is}\ z = 1 (\cos 120^\circ + i \sin 120^\circ)$ $z^n & = [r(\cos \theta + i \sin \theta)^n = r(\cos n\theta + i \sin n\theta)\\z^3 & = 1^3 [\cos 3 (120^\circ) + i \sin (120^\circ)] \\z^3 & = 1(\cos 360^\circ + i \sin 360^\circ) \\z^3 & = 1(1 + 0i) \\z^3 & = 1$ There are two other cube roots of $1$ in the complex plane. Can you find them and plot them on the complex plane? What do the three roots look like geometrically?
2. $r & = 2\ \text{and} \ \theta = 315^\circ \ \text{or} \ \frac{7 \pi} {4} \\z^n & = [r (\cos \theta + i\ \sin \theta)]^n = r^n (\cos n\theta + i\ \sin n\theta) \\z^3 & = 2^3 \left [ (\cos 3 \left (\frac{7\pi} {4}\right ) + i\ \sin 3 \left (\frac{7\pi} {4}\right ) \right ] \\z^3 & = 8\left (\cos \frac{21 \pi} {4} + i\ \sin \frac{21 \pi} {4}\right ) \qquad \qquad \frac{21\pi} {4}\ \text{is in the third quadrant so both are negative.}\\z^3 & = 8 \left (-\frac{\sqrt{2}} {2} - i \frac{\sqrt{2}} {2}\right ) \\z^3 & -4 \sqrt{2} - 4i \sqrt{2}$

## Date Created:

Feb 23, 2012

Dec 12, 2013
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