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6.12: DeMoivre's Theorem

Difficulty Level: At Grade Created by: CK-12
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Imagine that you are in math class one day, and you are given the following number:

$[5(\cos 135^\circ + i \sin 135^\circ)]^4$

Your instructor wants you to find this number. Can you do it? How long will it take you? Probably a long time, if you want to take a number to the fourth power, you'd have to multiply the number by itself, over and over again.

However, there is a shortcut. Read on, and by the end of this Concept, you'll be able to use DeMoivre's Theorem to simplify the calculation of powers of complex numbers.

Guidance

The basic operations of addition, subtraction, multiplication and division of complex numbers can all be carried out, albeit with some changes in form from what you may have seen with numbers having only real components. The addition and subtraction of complex numbers lent themselves best to numbers expressed 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

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

These examples suggest a general rule valid for all powers of $z$ , or $n$ . We offer this rule and assume its validity for all $n$ without formal proof, leaving that 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:

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

Where $z = r(\cos \theta + i \sin \theta)$ 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 $n$ th power; second, its angle is multiplied by $n$ . Conversely, the roots of a number have angles that are evenly spaced about the origin.

Example A

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:

$z^n=[r(\cos \theta+i \sin \theta)]^n &= r^n(\cos n\theta + i \sin n\theta)\\[2(\cos 120^\circ + i \sin 120^\circ)]^5 &= 2^5 \left[\cos 5\frac{2\pi}{3}+i \sin 5\frac{2\pi}{3}\right]\\&= 32\left (\cos \frac{10\pi}{3}+i \sin \frac{10\pi}{3} \right )\\&= 32 \left(-\frac{1}{2}+-\frac{i\sqrt{3}}{2}\right)\\&=-16-16i \sqrt{3}$

Example B

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

Solution: Change into polar form.

$r &= \sqrt{x^2+y^2} && \theta = \tan^{-1} \left(\frac{\sqrt{3}}{2} \cdot -\frac{2}{1} \right)=-\frac{\pi}{3} + \pi=\frac{2\pi}{3}\\r &= \sqrt{\left(\frac{-1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\\r &= \sqrt{\frac{1}{4}+\frac{3}{4}}\\r &= \sqrt{1}=1$

The polar form of $\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)$ is $1 \left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)$

Now use De Moivre’s Theorem:

$z^n = [r(\cos \theta + i \sin \theta)]^n &= r^n(\cos n\theta+i \sin n\theta)\\\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= 1^{10}\left[\cos 10\left(\frac{2\pi}{3}\right)+i \sin 10 \left(\frac{2\pi}{3}\right)\right]\\\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= 1 \left(\cos \frac{20\pi}{3}+i \sin \frac{20\pi}{3}\right)\\\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= -\frac{1}{2}+i \frac{\sqrt{3}}{2} \rightarrow \ \text{Standard Form}$

Example C

Find $[3(\cos 45^\circ + i \sin 45^\circ)]^3$

Solution: $\theta=45^\circ=\frac{\pi}{4} \ \text{rad}$ , using De Moivre’s Theorem:

$z^n=[r(\cos \theta+i \sin \theta)]^n &= r^n(\cos n\theta + i \sin n\theta)\\[3(\cos 45^\circ + i \sin 45^\circ)]^3 &= 3^3 \left[\cos 3\frac{\pi}{4}+i \sin 3\frac{\pi}{4}\right]\\&= 27\left (\cos \frac{3\pi}{4}+i \sin \frac{3\pi}{4} \right )\\&= 27 \left(-\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right)\\&=-13.5+13.5i \sqrt{2}$

Vocabulary

DeMoivres Theorem: DeMoivres theorem relates a complex number raised to a power to a set of trigonometric functions by stating that the complex number raised to a power is equal to the trigonometric representation of the number with the power times the angle under consideration as the argument for the trigonometric form.

Guided Practice

1. Evaluate: $\left[\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)\right]^8$

2. Evaluate: $\left [3 \left (\sqrt{3}-i\sqrt{3} \right ) \right ]^4$

3. Evaluate: $(\sqrt{5}-i)^7$

Solutions:

1. $\left[\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)\right]^8=\left(\frac{\sqrt{2}}{2}\right)^8 \left(\cos \frac{8\pi}{4}+i \sin \frac{8\pi}{4}\right)=\frac{1}{16}\cos 2\pi+ \frac{i}{16} \sin 2\pi = \frac{1}{16}$

2.

$\left[3(\sqrt{3}-i\sqrt{3})\right]^4 &= (3\sqrt{3}-3i\sqrt{3})^4\\r &= \sqrt{(3\sqrt{3})^2+(3\sqrt{3})^2}=3\sqrt{6}, \tan \theta=\frac{3\sqrt{3}}{3\sqrt{3}}=1 \rightarrow 45^\circ\\&= \left(3\sqrt{6} \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right)^4=(3\sqrt{6})^4 \left(\cos \frac{4\pi}{4}+i \sin \frac{4\pi}{4}\right)\\&= 81(36)[-1+i(0)]=-2936$

3.

$(\sqrt{5}-i)^7 \rightarrow r &= \sqrt{(\sqrt{5})^2+(-1)^2}=\sqrt{6}, \tan \theta=-\frac{1}{\sqrt{5}} \rightarrow \theta=335.9^\circ\\[\sqrt{6}(\cos 335.9^\circ+i \sin 335.9^\circ)]^7 &= (\sqrt{6})^7(\cos (7 \cdot 335.9^\circ)+i \sin (7 \cdot 335.9^\circ))\\&= 216 \sqrt{6}(\cos 2351.3^\circ+i \sin 2351.3^\circ)\\&= 216 \sqrt{6}(-0.981-0.196i)\\&= -519.04-103.7i$

Concept Problem Solution

The original problem was:

$[5(\cos 135^\circ + i \sin 135^\circ)]^4$

In this problem, $r = 5, \theta = \frac{3\pi}{4}$

Using $z^n = [r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)$ , you can substitute to get:

$[5(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})]^4\\ = 5^4(\cos 4\frac{3\pi}{4} + i \sin 4\frac{3\pi}{4})\\= 625(\cos(3\pi) + i\sin(3\pi)\\= 625(-1 + i0)\\= -625\\$

Practice

Use DeMoivre's Theorem to evaluate each expression. Write your answer in standard form.

1. $(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})^3$
2. $[2(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})]^2$
3. $[3(\cos\frac{3\pi}{2}+i\sin\frac{\pi}{2})]^5$
4. $(1+i)^5$
5. $(1-\sqrt{3}i)^3$
6. $(1+2i)^6$
7. $(\sqrt{3}-i)^5$
8. $(\frac{1}{2}+\frac{i\sqrt{3}}{2})^4$
9. $[3(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})]^5$
10. $(2-\sqrt{5}i)^5$
11. $(\sqrt{2}+\sqrt{2}i)^4$
12. $[2(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12})]^8$
13. $(-1+\sqrt{2}i)^6$
14. $[5(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})]^3$
15. $(3-4i)^6$

Sep 26, 2012

Feb 26, 2015

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