10.5 Polar form of complex numbers (Page 3/8)

Page 3 / 8

A General Note

Products of complex numbers in polar form

If $z_{1} = r_{1} (\cos θ_{1} + i \sin θ_{1})$ and $z_{2} = r_{2} (\cos θ_{2} + i \sin θ_{2}),$ then the product of these numbers is given as:

\begin{array}{l} \begin{array}{l} z_{1} z_{2} = r_{1} r_{2} [\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})] \\ z_{1} z_{2} = r_{1} r_{2} cis (θ_{1} + θ_{2}) \end{array} \end{array}

Notice that the product calls for multiplying the moduli and adding the angles.

Finding the product of two complex numbers in polar form

Find the product of $z_{1} z_{2},$ given $z_{1} = 4 (\cos (80°) + i \sin (80°))$ and $z_{2} = 2 (\cos (145°) + i \sin (145°)) .$

Follow the formula

\begin{array}{l} z_{1} z_{2} = 4 \cdot 2 [\cos (80° + 145°) + i \sin (80° + 145°)] \\ z_{1} z_{2} = 8 [\cos (225°) + i \sin (225°)] \\ z_{1} z_{2} = 8 [\cos (\frac{5 π}{4}) + i \sin (\frac{5 π}{4})] \\ z_{1} z_{2} = 8 [- \frac{\sqrt{2}}{2} + i (- \frac{\sqrt{2}}{2})] \\ z_{1} z_{2} = - 4 \sqrt{2} - 4 i \sqrt{2} \end{array}

Got questions? Get instant answers now!

Finding quotients of complex numbers in polar form

The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.

A General Note

Quotients of complex numbers in polar form

If $z_{1} = r_{1} (\cos θ_{1} + i \sin θ_{1})$ and $z_{2} = r_{2} (\cos θ_{2} + i \sin θ_{2}),$ then the quotient of these numbers is

\begin{array}{l} \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} [\cos (θ_{1} - θ_{2}) + i \sin (θ_{1} - θ_{2})], z_{2} \neq 0 \\ \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} cis (θ_{1} - θ_{2}), z_{2} \neq 0 \end{array}

Notice that the moduli are divided, and the angles are subtracted.

How To

Given two complex numbers in polar form, find the quotient.

Divide $\frac{r_{1}}{r_{2}} .$
Find $θ_{1} - θ_{2} .$
Substitute the results into the formula: $z = r (\cos θ + i \sin θ) .$ Replace $r$ with $\frac{r_{1}}{r_{2}},$ and replace $θ$ with $θ_{1} - θ_{2} .$
Calculate the new trigonometric expressions and multiply through by $r .$

Finding the quotient of two complex numbers

Find the quotient of $z_{1} = 2 (\cos (213°) + i \sin (213°))$ and $z_{2} = 4 (\cos (33°) + i \sin (33°)) .$

Using the formula, we have

\begin{array}{l} \frac{z_{1}}{z_{2}} = \frac{2}{4} [\cos (213° - 33°) + i \sin (213° - 33°)] \\ \frac{z_{1}}{z_{2}} = \frac{1}{2} [\cos (180°) + i \sin (180°)] \\ \frac{z_{1}}{z_{2}} = \frac{1}{2} [- 1 + 0 i] \\ \frac{z_{1}}{z_{2}} = - \frac{1}{2} + 0 i \\ \frac{z_{1}}{z_{2}} = - \frac{1}{2} \end{array}

Got questions? Get instant answers now!

Try It

Find the product and the quotient of $z_{1} = 2 \sqrt{3} (\cos (150°) + i \sin (150°))$ and $z_{2} = 2 (\cos (30°) + i \sin (30°)) .$

$z_{1} z_{2} = - 4 \sqrt{3}; \frac{z_{1}}{z_{2}} = - \frac{\sqrt{3}}{2} + \frac{3}{2} i$

Got questions? Get instant answers now!

Finding powers of complex numbers in polar form

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem . It states that, for a positive integer $n, z^{n}$ is found by raising the modulus to the $n th$ power and multiplying the argument by $n .$ It is the standard method used in modern mathematics.

A General Note

De moivre’s theorem

If $z = r (\cos θ + i \sin θ)$ is a complex number, then

\begin{array}{l} z^{n} = r^{n} [\cos (n θ) + i \sin (n θ)] \\ z^{n} = r^{n} cis (n θ) \end{array}

where $n$ is a positive integer.

Evaluating an expression using de moivre’s theorem

Evaluate the expression ${(1 + i)}^{5}$ using De Moivre’s Theorem.

Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write $(1 + i)$ in polar form. Let us find $r .$

\begin{array}{l} r = \sqrt{x^{2} + y^{2}} \\ r = \sqrt{{(1)}^{2} + {(1)}^{2}} \\ r = \sqrt{2} \end{array}

Then we find $θ .$ Using the formula $\tan θ = \frac{y}{x}$ gives

\begin{array}{l} \tan θ = \frac{1}{1} \\ \tan θ = 1 \\ θ = \frac{π}{4} \end{array}

Use De Moivre’s Theorem to evaluate the expression.

\begin{array}{l} {(a + b i)}^{n} = r^{n} [\cos (n θ) + i \sin (n θ)] \\ {(1 + i)}^{5} = {(\sqrt{2})}^{5} [\cos (5 \cdot \frac{π}{4}) + i \sin (5 \cdot \frac{π}{4})] \\ {(1 + i)}^{5} = 4 \sqrt{2} [\cos (\frac{5 π}{4}) + i \sin (\frac{5 π}{4})] \\ {(1 + i)}^{5} = 4 \sqrt{2} [- \frac{\sqrt{2}}{2} + i (- \frac{\sqrt{2}}{2})] \\ {(1 + i)}^{5} = - 4 - 4 i \end{array}

Got questions? Get instant answers now!

Finding roots of complex numbers in polar form

To find the n th root of a complex number in polar form, we use the $n th$ Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding $n th$ roots of complex numbers in polar form.

A General Note label

The n Th root theorem

To find the $n th$ root of a complex number in polar form, use the formula given as

z^{\frac{1}{n}} = r^{\frac{1}{n}} [\cos (\frac{θ}{n} + \frac{2 k π}{n}) + i \sin (\frac{θ}{n} + \frac{2 k π}{n})]

where $k = 0, 1, 2, 3, . . ., n - 1.$ We add $\frac{2 k π}{n}$ to $\frac{θ}{n}$ in order to obtain the periodic roots.

Questions & Answers

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3

Ken Reply

proof

AUSTINE

sebd me some questions about anything ill solve for yall

Manifoldee Reply

how to solve x²=2x+8 factorization?

Kristof Reply

x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)

Manifoldee

x=2x+8

Manifoldee

×=2x-8 minus both sides by 2x

Manifoldee

so, x-2x=2x+8-2x

Manifoldee

then cancel out 2x and -2x, cuz 2x-2x is obviously zero

Manifoldee

so it would be like this: x-2x=8

Manifoldee

then we all know that beside the variable is a number (1): (1)x-2x=8

Manifoldee

so we will going to minus that 1-2=-1

Manifoldee

so it would be -x=8

Manifoldee

so next step is to cancel out negative number beside x so we get positive x

Manifoldee

so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)

Manifoldee

so -1/-1=1

Manifoldee

so x=-8

Manifoldee

SO THE ANSWER IS X=-8

Manifoldee

so we should prove it

Manifoldee

x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India

mantu

lol i just saw its x²

Manifoldee

x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)

Manifoldee

I mean x²=2x+8 by factorization method

Kristof

I think x=-2 or x=4

Kristof

x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia

Mohamed

1KI POWER 1/3 PLEASE SOLUTIONS

Prashant Reply

hii

Amit

how are you

Dorbor

well

Biswajit

can u tell me concepts

Gaurav

Find the possible value of 8.5 using moivre's theorem

Reuben Reply

which of these functions is not uniformly cintinuous on (0, 1)? sinx

Pooja Reply

which of these functions is not uniformly continuous on 0,1

Basant Reply

solve this equation by completing the square 3x-4x-7=0

Jamiz Reply

X=7

Muustapha

mantu

x=7

mantu

3x-4x-7=0 -x=7 x=-7

x=-7

mantu

9x-16x-49=0 -7x=49 -x=7 x=7

mantu

what's the formula

Modress

-x=7

Modress

new member

siame

what is trigonometry

Jean Reply

deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent

Thomas

solve for me this equational y=2-x

Rubben Reply

what are you solving for

Alex

solve x

Rubben

you would move everything to the other side leaving x by itself. subtract 2 and divide -1.

Nikki

then I got x=-2

Rubben

it will b -y+2=x

Alex

goodness. I'm sorry. I will let Alex take the wheel.

Nikki

ouky thanks braa

Rubben

I think he drive me safe

Rubben

how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m

Pab Reply

More example of algebra and trigo

Stephen Reply

What is Indices

Yashim Reply

If one side only of a triangle is given is it possible to solve for the unkown two sides?

Felix Reply

cool

Rubben

kya

Khushnama

please I need help in maths

Dayo Reply

Okey tell me, what's your problem is?

Navin

<< Chapter < Page Page > Chapter >>

Practice Key Terms 4

Read also:

Get the best Algebra and trigonometry course in your pocket!

100% Free Android Mobile Application
Complete Textbook by OpenStax
Multiple Choices Questions (MCQ)
Essay Questions Flash Cards
Key-Terms Flash Cards

Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?

Ask

	4 Polar form of complex numbers By OpenStax Read Online Course
	2 Polar form of complex numbers By OpenStax Read Online Course
	4 Microbiology Final 4 By Madison Christian Start Quiz
	World Capitals Quiz By Yasser Ibrahim Start Quiz
	10 AP Key Terms 10 Muscle Tissue Key Terms By OpenStax Start Key Terms
	10 Neuroanatomy 10 Corticospinal Tract By M.D. Stephen Voron Start Quiz
	7 Arts Society: Theater 7 By Jonathan Long Start Quiz
	12 Dr Dowers Endocrinology By Brooke Delaney Start Exam
	Psychology By OpenStax Read Online Course
	Microeconomics Test 1 By IES Portal Start Test
	2 BOD - Neuropathology By Brooke Delaney Start Exam
©flickr: Abraham	Biology Exam 3 By Vanessa Soledad Start Exam