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

what is the answer to dividing negative index

Morosi Reply

In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.

Shivam Reply

give me the waec 2019 questions

Aaron Reply

the polar co-ordinate of the point (-1, -1)

Sumit Reply

prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x

Rockstar Reply

tanh`(x-iy) =A+iB, find A and B

Pankaj Reply

B=Ai-itan(hx-hiy)

Rukmini

what is the addition of 101011 with 101010

Branded Reply

If those numbers are binary, it's 1010101. If they are base 10, it's 202021.

Jack

extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero

archana Reply

the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve

Kc Reply

1+cos²A/cos²A=2cosec²A-1

Ramesh Reply

test for convergence the series 1+x/2+2!/9x3

success Reply

a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?

Lhorren Reply

100 meters

Kuldeep

Find that number sum and product of all the divisors of 360

jancy Reply

answer

Ajith

exponential series

Naveen

yeah

Morosi

prime number?

Morosi

what is subgroup

Purshotam Reply

Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1

Macmillan Reply

<< 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
	Biology By OpenStax Read Online Course
	2011 Dynamics CRM By Danielrosenberger Start Quiz
	Spanish Lesson 4 By Anonymous User Start Quiz
	Time and Stress Management MCQ By Dionne Mahaffey Start Quiz
©flickr: iClassical	Music Apperciation By Courntey Hub Start Test
	Spanish (Places) By Inderjeet Brar Start Flashcards
©flickr:	Biology 1 By Jill Zerressen Start Quiz
	9 BOD- Liver Quiz .... By Brooke Delaney Start Quiz
	2 Arts Society: Theater 2 By Jonathan Long Start Quiz
	11 Neuroanatomy 11 The Cerebellum By M.D. Stephen Voron Start Quiz