ma 1101 fall 2020 unit 1 complex numbers section 12.1

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MA 1101 Fall 2020 1 | Page Unit 1 Complex Numbers Section 12.1 Basic Definitions Definition 1: For any real number x (x β‰₯ 0), √ = 2 = . Example: √9 = 3 3 2 = 9. Note: If < 0, √ does not exist as a real number. Example: βˆšβˆ’9 is undefined (as a real number). Such numbers, however, do exist and have applications in certain fields (electrical engineering for example). We can work with such radicals with the following definition. Definition 2: The imaginary unit is denoted as j. In particular: = βˆšβˆ’ and = βˆ’ This definition, used in conjunction with the property below, enables us to evaluate and simplify negative square roots. Property of Radicals: √ Γ— = √ Γ— √ [provided at least one of a or b is positive] Example: βˆšβˆ’9 = √9 Γ— βˆ’1 = = Problem 1: Express each radical in terms of j. ) βˆšβˆ’25 ) βˆšβˆ’7 ) 3βˆšβˆ’20 ) βˆ’ βˆšβˆ’32 ) √ βˆ’3 16

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Page 1: MA 1101 Fall 2020 Unit 1 Complex Numbers Section 12.1

MA 1101 Fall 2020

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Unit 1 Complex Numbers

Section 12.1 Basic Definitions

Definition 1:

For any real number x (x β‰₯ 0), √π‘₯ = 𝑦 𝑖𝑓 𝑦2 = π‘₯.

Example:

√9 = 3 π‘π‘’π‘π‘Žπ‘’π‘ π‘’ 32 = 9.

Note:

If π‘₯ < 0, √π‘₯ does not exist as a real number.

Example:

βˆšβˆ’9 is undefined (as a real number).

Such numbers, however, do exist and have applications in certain fields (electrical engineering for

example). We can work with such radicals with the following definition.

Definition 2:

The imaginary unit is denoted as j. In particular:

𝒋 = βˆšβˆ’πŸ and π’‹πŸ = βˆ’πŸ

This definition, used in conjunction with the property below, enables us to evaluate and simplify negative

square roots.

Property of Radicals:

βˆšπ‘Ž Γ— 𝑏 = βˆšπ‘Ž Γ— βˆšπ‘ [provided at least one of a or b is positive]

Example:

βˆšβˆ’9 = √9 Γ— βˆ’1 =

=

Problem 1:

Express each radical in terms of j.

π‘Ž) βˆšβˆ’25 𝑏) βˆšβˆ’7 𝑐) 3βˆšβˆ’20

𝑑) βˆ’ βˆšβˆ’32 𝑒) βˆšβˆ’3

16

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Property of negative radicals:

βˆšβˆ’π‘Ž = 𝑗 βˆšπ‘Ž

Note:

The property βˆšπ’‚ Γ— βˆšπ’ƒ = βˆšπ’‚ Γ— 𝒃 can be used to combine a product of two radicals in many cases.

As noted on the previous page, however, this property may not be applied if both a and b are negative. In

such cases, first simplify each radical individually.

Problem 2:

Find and simplify each product.

π‘Ž) √4 Γ— √9 𝑏) βˆšβˆ’4 Γ— √9

𝑐) βˆšβˆ’4 Γ— βˆšβˆ’9 𝑑) βˆšβˆ’3 Γ— βˆšβˆ’4

𝑒) βˆ’ βˆšβˆ’2 Γ— βˆšβˆ’10 𝑓) βˆ’ (√(βˆ’2)(βˆ’5) ) (βˆšβˆ’10 )

𝑔) βˆšβˆ’5

3 Γ— √

4

15 β„Ž) √

βˆ’9

2 Γ— √

βˆ’16

25

Do #’s 5,7,9,11,13,17,19,21, p. 360 text

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Simplifying an expression of the form 𝑗𝑛 (where n is a whole number)

Example:

Simplify:

π‘Ž) 𝑗8 𝑏) 𝑗11

Useful strategy:

If the power n is even, rewrite 𝑗𝑛 as (𝑗2)π‘š. Use the definition of 𝑗2 = βˆ’1 to complete.

If the power n is odd, rewrite 𝑗𝑛 as (π‘—π‘›βˆ’1) 𝑗. Proceed as before.

Problem 3:

Simplify the following.

π‘Ž) 𝑗26 𝑏) 𝑗31 𝑐) 𝑗9 βˆ’ 𝑗7

𝑑) βˆ’ 𝑗4 𝑒) (βˆ’π‘—)4

Do #’s 25,27,29,31, p. 360 text

Complex Numbers

Definition 3:

A number of the form 𝐚 + 𝐛𝐣, where a and b are real numbers, is called a complex number.

Notes:

If a = 0, the resulting number bj is a pure imaginary number.

If b = 0, the resulting number a is a real number.

Conclusion:

Complex numbers include the set of real numbers as well as the set of pure imaginary

numbers.

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For example:

The pure imaginary number 2𝑗 could be rewritten as 0 + 2𝑗.

The real number -5 could be rewritten as βˆ’5 + 0𝑗.

Note:

The form 𝐚 + 𝐛𝐣 is known as the rectangular form of a complex number.

In this form:

β€œa” is the real part.

β€œb” is the imaginary part.

Problem 4:

Express each of the following complex numbers in rectangular form.

π‘Ž) βˆ’ √9 βˆ’ βˆšβˆ’9 𝑏)√54 βˆ’ βˆšβˆ’24

𝑐) 𝑗5 βˆ’ 4 𝑑) βˆšβˆ’25𝑗2 + βˆšβˆ’16

Do #’s 33,35,37,39,41,43, p. 360 text

Note:

Each complex number is unique. Therefore, two complex numbers (π‘₯ + 𝑦𝑗) π‘Žπ‘›π‘‘ (π‘Ž + 𝑏𝑗) are equal

only if π‘₯ = π‘Ž π‘Žπ‘›π‘‘ 𝑦 = 𝑏.

Example:

If π‘₯ + 𝑦𝑗 = 3 βˆ’ 2𝑗, then π‘₯ = 3 π‘Žπ‘›π‘‘ 𝑦 = βˆ’2.

Problem 5:

Find the values of x and y which satisfy the following equations.

π‘Ž) π‘₯ βˆ’ 𝑦𝑗 = βˆ’5 + 𝑗 𝑏) 4π‘₯ + 3𝑗𝑦 = 20 βˆ’ 6𝑗

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𝑐) 4 βˆ’ 3𝑗 + π‘₯ = 6𝑗 + 𝑗𝑦

𝑑) π‘₯ + 3π‘₯𝑗 + 3𝑦 = 5 βˆ’ 𝑗 βˆ’ 𝑗𝑦

Do #’s 49,51,53, p. 360 text

One final point:

The conjugate of the complex number (𝐚 + 𝐛𝐣) is (𝐚 βˆ’ 𝐛𝐣) and vice versa.

Example:

Write the conjugate of the following.

π‘Ž) 3 βˆ’ 5𝑗 𝑏) 6j 𝑐) βˆšβˆ’49 + 4𝑗2

Do #’s 45,47,55,57, pp. 360-361 text

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Section 12.2 Basic Operations with Complex Numbers (See 12.3 – 12.6, p. 361 text)

Case 1: The given numbers are in the form 𝐚 + 𝐛𝐣.

Problem 1:

Perform the indicated operation and simplify.

π‘Ž) (3 + 2𝑗) + (βˆ’5 βˆ’ 𝑗) 𝑏) (βˆ’1 βˆ’ 3𝑗) βˆ’ (5 βˆ’ 2𝑗)

𝑐) 1.5𝑗 (2𝑗 βˆ’ 1.4) 𝑑) (2 βˆ’ 𝑗)(5 + 3𝑗)

𝑒) (βˆ’1 βˆ’ 𝑗)2

Do #’s 5,7,9,13,15,23,25, p. 363 text

Problem 2:

Find the product of 2 + 𝑗 and its conjugate.

Do # 45, p. 363 text

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Division of Complex Numbers

To divide two complex numbers, use the same process as rationalizing the denominator of a rational expression:

multiply the numerator and the denominator by the conjugate of the denominator.

Example:

7βˆ’2𝑗

3+4𝑗

=7βˆ’2𝑗

3+4𝑗×

3βˆ’4𝑗

3βˆ’4𝑗

=21βˆ’28π‘—βˆ’6𝑗+8𝑗2

9βˆ’16𝑗2

=21βˆ’34π‘—βˆ’8

9+16

=13βˆ’34𝑗

25

=πŸπŸ‘

πŸπŸ“βˆ’

πŸ‘πŸ’

πŸπŸ“π’‹

Problem 3:

Simplify the following quotients. Express your final answer in the form 𝐚 + 𝐛𝐣.

a) 2βˆ’j

1+2j 𝑏)

4+3𝑗

2𝑗

𝑐) 2𝑗

3βˆ’π‘—+

1

2 𝑗 (𝑗 + 4)

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Extra Practice:

1. 12+10𝑗

6βˆ’8𝑗

2. 1

𝑗+

2

3+𝑗

3. (6𝑗+5)(2βˆ’4𝑗)

(5βˆ’π‘—)(4𝑗+1)

Do #’s 27,29,31,35,47, p. 363 text

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Case 2: The given numbers require some simplification before proceeding.

Problem 4:

Simplify.

π‘Ž) (3βˆšβˆ’16 + √9 ) βˆ’ (7 βˆ’ 2βˆšβˆ’4) 𝑏) (5 βˆ’ βˆšβˆ’36)(βˆ’βˆšβˆ’1)

𝑐) 𝑗7βˆ’π‘—

2π‘—βˆ’π‘—2

Do #’s 33,39, p. 363 text

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Miscellaneous Problems

1. Express π‘—βˆ’4 + 2π‘—βˆ’1 in rectangular form.

2. Write the reciprocal of 2 βˆ’ 3𝑗 in rectangular form.

3. Solve for x:

(π‘₯ + 5𝑗)2 = 11 βˆ’ 60𝑗

Solutions:

Do #’s 49,51, p. 363 text

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Section 12.3 Graphical representation of a complex number

Recall: Rectangular form of a complex number: π‘₯ + 𝑦𝑗

The complex plane:

A complex number 𝐱 + 𝐲𝐣 may be represented by the point (𝐱, 𝐲) in the complex plane

where:

π‘₯ = the real part 𝑦 = the imaginary part

Example:

Represent each number as a point in the complex plane.

π‘Ž) 2 + 𝑗 𝑏) βˆ’ 3 βˆ’ 2𝑗

Extending the concept: Drawing a line segment from the origin to the point (𝐱, 𝐲)in the plane gives an

alternate way to represent a complex number – as a vector.

Example: Represent each number as a vector in the complex plane.

π‘Ž) 3 + 𝑗 𝑏) βˆ’ 1 βˆ’ 4𝑗

Do #’s 3,5,7, p. 363 text

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Adding complex numbers graphically:

Step 1: Find the point in the plane corresponding to the first number. Draw a vector from the origin to this

point.

Step 2: Repeat this process for the second number.

Step 3: Complete a parallelogram with the lines drawn as adjacent sides. The resulting fourth vertex is the

point representing the sum.

Problem:

Perform the indicated operations graphically.

π‘Ž) (1 + 3𝑗) + (3 + 2𝑗) 𝑏)(4 βˆ’ 𝑗) + (βˆ’2 + 3𝑗)

𝑐) (3 βˆ’ 𝑗) βˆ’ (2 + 2𝑗)

Do #’s 9,11,13,15,17,21,23, p. 363 text

βˆ’ βˆ’ βˆ’

βˆ’

βˆ’

βˆ’

Imaginary

Real

βˆ’ βˆ’ βˆ’

βˆ’

βˆ’

βˆ’

Imaginary

Real

βˆ’ βˆ’ βˆ’

βˆ’

βˆ’

βˆ’

Imaginary

Real

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Section 12.4 Polar form of a complex number

Example:

Represent the number 3 + 4𝑗 as a vector in the complex plane.

Notes:

As a vector possesses both magnitude and direction, so too does a complex number.

The magnitude is simply the length of the vector.

The direction is the angle πœƒ formed between the positive real axis and the vector itself.

The form:

Consider the complex number 𝒙 + π’šπ’‹ represented as a vector in the complex plane.

Expressing x and y in terms of r using basic trigonometry we get:

cos πœƒ = π‘₯

π‘Ÿ β†’

sin πœƒ = 𝑦

π‘Ÿ β†’

Substituting for x and y, the number 𝒙 + π’šπ’‹ =

βˆ’ βˆ’ βˆ’

βˆ’

βˆ’

βˆ’

Imaginary

Real

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To convert the number 𝒙 + π’šπ’‹ to polar form, we need to find the values of β€œr” and β€œΞΈβ€ .

How to solve for β€œr” and β€œΞΈβ€:

π‘Ÿ2 = π‘₯2 + 𝑦2 (r > 0)

tan πœƒ = 𝑦

π‘₯ (0 ≀ πœƒ < 360Β°)

Problem 1:

Represent each of the following complex numbers graphically and determine its polar form.

π‘Ž)3 + 4𝑗 𝑏) βˆ’ 2 βˆ’ π‘—βˆš5

Do #’s 3,5,7,9,11,15,17, p. 368 text

βˆ’ βˆ’ βˆ’

βˆ’

βˆ’

βˆ’

Imaginary

Real

βˆ’ βˆ’ βˆ’

βˆ’

βˆ’

βˆ’

Imaginary

Real

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Alternate notation for polar form:

π‘Ÿβˆ πœƒ = π‘Ÿ(cos πœƒ + 𝑗 sin πœƒ) Examples:

π‘Ž) 5(π‘π‘œπ‘  53.1Β° + 𝑗𝑠𝑖𝑛 53.1Β°) =

𝑏) 1.82 ∠ 320Β° =

Problem 2:

Express each complex number in rectangular form.

π‘Ž) 2.57(π‘π‘œπ‘  112.3Β° + 𝑗 𝑠𝑖𝑛 112.3Β°) 𝑏) 1.82 ∠ 320Β°

Problem 3:

Represent the number 4(π‘π‘œπ‘ 270Β° + 𝑗𝑠𝑖𝑛270Β°) graphically and express it in rectangular

form.

Do #’s 19,21,25,27,29,33,35, p. 363 text

Multiplying and Dividing complex numbers in polar form

Basic Results:

𝑖) (π‘Ÿ1βˆ πœƒ1)(π‘Ÿ2βˆ πœƒ2) = π‘Ÿ1π‘Ÿ2 ∠ (πœƒ1 + πœƒ2)

𝑖𝑖) (π‘Ÿ1βˆ πœƒ1)

(π‘Ÿ2βˆ πœƒ2)=

π‘Ÿ1

π‘Ÿ2 ∠(πœƒ1 βˆ’ πœƒ2)

βˆ’ βˆ’ βˆ’

βˆ’

βˆ’

βˆ’

Imaginary

Real

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Problem 4:

Perform the indicated operation.

π‘Ž) (1.5 ∠ 230Β°)(2.4 ∠ 150Β°) 𝑏) 3.2∠67.3Β°

1.3∠154.1°

Do #’s 5,7,9,11,17,19,25,27,29, pp. 375-376 text

Consider the problem below:

Find the sum: 2 ∠ 30° + ∠ 60°

Notes:

β€’ Unlike multiplication and division of numbers in polar form, addition cannot be done.

β€’ The above sum can only be found by first converting each number to rectangular form,

adding the results and then converting this number to polar form.

Solution:

Do #’s 21,23, p. 375 text

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Applications:

a.) Given that the current in a circuit is 3.90 βˆ’ 6.04π‘—π‘šπ΄ and the impedance is 5.16 + 1.14π‘—π‘˜Ξ©, find the

magnitude of the voltage.

b.) Given that the voltage in a given circuit is 8.3 βˆ’ 3.1𝑗𝑉 and the impedance is 2.1 βˆ’ 1.1𝑗Ω, find the

magnitude of the current.

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c.) Two resistors have resistances that can be expressed as 𝑅1 = 5∠30π‘œ and 𝑅2 = 8∠120π‘œ. What is

the total resistance if the resistors are in series? Parallel?

Do # 55, p. 363 text

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Section 12.5 Exponential form of a complex number

Exponential form of a complex number is written as:

π’“π’†π’‹πœ½

As with polar form:

r is the magnitude of the representative vector

πœƒ is the angle formed by this vector (in radians)

Recall:

An angle is converted from degrees to radians by multiplying by πœ‹

180.

Example:

127.1Β° =127πœ‹

180 = 2.22

Probkems:

1. Express the complex number 5.24 ∠ 118.2° in:

a) exponential form b) rectangular form

2. Express 4.15 𝑒5.60𝑗 in polar form.

Summary of the three forms:

Rectangular: 𝒙 + π’šπ’‹

Polar: 𝒓(𝐜𝐨𝐬 𝜽 + 𝒋 𝐬𝐒𝐧 𝜽) = π’“βˆ πœ½ (𝜽 π’Šπ’ π’…π’†π’ˆπ’“π’†π’†π’”)

Exponential: π’“π’†π’‹πœ½ (𝜽 π’Šπ’ π’“π’‚π’…π’Šπ’‚π’π’”)

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3. Express βˆ’3 βˆ’ 2𝑗 in exponential form.

4. Simplify the following product and express the result in rectangular form.

(12.3 𝑒1.54𝑗)(5.9 𝑒3.17𝑗)

Do #’s 3,5,7,9,11,15,17,19,21,23,25,27,29,31,33 pp. 370-371 text

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Section 12.6 Powers and Roots of complex numbers

Topic 1:

Raising a complex number (in polar form) to a power

Demoivre’s Theorem:

(π’“βˆ πœ½)𝒏 = 𝒓𝒏 βˆ π’πœ½

Problems:

1. Simplify the following. Leave your answer in polar form.

π‘Ž) (0.5 ∠ 100Β°) 6 𝑏) [ 2(π‘π‘œπ‘  15Β° + 𝑗 𝑠𝑖𝑛 15Β°)]4

2. Change 2 βˆ’ 𝑗 to polar form. Using this result, simplify (2 βˆ’ 𝑗)5.

Do #’s 13,15,33,35, pp. 375-376 text

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Topic 2:

Using DeMoivre’s Theorem, find the β€œn” 𝒏𝒕𝒉 roots of a complex number (in polar form).

Basic Steps:

β€’ Express the given number in polar form.

β€’ Apply the appropriate fractional exponent (as per the root of interest) and determine

the first root…using DeMoivre’s Theorem.

β€’ Add 360Β° π‘‘π‘œ πœƒ and repeat step 2 to find the second root.

β€’ Continue adding 360Β° and repeating the process of step 2 until all required roots

have been found.

Example: Find the 3 cube roots of 3 βˆ’ 4𝑗

Step 1: 3 βˆ’ 4𝑗 = 5∠306.8699Β°

Step 2: cube root = exponent of 1

3

Step 3: (5∠306.8699°)1

3 = 51

3∠1

3(306.8699°) = 1.71∠102.3° (this is the first root)

Step 4: (5∠666.8699°)1

3 = 51

3∠1

3(666.8699°) = 1.71∠222.3° (second root)

(5∠1026.8699°)1

3 = 51

3∠1

3(1026.8699°) = 1.71∠342.3° (third root)

Example:

Find the cube roots of βˆ’1:

βˆ’1 = βˆ’1 + 0𝑗 = 1∠180Β°

First root: (1∠180°)1

3 = 11

3∠1

3(180Β°) = 1∠60Β° = 0.5 + 0.87𝑗

Second root: (1∠540°)1

3 = 11

3∠1

3(540Β°) = 1∠180Β° = βˆ’1

Third root: (1∠900°)1

3 = 11

3∠1

3(900Β°) = 1∠300Β° = 0.5 βˆ’ 0.87𝑗

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Example:

Find the square roots of 2𝑗. Give your answers in rectangular form.

Problem:

Find the cube roots of each of the following:

π‘Ž) 2 βˆ’ 𝑗

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b) 27

Do #’s 37,39,41,43,49, p. 376 text