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

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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}

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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}

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Try It

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

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

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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}

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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

By the definition, is such that 0!=1.why?

Unikpel Reply

(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)

Ajay Reply

hatdog

Mark

how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching

Shahid Reply

bsc F. y algebra and trigonometry pepper 2

Aditi Reply

given that x= 3/5 find sin 3x

Adamu Reply

remove any signs and collect terms of -2(8a-3b-c)

Joeval Reply

-16a+6b+2c

Will

is that a real answer

Joeval

(x2-2x+8)-4(x2-3x+5)

Ayush Reply

sorry

Miranda

x²-2x+9-4x²+12x-20 -3x²+10x+11

Miranda

x²-2x+9-4x²+12x-20 -3x²+10x+11

Miranda

(X2-2X+8)-4(X2-3X+5)=0 ?

master

The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer

master

The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer

master

X2-2X+8-4X2+12X-20=0 (X2-4X2)+(-2X+12X)+(-20+8)= 0 -3X2+10X-12=0 3X2-10X+12=0 Use quadratic formula To find the answer answer (5±Root11i)/3

master

Soo sorry (5±Root11* i)/3

master

x2-2x+8-4x2+12x-20 x2-4x2-2x+12x+8-20 -3x2+10x-12 now you can find the answer using quadratic

Mukhtar

2x²-6x+1=0

Ife

explain and give four example of hyperbolic function

Lukman Reply

What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?

Racelle Reply

y/y+10

Find nth derivative of eax sin (bx + c).

Anurag Reply

Find area common to the parabola y2 = 4ax and x2 = 4ay.

Anurag

y2=4ax= y=4ax/2. y=2ax

akash

A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden

Jhovie Reply

to find the length I divide the area by the wide wich means 1125ft/25ft=45

Miranda

thanks

Jhovie

What do you call a relation where each element in the domain is related to only one value in the range by some rules?

Charmaine Reply

A banana.

Yaona

a function

Daniel

a function

emmanuel

given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither

Snalo Reply

what are you up to?

Mark Reply

nothing up todat yet

Miranda

jai

hello

jai

Miranda Drice

jai

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jai

which language is that

Miranda

I am living in india

jai

good

Miranda

what is the formula for calculating algebraic

Propessor Reply

I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it

Miranda

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

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