Chapter 8 Chapter 8 - Complex Numbers and Polar Graphs

Chapter 8

Chapter 8 - Complex Numbers and Polar Graphs

8.1 Complex Numbers

8.2 Trigonometric Form of a Complex Number

8.3 Powers and Roots of a Complex Number (De Moivre’s Theorem) 8.4 Polar Coordinates

8.5 Polar Equations and Their Graphs 8.6 Parametric Equations and Their Graphs SE8 Sample Exam Questions

8.1 Complex Numbers

Learning Objectives

• Define the imaginary unit and complex numbers. • Add and subtract complex numbers.

• Multiply and divide complex numbers.

Complex numbers are used in many fields including electronics, engineering, physics, and mathematics. In this textbook, we will use them to better understand solutions to equations such as 2

4 0

x + = .

Objective: Define the imaginary unit and complex numbers.

We begin by defining the imaginary unit, i, as the square root of −1.

2

1 1

i= − and i = −

To express a square root of a negative number in terms of the imaginary unit i, we use the following property where a represents any non-negative real number:

1 1

a a a i a

− = − ⋅ = − ⋅ =

Using this property, we can write

9 1 9 1 9 i 3 3i

− = − ⋅ = − ⋅ = ⋅ =

If − = , then we would expect that 9 3i 3isquared will equal −9,

( )

2 ₂

( )

3i =9i = − = −9 1 9 

In this way, any square root of a negative real number can be written in terms of the imaginary unit. Such numbers are often called imaginary numbers.

Example: Rewrite in terms of the imaginary unit i: a. − b. 7 − c. 25 − 72

Solution:

a. − = − ⋅ = − ⋅7 1 7 1 7 = ⋅ = i 7 7i b. 25− = − ⋅1 25= − ⋅1 25= ⋅ = i 5 5i

c. 72− = − ⋅ ⋅ = − ⋅1 36 2 1 36⋅ 2= ⋅ ⋅i 6 2=6i 2

Notation Note: When an imaginary number involves a radical, we place i in front of the radical. Consider the following:

6i 2=6 2i

Since multiplication is commutative, these numbers are equivalent. However, in the form 6 2i

, the imaginary unit i is often misinterpreted to be part of the radicand. To avoid this confusion, it is a best practice to place the imaginary unit in front of the radical and use 6i 2.

Try this! Rewrite in terms of the imaginary unit: − 50

Answer: 5i 2

A complex number is any number of the form, z= +a bi

Here a and b are real numbers, where a is called the real part and b is called the imaginary

part. For example,

3 4

z= − i

In this case, the complex number z has a real part of 3 and an imaginary part of −4. It is important to note that any real number is also a complex number. For example, 7 is a real number; it can be written as 7+0i with a real part of 7 and an imaginary part of 0. Hence, the set of real numbers, denoted , is a subset of the set of complex numbers, denoted.

{

a bi a b| ,

}

= + ∈

 

Example: Write the complex number in standard form and state the real and imagery parts.

6+ − 8

Solution: The goal is to rewrite this complex number in the form a+bi .

6 8 6 1 4 2

6 2i 2 + − = + − ⋅ ⋅

= +

Answer: 6+2i 2; Real part: 6; Imaginary part: 2 2

Try this! Write the complex number in standard form and state the real and imagery parts:

5 16

− − −

Answer: − −5 4i ; Real part: −5; imaginary part: −4

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Objective: Add and subtract complex numbers.

Adding and subtracting complex numbers is performed by adding and subtracting like terms. We add or subtract the real parts and then the imaginary parts.

Example: Given z= −5 2i and w= +7 3i find: a. z+w b. z−w

Solution:

a. Add the real parts and add the imaginary parts.

(

5 2

) (

7 3

)

5 2 7 3 5 7 2 3 12 z w i i i i i i i + = − + + = − + + = + − + = +

b. To subtract complex numbers, we subtract the real parts and subtract the imaginary parts. This is consistent with the use of the distributive property.

(

5 2

) (

7 3

)

5 2 7 3 5 2 z w i i i i i − = − − + = − − − = − − 570

Answer: a. 12+i b. − −5i 2

In general, given real numbers a, b, c and d:

(

) (

) (

) (

)

(

) (

) (

) (

)

a bi c di a c b d i a bi c di a c b d i + + + = + + + + − + = − + −

Adding or subtracting complex numbers always results in a complex number.

Example: Simplify:

(

5+ + −i

) (

2 3i

) (

− −4 7i

)

.

Solution: Add all of the real parts and then add the imaginary parts.

(

5

) (

2 3

) (

4 7

)

5 2 3 4 7 3 5 i i i i i i i + + − − − = + + − − + = + Answer: 3 5i+

Try this! Simplify:

(

4 5+ i

) (

− −2 3i

) (

+ − 8 i

)

Answer: 10+7i

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Objective: Multiply and divide complex numbers.

Multiplying complex numbers is performed in the same way as multiplying polynomials – the distributive property applies. Furthermore, we make use of the fact that i2 = −1 to simplify the result into the standard form a+bi.

Example: Given z= −6i and w= −2 3i find z w⋅ .

Solution: We begin by applying the distributive property.

(

)

( )

2 2 Distribute. Substitute 1 6 2 3 . Simplif 12 18 12 18 1 12 1 1 y. 8 18 2 i z w i i i i i i i ⋅ = − − = − + = − + − = − − = − − = − Answer: − −18 12i

In general, given real numbers a, b, c and d:

(

)(

)

( )

(

)

(

) (

)

2 1

a bi c di ac adi bci bdi ac adi bci bd ac ad bc i bd ac bd ad bc i + + = + + + = + + + − = + + − = − + +

Multiplying complex numbers always results in a complex number.

Try this! Given z= −3 2i find z . 2

Answer: 5 12i−

Given a complex number z= +a bi, its complex conjugate is z = −a bi. The bar above the variable used for a complex number denotes the complex conjugate.

Example: Given z= +5 2i find z z⋅ .

Solution:

(

)(

)

( )

2 5 2 5 2 25 10 10 4 25 4 1 25 4 29 z z i i i i i ⋅ = + − = − + − = − − = + = Answer: 29

In general, the product of complex conjugates follows:

(

)(

)

( )

2 2 2 2 2 2 2 1 z z a bi a bi a abi abi b i a b a b ⋅ = + − = − + − = − − = +

Note that result does not involve the imaginary unit; hence, it is real. This leads us to the very useful property,

2 2

z z⋅ =a +b

Dividing complex numbers is performed using the technique used for rationalizing denominators involving square roots. This typically involves multiplying the numerator and denominator by the conjugate of the denominator. The result can then be simplified into standard form a+bi.

Example: Given z= −2 3i find 1 z.

Solution: In this example, the conjugate of the denominator is z = +2 3i. Therefore, we will multiply by 1 in the form

(

)

(

)

2 3 2 3 i i + + .

(

)

(

(

)

)

(

)

2 2 1 1 2 3 2 3 2 3 2 3 1 2 2 3 3 3 z i i i i i = ⋅ − + = + + = + +

To write this complex number in standard form, we make use of the fact that 13 is a common denominator. 2 3 2 3 13 13 13 2 3 13 13 i i i + _{= +} = + Answer: 2 3 13+13i 573

Example: Given z= −1 5i and w= +4 i find z w . Solution:

(

)

(

)

(

(

)

)

( )

2 2 2 1 5 4 4 20 5 4 1 4 20 5 1 17 1 21 4 4 17 i z w i i i i i i i i i − = ⋅ + − − + = + − − + − = − − = − − Answer: 1 21 17 17i − −

Dividing two complex numbers always results in a complex number.

Try this! Given z= +3 2i and w= −1 i find z w .

Answer: 1 5

2+2i

When multiplying and dividing complex numbers we must take care to understand that the product and quotient rules for radicals require that a and b are not both negative. In this case, we have the following rules.

Product rule for radicals: n _{a b}⋅ = n _a⋅n_b Quotient rule for radicals:

n n

n

a a

b = b

For example, we can demonstrate that the product rule is true when a and b are both positive as follows:

4 9 36 2 3 6 6 6 ⋅ = ⋅ = = 

However, when a and b are both negative the property is not true.

? 2 4 9 36 2 3 6 6 6 6 6 i i i − ⋅ − = ⋅ = = − = 

Here −4 and − both are not real numbers and the product rule for radicals fails to produce 9 a true statement. Therefore, to avoid some common errors associated with this technicality, ensure that any complex number is written in terms of the imaginary unit i before performing any operations.

Example: Multiply: − ⋅ − . 6 15

Solution: Begin by writing the radicals in terms of the imaginary unit i.

6 15 i 6 i 15

− ⋅ − = ⋅

Now the radicands are both positive and the product rule for radicals applies.

( )

( )

2 6 15 6 15 90 1 9 10 1 3 10 3 10 i i i − ⋅ − = ⋅ = = − ⋅ = − ⋅ ⋅ = − Answer: 3 10−

Try this! Multiply: −2 3

(

− − . 6

)

Answer: 2 3+3i 2

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Key Takeaways ***** ***** *****

• The imaginary unit i is defined to be the square root of negative one. In other words,

1

i= − and i2 = − . 1

• Complex numbers have the form z= +a bi where a and b are real numbers. • The set of real numbers is a subset of the complex numbers.

• The result of adding, subtracting, multiplying, and dividing complex numbers is a complex number.

• The product of complex conjugates, z= +a bi and z = −a bi, is a real number. Use this fact to divide complex numbers. Multiply the numerator and denominator of a fraction by the complex conjugate of the denominator and then simplify.

• Ensure that any complex number is written in terms of the imaginary unit i before performing any operations.

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Section Exercises Part A:

Rewrite in terms of the imaginary unit i.

1. − 25 2. − −4 3. − − 36 4. − 20 5. 2 9 − 6. −1.44

Write the complex number in standard form a+bi.

7. 5 2− −4 8. 3 5− − 9 9. − +2 3 − 8 10. 4 2− − 18 11. 3 24 6 − − 12. 2 75 10 + − 13. 63 5 12 − − − 14. 72 8 24 − − + −

Given that i2 = −1 compute the following powers of i . 15. i3 16. i4 17. i5 18. i6 19. i15 20. i24 576

Part B:

Given z and w find z+w and z−w.

21. z= +3 5i and w= −7 4i 22. z= − −8 3i w= +5 2i 23. z= − +10 15i and w=15 20− i 24. z= +5 2i and w= −8 3i 25. 1 3 2 4 z= + i and 1 1 6 8 w= − i 26. 3 1 8 3 z= − i and 1 1 2 2 w= − i 27. z=0.25 0.75− i and 1.25 0.75 w= − + i 28. z= −1.2+2.4i and 3.8 1.6 w= − i

Perform the operations.

29.

(

2− + +i

) (

3 4i

) (

− −6 5i

)

30.

(

7 2+ i

) (

− − − −6 i

) (

3 4i

)

31. 1 1 1 1 1 3 i 2i 6 6i  _{− − −}   _{− +}              32. 1 3 5 1 5 4i 2 i 4 8i  ₋  _{+ + −}  ₋              33.

(

5 3− i

) (

− +2 7i

) (

− −1 10i

)

34.

(

6 11− i

) (

+ +2 3i

) (

− −8 4i

)

Simplify and write in standard form (a+bi).

35. −16− − −

(

3 1

)

36. −100+

(

− +9 7

)

37.

(

1+ − − − −1

) (

1 1

)

38.

(

3− −81

) (

− −5 3 −9

)

39.

(

5 2− −25

) (

− − +3 4 −1

)

40.

(

− − − − − −12 1

) (

3 49

)

Part C:

Given z and w find z w⋅ and z

w. 41. z=15 and w= +1 2i 42. z=2i and 1 3i− 43. z= +2 i and w= −2 3i 44. z= −1 5i and w= +3 4i 45. z= − −4 2i and w= −1 i 46. z= +6 12i and w= −3 i 47. z= − −5 2i and w= − +5 2i 48. z= +4 2i and w= −4 2i Given z find z z⋅ . 49. z= −4 5i 50. z= − +3 7i 51. z= − +5 i 52. z= −6 10i 53. z= 3− 2i 54. z= 2−i 5 577

Given that i n 1_n i

− ₌

compute the following powers of i . 55. i−1

56. i−2

57. i−3 58. i−4

Perform the operations.

59. −2

(

− −2 6

)

60. −1

(

− +1 8

)

61.

(

2 3− −2

)(

2 3+ −2

)

62.

(

1+ −5 1

)(

− −5

)

63. 1 1 1 1 − − + − 64. 6 18 4 − − + −

Perform the operations and simplify.

65. 2 1 2

(

− i

)

2+3i 66.

(

1−i

)

2−2 1

(

− +i

)

2 67.

(

2−i

) (

2− 2+i

)

2 68.

(

i 3 1+

) (

2− 4i 2

)

2 69.

( )

1 i+ 3 70.

(

2 i−

)

3 71. 1 a bi− 72. a bi c di + + 73.

(

a bi−

) (

2− a bi+

)

2 74.

(

2

)(

2

)

1 1 a + +ai a − +ai

75. Show that both ±2i satisfy x2+ =4 0.

76. Show that both 3 2i− and 3 2i+ satisfy x2−6x+13=0.

77. Show that 3, −2i, and 2i are all solutions to x3−3x2 +4x−12=0. 78. Show that −2, 1 i− , and 1 i+ are all solutions to x3−2x+ =4 0.

Answers to Section Exercises Part A: 1. 5i 2. −2i 3. −6i 4. 2i 5 5. 2 3 i 6. 1.2i 7. 5 4i− 8. 3 15i− 9. − +2 6i 2 10. 4 6− i 2 11. 1 6 2− 3 i 12. 1 3 5+ 2 i 13. 5 7 12 − 4 i 14. 2 2 12 4 i − + 15. − i 16. 1 17. i 18. −1 19. − i 20. 1 Part B: 21. 10+i; − +4 9i 22. − −3 i; − −13 5i 23. 5 5i− ; − +25 35i 24. − +3 5i; − +3 5i 25. 2 5 3+8i; 1 7 3+8i 26. 7 5 8−6i; 1 1 8 6i − + 27. −1; 1.5 1.5i− 28. 2.6+0.8i; 5.0 4.0i − + 29. − +1 8i 30. − +2 7i 31. 5 2 6 3i − − 32. 13 7 4 +8i 33. 2 34. −4i 35. − +3 5i 36. 7 13i+ 37. 2i 38. −2 39. 8 14i− 40. − +15 6i Part C: 41. 15 30i+ ; 3 6i− 42. 6+2i; 3 1 5 5i − + 43. 7−4i; 1 8 13 13+ i 44. 23 11i− ; 17 19 25 25i − − 45. − +6 2i; − −1 3i 46. 30 30i+ ; 3 21 5+ 5 i 47. 29; 21 20 29+29i 48. 20; 3 4 5+5i 49. 41 50. 58 51. 26 52. 136 53. 7 54. 7 55. − i 56. − 1 57. i 58. 1 59. 2 2− − i 3 60. − +1 2i 2 61. 22 62. 6 63. − i 64. 6 3 3 11 11 i − − 65. − −6 5i 66. 0 67. −4i 2 68. 30 2+ i 3 69. − +2 2i 70. 2 11i− 579

71. a bi_{2 2} a b + + 72. ac bd_{2 2} bc₂ ad₂ i c d c d + −   ₊   ₊   ₊      73. −4abi 74. 4 2 3 1 a + a + 75. Proof. 76. Proof. 77. Proof. 78. Proof. 580

8.2 Trigonometric Form of a Complex Number

Learning Objectives

1. Define and graph complex numbers in the complex plane. 2. Define trigonometric form of a complex number.

3. Products and Quotients of complex numbers in trigonometric form.

There is a close connection between vectors and complex numbers. In this section, we explore a method for defining complex numbers as vectors and define them in terms of trigonometric functions. Doing this will streamline calculations requiring algebraic operations.

Objective: Define and graph complex numbers in the complex plane.

A complex number can be viewed as a position vector extending from the origin, where the real part represents the magnitude of the horizontal component and the imaginary part represents the magnitude of the vertical component. In this way, every complex number is uniquely associated with a position in the complex plane. The complex plane looks similar to the rectangular coordinate plane, where the x-axis represents the real part of the complex number and is called the real axis (Re), and the y-axis represents the imaginary part of the complex number and is called the imaginary axis (Im). Furthermore, a complex number written as z= + is said to be in rectangular form. a bi

For example, consider the following complex numbers and their corresponding graphs in the complex plane.

4 3 3 2 i i − + − − 3 0 3 5 5 0 i i i = + = +

Real numbers lie on the x-axis because the imaginary part is 0. When the real part is 0, the complex number lies on the y-axis and is called a pure imaginary number.

Example: Graph the set of complex numbers in the same complex plane:

{

−i i, , − − − + 1 i, 1 i

}

Solution: Begin by determining the real and imaginary parts of each complex number.

1 1 i i − + − − 0 0 i i i i = + − = −

Choose an appropriate scale for the complex plane and graph the corresponding vectors.

Answer:

Notice the symmetry between complex conjugates.

Try this! Graph the set of complex numbers in the same complex plane:

{

−2, 1, − +3 2 , 3 2i + i

}

Answer:

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Objective: Define trigonometric form of a complex number.

Using the fact that a complex number can be considered as a vector in the complex plane, we next apply trigonometry.

The vector that represents the complex number z= + has magnitude r and forms an angle x iy θ with the positive real axis. In the language of complex numbers, the magnitude of a complex number is called the modulus, denoted z . Therefore,

2 2

r= =z x +y

The angle that the vector makes, in standard position, is called its argument. We have,

cos cos x r x r θ θ = = sin sin y r y r θ θ = =

Now substituting rcosθ for x and rsinθ for y and then factoring out r, we have

(

)

cos sin cos sin x iy r i r r i θ θ θ θ + = + ⋅ = +

Which gives us trigonometric form of a complex number z,

(

cos sin

)

cis z x iy r i r θ θ θ = + = + =

Sometimes the quantity “cosine theta plus i sine theta” is abbreviated cisθ . Furthermore, the

following facts are used to convert from rectangular form to trigonometric form.

2 2 2

r =x + and y tan y

x θ =

The steps for finding trigonometric form of any complex number z= + follow: x iy

Step 1: Determine the modulus using r= z = x2+y2 .

Step 2: Determine the argument, angle θ, formed with the real axis using tan y

x θ = .

Step 3: Substitute r and angle θ into z=r

(

cosθ +isinθ

)

. Sketching the complex number as a vector helps.

Example: Write in trigonometric form: z= 3+ i

Solution: Here, the real part x= 3 and the imaginary part y=1.

Find r and θ as follows:

( )

( )

2 2 2 ₂ 3 1 4 2 r= z = x +y = + = = 1 tan 1 3 1 tan 30 3 y x θ θ − = =   = _{ }= °  

Now substituting, r= and 2 θ =30°

(

)

(

)

cos sin 2 cos 30 sin 30 2 cis 30 x iy r i i θ θ + = + = ° + ° = °

Answer: z=2 cos 30

(

° +isin 30°

)

As a check, simplifying trigonometric form should lead to the original complex number.

(

)

3 1 2 cos 30 sin 30 2 2 2 3 i i i   ° + ° = _ + _   = + 

The argument can be expressed using radians, the process is the same.

Example: Write in trigonometric form (use radians): z= −1 i

Solution: Here, the real part x=1 and the imaginary part y= −1.

Find r and θ as follows:

( ) ( )

2 2 2 2 1 1 2 r= z = x +y = + − =

( )

1 1 tan 1 1 ˆ _{tan 1} 4 y x θ π θ − − = = = − = − = QIV: 2 7 4 4 π π θ = π− =

Now substituting, r= 2 and 7

4 π θ =

(

cos sin

)

7 7 2 cos sin 4 4 7 2 cis 4 x iy r i i θ θ π π π + = +   = _ + _   =

Answer: 2 cos7 sin7

4 4

z= _ π +i π _

 

Try this! Write 2i in trigonometric form.

Answer: 2 cos 90

(

° +isin 90°

)

The argument can be expressed using decimal degrees.

Example: Write in trigonometric form: w= − −3 4i

Solution: Here, the real part is x= −3 and the imaginary part is y= −4.

Find r and θ as follows: 586

( ) ( )

2 2 2 2 3 4 25 5 r= z = x +y = − + − = = 1 4 4 tan 3 3 4 ˆ _tan 3 53.1 θ θ − − = = −   = _{ }   = ° QIII: θ =180° +53.1° =233.1°

Now substituting, r=5 and θ =233.1°

(

)

(

)

cos sin 5 cos 233.1 sin 233.1 5 cis 233.1 x iy r i i θ θ + = + = ° + ° = °

Answer: w=5 cos 233.1

(

° +isin 233.1°

)

Try this! Write − +5 12i in trigonometric form.

Answer: 13 cos112.6

(

° +isin112.6°

)

Sometimes we will be asked to convert a complex number written in trigonometric form to rectangular form. To do this, replace the appropriate sine and cosine values and then simplify.

Example: Write 2 cos150

(

° +isin150° in rectangular form. Use exact values.

)

Solution: Substitute in the appropriate sine and cosine values and then simplify.

(

)

3 1 2 cos150 sin150 2 2 2 2 i  i  ° + ° = _− + _   = 3 2 − 2   +       1 2 i 3 i       = − +

Answer: 2 cos150

(

° +isin150° = −

)

3−i

Example: Write 5 cos 255

(

° +isin 255° in rectangular form. Approximate to 4 decimal

)

places.

Solution: Substitute in the appropriate sine and cosine values and then simplify.

(

)

(

)

5 cos 255 sin 255 5 cos 255 5sin 255

1.2941 4.8296 1.2941 4.8296 i i i i ° + ° = ° + ⋅ ° ≈ − + − = − −

Answer: 5 cos 255

(

° +isin 255° ≈ −

)

1.2941 4.8296− i

Try this! Write 6 cos135

(

° +isin135° in rectangular form. Use exact values.

)

Answer: −3 2+3i 2

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Objective: Products and quotients of trigonometric form.

In this section, we streamline the process of multiplying complex numbers using trigonometric form. For example, suppose

1 2 3 2cis 30 1 3 2cis 60 z i z i = + = ° = + = °

To multiply z and ₁ z in rectangular form, distribute ₂

(

)(

)

1 2 2 3 1 3 3 9 3 3 3 3 4 z z i i i i i i i i = + + = + + + = + + − =

This process can be simplified using trigonometric form. In general, let z and ₁ z represent ₂ two complex numbers written in trigonometric form,

(

)

(

)

1 1 1 1 1 1

2 2 2 2 2 2

cos sin cis

cos sin cis

z r i r z r i r θ θ θ θ θ θ = + = = + = Multiplying, 588

(

) (

)

(

)

(

)

(

)

1 2 1 1 1 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2

cos sin cos sin

cos cos cos sin sin cos sin sin

cos cos cos sin sin cos sin sin

cos cos sin sin sin cos cos sin

cos cos z z r i r i r r i i i r r i i r r i i r r θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ = + + = + + + = + + − = − + + =

(

[

] [

]

)

(

)

(

)

(

)

1 2 1 2 1 2 1 2 1 2 1 2

sin sin sin cos cos sin

cos sin i r r i θ θ θ θ θ θ θ θ θ θ − + + = + + +

That is, to multiply two complex numbers in trigonometric form, multiply moduli and add their arguments. This is called the product theorem,

(

)

(

)

(

)

(

)

1 2 1 2 1 2 1 2 1 2 1 2 cos sin cis z z r r i r r θ θ θ θ θ θ = + + + = +

In our opening example,

(

)

1 2 2cis 30 2cis 60 2 2cis 30 60 4cis 90 z z = °⋅ ° = ⋅ ° + ° = °

Convert this to standard form to verify the result.

(

)

(

)

4cis 90 4 cos 90 sin 90 4 0 4 i i i ° = ° + ° = + = 

Example: Find the product of 6cis70° and 2cis35°.

Solution: Applying the product theorem, we have

(

)

6cis 70 2cis 35 6 2cis 70 35

12cis105

°⋅ ° = ⋅ ° + °

= °

Answer: 12cis105°

Try this! Multiply: 5cis 2cis5

6 3

π π

⋅

Answer: 10cis11 6

π

Let z and ₁ z represent two complex numbers written in trigonometric form and divide. ₂ Simplifying this to the quotient theorem,

(

)

(

)

(

)

(

)

1 1 1 2 1 2 2 2 1 1 2 2 cos sin cis z r i z r r r θ θ θ θ θ θ = − + − = −

That is, to divide two complex numbers in trigonometric form, divide their moduli and subtract their arguments. The proof of the quotient theorem is similar to the proof of the product theorem and is left as an exercise.

Example: Find the quotient of 6cis70° and 2cis35°.

Solution: Applying the quotient theorem, we have

(

)

6cis 70 6 cis 70 35 2cis 35 2 3cis 35 °_{= ° − °} ° = ° Answer: 3cis 35°

Try this! Divide:

5 8cis 6 12cis 3 π π Answer: 2cis 3 2 π ***** ***** ***** ***** ***** 590

Key Takeaways ***** ***** *****

• Graph complex numbers in the complex plane where the x-axis represents the real part and the y-axis represents the imaginary part.

• Every complex number has a modulus r and argument θ. These values are used to write the complex number in trigonometric form: z= + =x iy r

(

cosθ+isinθ

)

=rcisθ . • The process of multiplying and dividing complex numbers is streamlined when they

are written in trigonometric form.

(

)

1 2 1 2cis 1 2 z z =r r θ θ+

(

)

1 1 1 2 2 2 cis z r z =r θ θ− ***** ***** ***** ***** ***** ***** Section Exercises Part A

Graph the set of complex numbers in the same complex plane.

1.

{

1 2 , 1 2 , 2+ i − i −

}

2.

{

− + − −3 i, 3 i,1

}

3.

{

− +4 3 , 5 2 , 3 3i + i − i

}

4.

{

− −5 3 , 5 3 , 3i − i i

}

5.

{

− +4 i, 3 3 ,+ i −3i

}

6.

{

− −5, 2 , 5 2i + i

}

7.

{

4 3 ,± i − ±5 2i

}

8.

{

− ±3 4 , 6 2i ± i

}

9.

{

− +10 2 , 8 6 ,i + i ±6i

}

10.

{

± − −8, 10 6 , 12 4i − i

}

Give rectangular form of the complex number represented in each graph.

11. 12.

13. 14. 15. 16. Part B

Write the complex number in rectangular form. Use exact values.

17. 2 cos 30

(

° +isin 30°

)

18. 2 cos 60

(

° +isin 60°

)

19. 4 cos 45

(

° +isin 45°

)

20. 6 cos 45

(

° +isin 45°

)

21. 3 cos 90

(

° +isin 90°

)

22. 5 cos 0

(

° +isin 0°

)

23. 2 cos120

(

° +isin120°

)

24. 2 cos 210

(

° +isin 210°

)

25. 2 2 cos 315

(

° +isin 315°

)

26. 2 cos 225

(

° +isin 225°

)

27. 7 cos180

(

° +isin180°

)

28. 5 cos 270

(

° +isin 270°

)

29. 10cis5 6 π 30. 8cis4 3 π 31. 9cis3 2 π 32. 6cis7 4 π 33. 5 cis3 4 π 34. 3 cis7 6 π 592

Write the complex number in trigonometric form where the argument 0≤ <θ 360°. 35. 1+i 3 36. 3+i 37. −2 3+2i 38. −2 2−2i 2 39. − −3 3i 3 40. 3 3−3i 41. − +1 i 42. 3−3i 43. − 6−i 2 44. 3−i 3 45. 1 3 2+ 2 i 46. 3 1 2 2i − + 47. 16i 48. −8i 49. 7 50. −1

Write each complex number in trigonometric form. Round the modulus and argument to the nearest tenth.

51. 2 3i+ 52. − −1 4i 53. 5 2i− 54. − +3 5i 55. − −7 11i 56. − +9 i

Write each complex number in rectangular form. Approximate to 4 decimal places.

57. 5 cos 95.5

(

° +isin 95.5°

)

58. 6 cos125.4

(

° +isin125.4°

)

59. 10 cis190° 60. 15 cis 335° 61. 3 cis11 6 π 62. 6 cis2 3 π Part C

Find the product. Give the exact value in trigonometric form and rectangular form.

63. 5 cis10 2 cis35°⋅ ° 64. 4 cis35 3cis25°⋅ ° 65. 6cis75 5 cis15°⋅ ° 66. 7 cis95 4 cis55°⋅ ° 67. 10 cis165 3cis75°⋅ ° 68. 3cis220 6 cis95°⋅ ° 69. cis52 cis68°⋅ ° 70. 3cis38 cis97°⋅ ° 71. 2 cis 3cis 4 12 π π ⋅ 72. 4 cis 5 cis 12 6 π _⋅ π 73. 2 cis7 2 cis13 15 15 π _⋅ π 74. cis7 8 cis13 12 12 π _⋅ π 593

Find the product. Write the result in trigonometric form and rectangular form. Approximate to 4 decimal places.

75. 2.5 cis53 5.1cis22°⋅ ° 76. 6.3cis88 2.5 cis39°⋅ ° 77. 1.5 cis125 3.4 cis83°⋅ ° 78. 6 cis167 8.3cis105°⋅ ° 79.

(

2.5 cis25°

)

2 80.

(

4.1cis63°

)

2

Find the quotient. Give the exact value in trigonometric form and rectangular form.

81. 30 cis105 5 cis75 ° ° 82. 28 cis335 7 cis110 ° ° 83. 6 cis355 3cis55 ° ° 84. 32cis123 4 cis63 ° ° 85. 16 cis256 12 cis166 ° ° 86. 28 cis347 24 cis77 ° ° 87. 3 14 cis 7 5 2 cis 28 π π 88. 7 cis 8 5 3cis 24 π π 89. 11 cis 12 3 5 cis 4 π π 90. 3cis 2 2 cis 4 π π

Write each number in trigonometric form and rectangular form. Approximate to 4 decimal places. 91. 3.4 cis57 2 cis22 ° ° 92. 7.7 cis124 2.2 cis32 ° ° 93. 11.2 cis256 3.2 cis105 ° ° 94. 4.25 cis331 8.5 cis213 ° ° 95. 1 4 cis25° 96. 5 cis33°

97. Show that x=cis60° is a solution to 2

1 0

x − + =x .

98. Show that x=cis45° is a solution to x2− 2x+ = . 1 0 99. Show that x=2cis135° is a solution to x2+2 2x+ = . 4 0 100. Show that x=4cis150° is a solution to x2+4 3x+16= . 0

Answers to Section Exercises Part A 1. 2. 3. 4. 5. 6. 595

7. 8. 9. 10. 11. 3 2i− 12. − +2 2i 13. 5 3i− 14. − −5 4i 15. 4i 16. −5 Part B 17. 3+ i 18. 1+i 3 19. 2 2+2i 2 20. 3 2 3+ i 2 21. 3i 22. 5 23. 1− +i 3 24. − 3 i− 25. 2−2i 26. − −1 i 27. −7 28. −5i 29. 5 3− + 5i 30. − −4 4i 3 31. −9i 32. 3 2 3− i 2 33. 5 2 5 2 2 2 i − + 34. 3 3 2 2 i − − 35. 2 cos 60

(

° +isin 60°

)

36. 2 cos 30

(

° +isin 30°

)

37. 4 cos150

(

° +isin150°

)

596

38. 4 cos 225

(

° +isin 225°

)

39. 6 cos 240

(

° +isin 240°

)

40. 6 cos 330

(

° +isin 330°

)

41. 2 cos135

(

° +isin135°

)

42. 2 3 cos 300

(

° +isin 300°

)

43. 2 2 cos 210

(

° +isin 210°

)

44. 6 cos 315

(

° +isin 315°

)

45. cos 60° +isin 60° 46. cos150° +isin150° 47. 16 cos 90

(

° +isin 90°

)

48. 8 cos 270

(

° +isin 270°

)

49. 7 cos 0

(

° +isin 0°

)

50. 12 cos180

(

° +isin180°

)

51. 3.6 cos 56.3

(

° +isin 56.3°

)

52. 4.1 cos 256.0

(

° +isin 256.0°

)

53. 5.4 cos 338.2

(

° +isin 338.2°

)

54. 5.8 cos120.0

(

° +isin120.0°

)

55. 13.0 cos 237.5

(

° +isin 237.5°

)

56. 9.1 cos173.7

(

° +isin173.7°

)

57. −0.4792+4.9770i 58. −3.4757+4.8908i 59. −9.8481 1.7365i− 60. 13.5946 6.3393i− 61. 1.5 0.8660i− 62. −1.2247+2.1213i Part C 63. 10 cis45° =5 2+5i 2 64. 12 cis60° = +6 6i 3 65. 30 cis90° =30i 66. 28 cis150° = −14 3 14i+ 67. 30 cis240° = − −15 15i 3 68. 18 cis315° =9 2−9i 2 69. cis120 1 3 2 2 i ° = − + 70. 3cis135 3 2 3 2 2 2 i ° = − + 71. 6 cis 3 3 3 3 i π _{= +} 72. 20 cis 10 2 10 2 4 i π = + 73. 4 cis4 2 2 3 3 i π _{= − −} 74. 8 cis5 4 4 3 3 i π _{= −} 75. 12.75 cis75° ≈3.2999 121.3156i+ 76. 15.75 cis127° ≈ −9.4786 12.5785i+ 77. 5.1cis208° ≈ −4.5030 2.3943i− 78. 49.8 cis272° ≈1.7380 49.7697i− 79. 6.25 cis50° ≈4.0174 4.7878i+ 80. 16.81cis126° ≈ −9.8807 13.5996i+ 81. 6 cis30° =3 3+3i 82. 4 cis225° = −2 2−2i 2 83. 2 cis300° = −1 i 3 84. 8 cis60° = +4 4i 3 85. 4cis90 4 3 ° =3i 86. 7cis270 7 6 ° = −6i 87. 7 cis 7 2 7 2 4 2 2 i π _{= +} 88. 1cis2 1 3 3 3 6 6 i π _{= − +} 89. 1cis 3 1 5 6 10 10i π = + 597

90. 3cis 3 2 3 2 2 4 4 4 i π _{= +} 91. 1.7 cis35° ≈1.3926 0.9751i+ 92. 3.5 cis92° ≈ −0.1221 3.4979i+ 93. 3.5 cis151° ≈ −3.0612 1.6968i+ 94. 0.5 cis118° ≈ −0.2347 0.4415i+ 95. 0.25 cis335° ≈0.2266 0.1057i− 96. 5 cis327° ≈4.1934 2.7232i− 97. Proof 98. Proof 99. Proof 100. Proof 598

8.3 Powers and Roots of a Complex Number

Learning Objectives

1. Powers of complex numbers using De Moivre’s Theorem 2. Finding roots of complex numbers.

3. Solving equations with complex roots.

Connections between complex numbers and trigonometry streamline many calculations that would otherwise prove to be very tedious. In fact, we will extend the ideas presented in this section to find the complex roots of certain polynomials.

Objective: Powers of complex numbers using De Moivre’s Theorem.

In this section, we introduce a powerful theorem that will allow us to streamline the process of raising complex numbers to powers. For example, to raise z= 2+i 2=2cis 45°to the fourth power in the traditional manner we would begin as follows:

(

)

(

)(

)(

)(

)

4 4 2 2 2 2 2 2 2 2 2 2 z i i i i i = + = + + + +

As you can see, this would be quite the cumbersome calculation. However, if we convert it to trigonometric form, then we could apply the product theorem. For example, let

(

cos sin

)

cis

z=r θ +i θ =r θ and square it as follows:

(

)

2 2 2 cis cis cis cis 2 z r r r r θ θ θ θ θ = ⋅ = + =

Now multiplying by another factor of z and applying the product theorem again we have,

(

)

3 2 3 3 cis cis 2 cis 2 cis 3 z r r r r θ θ θ θ θ = ⋅ = + = 599

There is a pattern; raise the magnitude r to the power and use the power as a multiple of the argument θ. This is true in general and leads to De Moivre’s theorem which states that given any real number n,

(

)

(

)

cos sin cos sin cis n n n n z r i r n i n r n θ θ θ θ θ =_ + _ = + =

Applying this to the opening problem, raising z= 2+i 2 =2cis 45° the fourth power we have,

(

)

(

)

(

)

4 4 4 2 cos 45 sin 45 2 cos 45 sin 45 16 cos180 sin180 4 4 z i i i =_ ° + ° _ = ⋅ ° + ⋅ ° = ° + °

Now it’s just a matter of replacing the cosine and sine values and simplifying.

(

)

(

)

4 16 cos180 sin180 16 1 0 16 z i i = ° + ° = − + ⋅ = −  Example: Find

(

)

5 1 i 3 − + .

Solution: Begin by finding trigonometric form of the complex number z= − +1 i 3.

1 3 2cis120

z= − +i = °

Next, raise this to the fifth power and apply De Moivre’s theorem.

[

]

5 5 5 2cis120 2 cis 120 32cis 600 5 z = ° = ⋅ ° = °

Now simplify to find the result.

(

)

(

)

5 32 cos 600 sin 600 32 cos 240 sin 240 1 3 32 16 16 3 2 2 z i i i i = ° + ° = ° + °    = _− + −_ __= − −     Answer: 16 16− − i 3

The theorem can be applied when the argument is expressed using radians.

Example: Find

(

1 i−

)

4.

Solution: Begin by finding trigonometric for the complex number z= −1 i. In this case, we choose to express the argument using radians.

7

1 2 cis

4

z= − =i π

Next, raise this to the fourth power and apply De Moivre’s theorem.

4 4 4 7 2cis 4 7 2 cis 4 4cis 4 7 z π π π   = _ _ = ⋅ = Now simplify to find the result.

(

)

(

)

(

)

4 4 cos 7 sin 7 4 cos sin 4 1 0 4 z i i i π π π π = + = + = − + ⋅ = − Answer: −4

Try this! Find

(

1+i 3

)

8.

Answer: 128 128− + i 3

***** ***** ***** ***** *****

Objective: Finding roots of complex numbers.

Recall from algebra that an nth root of a number is another number that when raised to the nth power gives the number and is denoted a1/ n = na. Some examples follow,

1/ 2 2 1/3 3 3 1/ 4 4 4 25 25 5 because 5 25 27 27 3 because 3 27 16 16 2 because 2 16 = = = = = = = = =

Actually, there are two square roots of 25. Because

( )

−5 2 =25, we can say −5 is a square root

of 25 as well. In fact, there are n distinct nth roots for any complex number. We can see that 3 is a cube root of 27, the other two cube roots are complex numbers. De Moivre’s theorem helps us find all of the nth roots of any complex number. Let’s consider 27 as a complex number,

(

)

27=27 0+ =i 27 cos 0° +isin 0°

We are looking for numbers w_k =r

(

cosα+isinα

)

such that when cubed gives 27. In other words,

(

)

(

)

(

)

(

)

3 3 3 27

cos sin 27 cos 0 sin 0

cos 3 sin 3 27 cos 0 sin 0

k w r i i r i i α α α α = + = ° + °     + = ° + °

Equating the resulting complex numbers, we have r =3, cos 3α =cos 0° , and sin 3α =sin 0°. Here, 3α must be coterminal to 0°,

3 0 360 0 360 3 k k α α = ° + °⋅ ° + °⋅ =

In this case, k is an integer and we can substitute as follows:

0 k= 0 360 3 0 0 α = ° + °⋅ = ° 1 k= 0 360 1 0 3 1 2 α = ° + °⋅ = ° 2 k= 0 360 2 0 3 2 4 α = ° + °⋅ = ° 602

When k =3 the angle is coterminal with 360° and the results begin to repeat. These are the angles that give us the 3 distinct cube roots of 27.

(

)

(

)

(

)

0 1 2 3 cos 0 sin 0 3 3 3 3 3 cos120 sin120 2 2 3 3 3 3 cos 240 sin 240 2 2 w i w i i w i i = ° + ° = = ° + ° = − + = ° + ° = − −

Notice that the two complex cube roots of 27 are conjugates. This process can be worked in general, which leads to the nth root theorem. If z=r

(

cosθ+isinθ

)

=rcisθ is any complex number, then the n distinct complex roots w , where k is an integer, are _k

(

)

1/ 1/ 1/ cos sin 360 360 cos sin 360 cis n k n n w r i k k r i n n k r n θ θ θ θ θ =_ + _   + °⋅   + °⋅  = _{  }+ _{ }       + °⋅   = _{ }  

Note that the formula is very similar to De Moivre’s theorem, with only an adjustment to the argument.

Example: Find the two square roots of 1+i 3.

Solution: Begin by expressing the complex number in trigonometric form.

(

)

1+i 3=2 cos 60° +isin 60° Next, apply the nth root theorem where n=2

(

)

1/ 2 1/ 2 2 cos 60 sin 60 60 360 60 360 2 cos sin 2 2 k w i k k i =_ ° + ° _   ° + °⋅   ° + °⋅  = _{  }+ _{ }      

Substitute 0 and 1 in for k.

0 k = 60 360 ₃ 2 0 0 ° + °⋅ _{= °} 603

1

k= 60 360 ₂₁₀

2

1

° + °⋅ _{= °}

These are the angles that allow us to find the two square roots.

(

)

(

)

1/ 2 0 1/ 2 1 3 1 6 2 2 cos 30 sin 30 2 2 2 2 2 1 3 2 6 2 cos 60 sin 60 2 2 2 2 2 w i i i w i i i   = ° + ° = _ + _= +     = ° + ° = _ + _= +   Answer: ₀ 6 2 2 2 w = + i, ₁ 2 6 2 2 w = + i

Sometimes the complex number will be given in trigonometric form.

Example: Find the three cube roots z=8 cos 45

(

° +isin 45° . Leave your answers in

)

trigonometric form.

Solution: In this case, we will apply the nth root theorem where n=3

(

)

1/3 1/3 8 cos 45 sin 45 45 360 45 360 8 cos sin 3 3 k w i k k i =_ ° + ° _   ° + °⋅   ° + °⋅  = _{  }+ _{ }      

Substitute 0, 1, and 2 for k,

0 k = 45 360 1 3 0 5 ° + °⋅ = ° 1 k = 45 360 135 2 1 ° + °⋅ _{= °} 2 k = 45 360 255 3 2 ° + °⋅ _{= °}

These are the angles that allow us to find the three cube roots.

(

)

(

)

(

)

1/3 0 1/3 1 1/3 2

8 cos15 sin15 2 cis15 8 cos135 sin135 2 cis135

8 cos 255 sin 255 2 cis 255

w i w i w i = ° + ° = ° = ° + ° = ° = ° + ° = ° 604

Answer: w₀ =2 cis15° , w₁=2 cis135° , w₂ =2 cis 255°

Try this! Find the four 4th roots of i in trigonometric form.

Answer: w₀ =cis 22.5° , w₁ =cis112.5° , w₂ =cis 405° , w₃ =cis 292.5° ***** ***** ***** ***** *****

Objective: Solving equations with complex roots.

The fundamental theorem of algebra states that for any degree-n polynomial with complex coefficients, there are exactly n roots and the nth root theorem can help find them. For example, consider the polynomial equation

2 2 0 x i x i − = =

The solutions are the two square roots of the imaginary unit, i. Writing the imaginary unit in trigonometric form we have,

(

)

1 cos 90 sin 90

i= ° +i °

There are two square roots, so apply the nth root theorem where n=2.

(

)

1/ 2 1/ 2 1 cos 90 sin 90 90 360 90 360 1 cos sin 2 2 k w i k k i =_ ° + ° _   ° + °⋅   ° + °⋅  = _{  }+ _{ }      

Substitute 0 and 1 for k.

0 k= 90 360 4 2 0 5 ° + °⋅ = ° 1 k = 90 360 225 2 1 ° + °⋅ _{= °}

These are the angles that allow us to find the two square roots.

(

)

(

)

1/ 2 0 1/ 2 1 2 2

1 cos 45 sin 45 cis 45

2 2

2 2

1 cos 225 sin 225 cis 225

2 2 w i i w i i = ° + ° = ° = + = ° + ° = ° = − − 605

Therefore we can say, 2 2 2 2 i = + i because 2 2 2 2 2 i i   + =      

Certainly, you could check by multiplying.

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 4 1 1 2 2 i i i i i i i         + = + ⋅ +                         = + + = + − = 

The second square root found here is left to the reader to check.

Example: Solve: x5− = . Round to 4 decimal places. 1 0

Solution: Since the degree is 5, we expect five solutions. 5 5 1 0 1 x x − = =

In other words, we are trying to find the five roots of 1. Writing 1 in trigonometric form, we have

(

)

1 1 cos 0= ° +isin 0° Apply the nth root theorem where n=5

(

)

1/5 1/5 1 cos 0 sin 0 0 360 0 360 1 cos sin 5 5 k w i k k i =_ ° + ° _   ° + °⋅   ° + °⋅  = _{  }+ _{ }       Substitute 0,1,2,3,4 in for k. 0 k = 0 360 0 0 5 ° + °⋅ _{= °} 1 k= 0 360 72 5 1 ° + °⋅ = ° 606

2 k = 0 360 144 3 2 ° + °⋅ _{= °} 3 k= 0 360 216 5 3 ° + °⋅ _{= °} 4 k = 0 360 288 5 4 ° + °⋅ = °

These are the angles that allow us to find the five cube roots.

Answer: [ WolframAlpha Solution ]

0 1 2 3 4 cos 0 sin 0 1 0 cos 72 sin 72 0.3090 0.9511 cos144 sin144 0.8090 0.5878 cos 216 sin 216 0.8090 0.5878 cos 288 sin 288 0.3090 0.9511 w i i w i i w i i w i i w i i = ° + ° = + = ° + ° = + = ° + ° = − + = ° + ° = − − = ° + ° = −

The above example is of particular importance in higher level mathematics. Any complex number such that

1

n

z =

is called a root of unity or a De Moivre number. Furthermore, these numbers have interesting symmetries on the unit circle. The five roots of 1 found in the previous example are plotted in the complex plane below.

Notice that they come in conjugate pairs and are evenly spaced around the unit circle.

Try this! Solve: x3+ =1 0.

Answer: −1, 1 3

2+ 2 ,

1 3

2− 2

***** ***** ***** ***** *****

Key Takeaways ***** ***** *****

• De Moivre’s theorem streamlines the process of raising complex number to a large power.

cis

n n

z =r nθ

• De Moivre’s theorem leads to the nth root theorem, which allows us to find roots of complex numbers. 1/ 360 cis n k k w r n θ+ °⋅   = _{ }  

• The fundamental theorem of algebra states that there are exactly n roots for any nth degree polynomial with complex coefficients. The nth root theorem gives us a method that we can use to solve polynomials.

***** ***** ***** ***** ***** *****

Section Exercises Part A

Simplify and write your answer in rectangular form. Use exact values.

1. _3 cos 60

(

° +isin 60°

)

_2 2. _5 cos 30

(

° +isin 30°

)

_2 3. _4 cos 240

(

° +isin 240°

)

_3 4. _2 cos150

(

° +isin150°

)

_3 5. 4 4 cos sin 4 i 4 π π   ₊     _{ }   6. 4 3 3 5 cos sin 2 i 2 π π   ₊     _{ }   7.

(

1 i+

)

6 8.

( )

1 i− 6 9.

(

1−i 3

)

4 10.

(

3−i

)

4 11.

(

)

6 2−i 2 12.

(

2+i 2

)

8 13.

(

− 3+i

)

6 14.

(

)

6 3+i 15. 10 3 1 2 2i   − −       608

16. 10 1 3 2 2 i   +       17.

(

6cis300°

)

3 18.

(

2cis225°

)

5 19. 6 5 3cis 4 π       20. 4 7 2cis 3 π       21.

(

2cis40°

)

6 22.

(

cis15°

)

8 23. Show that x=cis60° is a solution to x3+ =1 0.

24. Show that x=cis45° is a solution to x4+ =1 0.

Part B

Find the two square roots of the following numbers. Use exact values and express your answer in standard form.

25. 4 cos120

(

° +isin120°

)

26. 25 cos 60

(

° +isin 60°

)

27. 9 cos 240

(

° +isin 240°

)

28. 16 cos 90

(

° +isin 90°

)

29. 3 cos 270

(

° +isin 270°

)

30. 2 cos 300

(

° +isin 300°

)

31. 2 2− i 3 32. 2 2 3− − 33. 16i 34. −16i

Find the three cube roots of the following numbers. Leave answers in trigonometric form. 35. 8 cos 30

(

° +isin 30°

)

36. 8 cos 60

(

° +isin 60°

)

37. 27 cos 45

(

° +isin 45°

)

38. 27 cos150

(

° +isin150°

)

39. 3cis 72° 40. 2cis48° 41. 4 2−4i 2 42. − +4 4i 3 43. 1 i+ 44. 1 i−

Find the three cube roots. Use exact values and give answers in rectangular form.

45. 8i 46. −64i 47. −8 48. 27 49. 1 50. −1 51. − i 52. i 609

Part C

Solve. Give exact solutions.

53. x3− =1 0 54. x3− =8 0 55. 3 27 0 x + = 56. 3 27 0 x − = 57. x4− =1 0 58. x4+ =1 0 59. 3 0 x − = i 60. 3 0 x + = i

Solve. Round to 4 decimal places.

61. x5+ =1 0 62. x5+ =i 0

63. x4+ =i 0 64. x4− =i 0

Solve. Round to 4 decimal places. (Hint: Use the quadratic formula to solve for x^2 and then find the square roots of the result.)

65. x4 −2x2 + =2 0 66. x4−4x2+ =5 0

67. x4−2x2+ =4 0 68. x4−2x+ =3 0

Answers to Section Exercises Part A 1. 9 9 3 2 2 i − + 2. 25 25 3 2 + 2 i 3. 64 4. 8i 5. −256 6. 625 7. −8i 8. 8i 9. − +8 8i 3 10. 8 8− − i 3 11. 64i 12. 256 13. −64 14. −64 15. 1 3 2− 2 i 16. 1 3 2 2 i − − 17. −216 18. 16 2 16+ i 2 19. −729i 20. − −8 8i 3 21. − −32 32i 3 22. 1 3 2 2 i − + 23. Proof 24. Proof Part B 25. 1+i 3, − −1 i 3 26. 5 3 5 2 +2i, 5 3 5 2 2i − − 27. 3 3 3 2 2 i − + , 3 3 3 2− 2 i 28. 2 2+2i 2, −2 2−2i 2 29. 6 6 2 2 i − + , 6 6 2 − 2 i 30. 6 2 2 2 i − + , 6 2 2 − 2 i 31. − 3 i+ , 3 i− 32. 1−i 3, 1− +i 3 33. 2 2+2i 2, −2 2−2i 2 34. −2 2+2i 2, 2 2−2i 2

35. 2cis10°, 2cis130°, 2cis250° 36. 2cis20°, 2cis140°, 2cis260° 37. 3cis15°, 3cis135°, 3cis255° 38. 3cis50°, 3cis170°, 3cis290°

39. 33 cis24° , 33 cis144° , 33 cis264° 40. 3 2 cis16°_, 3 2 cis136°_, 3 2 cis 256°

41. 2cis105°, 2cis225°, 2cis345° 42. 2cis60°, 2cis160°, 2cis280° 43. 62 cis15°_, 62 cis135°_, 6 2 cis255° 44. 6 2 cis105° , 62 cis225° , 6 2 cis345° 45. 3 i+ , − 3 i+ , −2i 46. 4i, 2 3− − , 2 3 2i2i − 47. 1+i 3, 1−i 3, − 2 48. 3, 3 3 3 2 2 i − + , 3 3 3 2 2 i − − 49. 1, 1 3 2 2 i − + , 1 3 2 2 i − − 611

50. − , 1 1 3 2+ 2 i, 1 3 2− 2 i 51. i , 3 1 2 2i − − , 3 1 2 −2i 52. − , i 3 1 2 +2i, 3 1 2 2i − + Part C 53. 1, 1 3 2 2 i − + , 1 3 2 2 i − − 54. 2, 1− + 3i, 1− − 3i 55. −3, 3 3 3 2+ 2 i, 3 3 3 2− 2 i 56. 3, 3 3 3 2 2 i − + , 3 3 3 2 2 i − − 57. ± , i1 ± 58. 2 2 2 ± 2 i, 2 2 2 2 i − ± 59. − , i 3 1 2 +2i, 3 1 2 2i − + 60. i , 3 1 2 2i − − , 3 1 2 −2i 61. [ wolframAlpha ] 62. [ wolframAlpha ] 63. [ wolframAlpha ] 64. [ wolframAlpha ] 65. [ wolframAlpha ] 66. [ wolframAlpha ] 67. [ wolframAlpha ] 68. [ wolframAlpha ] 612

8.4 Polar Coordinates

Learning Objectives

1. Introduce the polar coordinate system and plot points.

2. Convert between rectangular coordinates and polar coordinates. 3. Find equivalent equations using polar coordinates.

In mathematics, position in two-dimensional space is usually expressed using rectangular (Cartesian) coordinates. However, the rectangular coordinate system is not the only way to identify points in the plane. In this section we define a new positional system that can be used to easily represent relations that are not functions.

Objective: Introduce the polar coordinate system and plot points.

The polar coordinate system is a positional system based on a point, called the pole, and a ray called the polar axis. Given any other point in a plane, another ray can be constructed from the pole through that given point. Then the point can be described using coordinates that consist of a directed distance r from the pole to the point along the ray, and angle θ from the polar axis to the ray. That is, points are expressed as ordered pairs in the form

( )

r,θ .

The polar axis is drawn to coincide with the positive x-axis where any point in the plane can be used to form an angle in standard position. Furthermore, it is important to note that the polar coordinate system allows for r to be negative. When the directed distance r is negative, the terminating ray is extended 180° in the opposite direction and the point is located r units from the pole in this opposite direction.

In the same way rectangular graph paper aids in graphing with rectangular coordinates, polar graph paper aids in graphing using polar coordinates and consists of concentric circles and lines at the following special angles. [Link to polar graph paper.]

Use the above construction to graph points expressed with polar coordinates.

Example: Plot the following points using the polar coordinate system.

a.

(

5, 45°

)

b.

(

−4, 60°

)

c.

(

3, 135− °

)

Solution:

a. Here r=5 units and θ =45°.

b. Since r is negative, mark the point 4 units in the opposite direction as the terminal ray indicated by θ =60°.

c. In this case, mark a point 3 units along the terminal side of θ = −135° drawn in standard position.

It is not uncommon for the angle in polar coordinates to be expressed using radian measure.

Try this! Plot 5,5 6 π ₋      Answer:

Points expressed using the rectangular coordinate system are unique, that is, there is only one way to express any given point

( )

x y . This is not the case with the polar coordinate system; , points can be expressed in many different ways. In other words, polar coordinates are not unique.

Example: Give three equivalent points using polar coordinates:

(

4, 75° .

)

Solution: A conterminal angle in standard position can be calculated by subtracting 360°.

75° −360° = −285°

Therefore, the point

(

4, 285− ° locates the same point.

)

Certainly, we could obtain many more coordinate pairs locating the same point simply by adding and subtracting multiples of 360° to 75°. However, we can also express this point using

a negative value for r where the angle terminates on the opposite ray that lies along the same line. To do this, add and subtract 180° to 75° and user= − . 4

75 180 255

75 180 105

° + ° = °

° − ° = − °

Answer:

(

4, 285− ° ,

)

(

− −4, 105° ,

)

(

−4, 255°

)

Try this! Give three equivalent sets of polar coordinates:

(

5, 315° .

)

Answer:

(

5, 45− ° ,

)

(

−5,135° ,

)

(

− −5, 225°

)

***** ***** ***** ***** *****

Objective: Convert between polar coordinates and rectangular coordinates.

Any point in a plane can be expressed using polar coordinates or rectangular coordinates. To convert between systems, we use the relationships given by the general definition of the trigonometric functions. Recall that, cos x r θ = sin y r θ =

Multiplying both sides by r we have,

cos

x=r θ y=rsinθ

These equations can be used to convert from polar coordinates to rectangular coordinates.

Example: Convert the given polar coordinates to rectangular coordinates.

a.

(

2, 60°

)

b.

(

−4,150°

)

c.

(

2, 315°

)

Solution: a. Here r= and 2 θ =60°, 1 cos 2 cos 60 2 1 2 3 sin 2 sin 60 2 3 2 x r y r θ θ   = = ° = _{ }=     = = ° = _{ }=  

Therefore,

( )

1, 3 is the rectangular equivalent to

(

4, 60° .

)

b. Here r= − and 4 θ =150°,

(

)

(

)

3 cos 4 cos 150 4 2 3 2 1 sin 4 sin 150 4 2 2 x r y r θ θ   = = − ° = − −_{ }=     = = − ° = − _{ }= −  

Therefore,

(

2 3, 2−

)

is the rectangular equivalent to

(

−4,150° .

)

c. Here r= 2 and θ =315°,

(

)

(

)

1 cos 2 cos 315 2 1 2 1 sin 2 sin 315 2 1 2 x r y r θ θ   = = ° = _{ }=     = = ° = _− _= −  

Therefore,

(

1, 1− is the rectangular equivalent to

)

(

2, 315°

)

.

Answer: a.

( )

1, 3 b.

(

2 3, 2−

)

c.

(

1, 1−

)

Try this! Convert

(

−6, 270° to rectangular coordinates.

)

Answer:

( )

0, 6

Furthermore, from the general definition of the trigonometric functions we can find that,

tan y

x

θ = 2 2 2

r =x + y

These equations can be used to convert rectangular coordinates to polar coordinates. Take care to identify the quadrant of the give point because that helps determine the angle θ in standard position.

Example: Convert the given rectangular coordinates to polar coordinates.

a.

( )

1,1 b.

(

− 3,1

)

c.

(

0, 5−

)

Solution: After applying the square root property to r2 =x2+ we have, y2

2 2

r= ± x +y

This allows us to determine the directed distance r using a positive or negative value. However, for simplicity, we will choose r to be positive using 2 2

r= x +y and determine the angle θ accordingly.

a. Here x=1 and y=1, which terminates in QI.

( ) ( )

2 2 2 2 1 1 2 r= x +y = + =

( )

1 tan 1 tan 1 tan 1 45 y x θ θ θ − = = = = ° Therefore,

(

2, 45°

)

is equivalent to

( )

1,1 . b. Here x= − 3 and y=1, which lies in QII.

( )

( )

2 2 2 ₂ 3 1 4 2 r= x +y = − + = = 1 tan 1 tan 3 1 ˆ _{tan 30} 3 y x θ θ θ − = = −   = _{ }= °   QII: θ =180° − ° =30 150°

Therefore,

(

2,150° is the equivalent to

)

(

− 3,1

)

.

c. Here x=0 and y= −5, which lies on the negative y-axis.

( ) ( )

2 2 2 2 0 5 25 5 r= x +y = + − = = tan 5 tan undefined 0 270 y x θ θ θ = − = = °

Therefore,

(

5, 270° is the equivalent to

)

(

0, 5− .

)

Answer: a.

(

2, 45°

)

b.

(

2,150° c.

)

(

5, 270°

)

Try this! Convert

( )

3, 4 to polar coordinates.

Answer:

(

5, 53.1°

)

***** ***** ***** ***** *****

Objective: Find equivalent equations using polar coordinates.

In algebra, we study equations that involve the variables x and y, where y is often expressed as a function of x:

(

)

2

2 5

y= x− + y= − x+3 y=log₂

( )

x