23.1 Polar Coordinate System Problem Set

POLAR COORDINATE SYSTEM | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: The polar coordinates of a point located at the rectangular coordinates of (4,6) is most close to: A. (4, 6°) B. (4, 56.3°) C. (7.21, 33.7°) D. (7.21, 56.3°)

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PROBLEM 2: The polar form of the complex number 𝑧 = 3 + 4𝑖 is best represented as: A. (3)(cos 36.87° + 𝑖 sin 36.87°) B. (3)(cos 53.15° + 𝑖 sin 36.87°) C. (4)(cos 53.15° + 𝑖 sin 53.15°) D. (5)(cos 53.15° + 𝑖 sin 53.13°)

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PROBLEM 3: Given the polar coordinates of a point are (4, 120°), the equivalent rectangular coordinates are best represented as: A. (2, 3.46) B. (3.46, 2) C. (−2, 3.46) D. (−2, −3.46)

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PROBLEM 4: If we are given a point with rectangular coordinates (−3, −5.2), the polar coordinates of this point are best represented as: A. (−6, −120°) B. (6, 60°) C. (6, 120°) D. (6, −150°)

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PROBLEM 5: The quotient of the following set of complex numbers, represented in polar form, is most close to: 𝑧1 = 5(cos 330° + 𝑗 sin 330°) 𝑧2 = 9(cos 90° + 𝑗 sin 90°) 5

A. 9 (sin 240° + 𝑗 cos 240°) 5

B. 9 (sin 240° − 𝑗 cos 240°) 5

C. 9 (cos 240° + 𝑗 sin 240°) 5

D. 9 (cos 240° − 𝑗 sin 240°)

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PROBLEM 6: Using Euler’s Identity, sin2 𝜃 written in exponential terms is best represented as:

A. sin2 𝜃 =

1 4𝑗 2

(𝑗 sin 2𝜃 − 1)

1

B. sin2 𝜃 = 4𝑗 2 (𝑗 sin 2𝜃 + 1) 1

C. sin2 𝜃 = 2𝑗 2 (𝑗 sin 2𝜃 − 1) 1

D. sin2 𝜃 = 2𝑗 2 (𝑗 sin 2𝜃 + 1)

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PROBLEM 7: The result of the stated complex number, in standard form, is most close to: 10

1 𝑗√3 (− + ) 2 2

1

A. − 2 − 1

B. − 2 + 1

𝑗 √3 2 𝑗 √3

C. 2 −

𝑗 √3

1

𝑗 √3

D. 2 +

2

2 2

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PROBLEM 8: 3

Determine the third root of the expression (−2 − 2𝑗√3) : 16𝜋

3

A. 𝑧3 = √4 [− cos ( 3

B. 𝑧3 = √4 [−cos (

9

16𝜋 9

C. 𝑧3 = √4 [cos (

13𝜋

3

13𝜋

3

D. 𝑧3 = √4 [cos (

9 9

) − 𝑗 sin (

) + 𝑗 sin (

) − 𝑗 sin ( ) + 𝑗 sin (

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16𝜋 9

16𝜋 9

13𝜋 9 13𝜋 9

)]

)]

)] )]

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POLAR COORDINATE SYSTEM | SOLUTIONS SOLUTION 1: The TOPIC of a POLAR COORDINATE SYSTEM can be referenced under the subject of MATHEMATICS subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The radius and angle in POLAR FORM can be determined using the “𝑥” and “𝑦” rectangular coordinates, using the formulas: • 𝑟 = |𝑥 + 𝑗𝑦| = √𝑥 2 + 𝑦 2 𝑦

• 𝜃 = arctan (𝑥 ) Where: • 𝑟 is the distance from the origin to the point • 𝑥 is the x-coordiante of the rectangular coordinates • 𝑦 is the y-coordinate of the rectangular coordinates • 𝜃 is the angle between the radius “𝑟” and the positive 𝑥 − 𝑎𝑥𝑖𝑠 In our problem, we are given the RECTANGULAR COORDINATES of (4,6).

Plugging these values in to our general formulas given for a complex expression, we find that: Made with

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𝑟 = √(4)2 + (6)2 = 7.211 And 6

𝜃 = arctan (4) = 56.3° The correct answer choice is D. (𝟕. 𝟐𝟏, 𝟓𝟔. 𝟑°)

SOLUTION 2: The FORMULAS to CONVERT BETWEEN POLAR AND RECTANGULAR COORDINATES can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The radius and angle of a complex number in POLAR FORM can be determined using the “𝑥” and “𝑦” coordinates, as defined in the general formulas: • 𝑟 = |𝑥 + 𝑗𝑦| = √𝑥 2 + 𝑦 2 𝑦

• 𝜃 = arctan (𝑥 )

Where:

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• 𝑟 is the distance from the origin to the point • 𝑥 is the x-coordiante of the rectangular coordinates • 𝑦 is the y-coordinate of the rectangular coordinates • 𝜃 is the angle between the radius “𝑟” and the positive 𝑥 − 𝑎𝑥𝑖𝑠 In our problem, we are given a complex expression in RECTANGULAR FORM, which is: 𝑧 = 3 + 4𝑖 Knowing the standard RECTANGULAR FORM for a complex expression is written in the form: 𝑥 + 𝑗𝑦 This tells us that: • x=3 • y=4 Therefore, plugging these values in to our general formulas, we find that:

𝑟 = √(3)2 + (4)2 = 5

And: Made with

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4

𝜃 = arctan (3) = 53.13° The correct answer choice is D. (𝟓)(𝐜𝐨𝐬 𝟓𝟑. 𝟏𝟑° + 𝒊 𝐬𝐢𝐧 𝟓𝟑. 𝟏𝟑°)

SOLUTION 3: The FORMULAS to CONVERT BETWEEN POLAR AND RECTANGULAR COORDINATES can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The “𝑥” and “𝑦” coordinates in RECTANGULAR FORM can be determined using the defined radius “𝑟” and the angle 𝜃. The formulas that will allow us to do this are: • 𝑥 = 𝑟 cos 𝜃 • 𝑦 = 𝑟 sin 𝜃 𝑦

• tan 𝜃 = 𝑥

The third equation of this bunch tells us that we can develop a relationship between the “𝑥” and “𝑦” coordinates, such that: 𝑦 = 𝑥 tan 120° Or: Made with

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𝑦 = −1.7321𝑥 We can then use the formula relating the radius and rectangular coordinates to solve for the value of the x-coordinate, given that:

𝑟 = |𝑥 + 𝑗𝑦| = √𝑥 2 + 𝑦 2 Plugging in what we have developed up to this point, in terms of the radius and the x-y relationship, we find that: (4)2 = 𝑥 2 + (−1.7321𝑥)2 NOTE: We squared both sides to get rid of the square root. We are now able to solve for x and find: 𝑥 = 4𝑥 2 = ±2 We don’t know where this point falls, on the plus or minus side of the y-axis…but we do know one thing…the angle of the radius, which is said to be 120°. This tells us that our x-coordinate will fall in the second quadrant, which means: 𝑥 = −2

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Rallying back up to the x-y relationship we developed previously, we can plug this value in for x and define our y-coordinate, which is: 𝑦 = −1.7321𝑥 Plugging in our value for x, we get: 𝑦 = (−1.7321)(−2) = 3.46 The correct answer choice is C. (−𝟐, 𝟑. 𝟒𝟔)

SOLUTION 4: The TOPIC of a POLAR COORDINATE SYSTEM can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The radius and angle in POLAR FORM can be determined using the 𝑥 and 𝑦 coordinates as shown in the relationships: • 𝑟 = |𝑥 + 𝑗𝑦| = √𝑥 2 + 𝑦 2 𝑦

• 𝜃 = arctan (𝑥 ) Where: • 𝑟 is the distance from the origin to the point Made with

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• 𝑥 is the x-coordiante of the rectangular coordinates • 𝑦 is the y-coordinate of the rectangular coordinates • 𝜃 is the angle between the radius “𝑟” and the positive 𝑥 − 𝑎𝑥𝑖𝑠 In our problem, we are given the RECTANGULAR COORDINATES: (−3, −5.2) Plugging these values in to our general formulas, we find that:

𝑟 = √(−3)2 + (−5.2)2 = 6 And: −5.2

𝜃 = arctan ( −3 ) = 60.018° The correct answer choice is B. (𝟔, 𝟔𝟎°)

SOLUTION 5: We are given two complex numbers in polar form: • 𝑧1 = 5(cos 330° + 𝑗 sin 330°) • 𝑧2 = 9(cos 90° + 𝑗 sin 90°)

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The FORMULA to CALCULATE THE QUOTIENT OF TWO COMPLEX NUMBERS IN POLAR REPRESENTATION can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We can calculate the quotient of two complex numbers in polar representation using the general formula: 𝑟1 (cos 𝜃1 + 𝑗 sin 𝜃1 ) 𝑟1 = [cos(𝜃1 − 𝜃2 ) + 𝑗 sin(𝜃1 − 𝜃2 )] 𝑟2 (cos 𝜃2 + 𝑗 sin 𝜃2 ) 𝑟2 Where: • 𝑟 is the absolute value of modulus of the complex number • 𝑥 is the x-coordinate of the rectangular coordinates • 𝑦 is the y-coordinate of the rectangular coordinates • 𝜃 is the angle between the radius "𝑟” and the positive 𝑥 − 𝑎𝑥𝑖𝑠 With the complex expressions already defined for us, we just simply plug and play, such that: 𝑧1 5(cos 330° + 𝑗 sin 330°) = 𝑧2 9(cos 90° + 𝑗 sin 90°)

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We are provided a means to simplify these quotients further given the formula: 𝑟1 (cos 𝜃1 + 𝑗 sin 𝜃1 ) 𝑟1 = [cos(𝜃1 − 𝜃2 ) + 𝑗 sin(𝜃1 − 𝜃2 )] 𝑟2 (cos 𝜃2 + 𝑗 sin 𝜃2 ) 𝑟2 Where: • 𝑟 is the absolute value of modulus of the complex number • 𝑥 is the x-coordinate of the rectangular coordinates • 𝑦 is the y-coordinate of the rectangular coordinates • 𝜃 is the angle between the radius "𝑟” and the positive 𝑥 − 𝑎𝑥𝑖𝑠 We can define all of the variables called for in this formula: • 𝑟1 = 5 • 𝑟2 = 9 • 𝜃1 = 330° • 𝜃2 = 90° Plugging in the values from the given complex expressions, we find the quotient of the two complex numbers can be expressed as: 𝑧1 5 = [cos(330° − 90°) + 𝑗 sin(330° − 90°)] 𝑧2 9

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Simplifying this expression, we get: 𝑧1 5 = (cos 240° + 𝑗 sin 240°) 𝑧2 9 𝟓

The correct answer choice is C. 𝟗 (𝐜𝐨𝐬 𝟐𝟒𝟎° + 𝐣 𝐬𝐢𝐧 𝟐𝟒𝟎°)

SOLUTION 6: The FORMULAS and VARIOUS FORMS OF EULER’S IDENTITY can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We are given a function written in trigonometric terms and asked to convert the expression in to exponential terms using Euler’s Identity. Given: sin2 𝜃 Euler’s Identity allows us to rewrite exponential terms in terms of trigonometric functions, or vice versa, to rewrite trigonometric functions in terms of exponential terms.

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We are given a sine function, so we will use Euler’s Identity for sine, which is: 𝑒 𝑗𝜃 − 𝑒 −𝑗𝜃 sin 𝜃 = 2𝑗 The trick with this question is to realize that the sine term is squared in our original expression, therefore, we must also square the exponential side of the equation, this presents to us: 2

𝑒 𝑗𝜃 − 𝑒 −𝑗𝜃 2 sin 𝜃 = ( ) 2𝑗

Foiling the terms, we can rewrite the exponential expression as: (𝑒 𝑗𝜃 )(𝑒 𝑗𝜃 ) + (𝑒 𝑗𝜃 )(−𝑒 −𝑗𝜃 ) + (−𝑒 −𝑗𝜃 )(𝑒 𝑗𝜃 ) + (−𝑒 −𝑗𝜃 )(−𝑒 −𝑗𝜃 ) sin 𝜃 = 22 𝑗 2 2

We can then simply the exponential expression using the laws of exponents: (𝑒 2𝑗𝜃 ) + (−𝑒 0 ) + (−𝑒 0 ) + (−𝑒 −2𝑗𝜃 ) sin 𝜃 = 4𝑗 2 2

Grouping the terms, we can then simply the expression again as:

sin2 𝜃 =

𝑒 2𝑗𝜃 − 2(1)−𝑒 −2𝑗𝜃 1 = (𝑒 2𝑗𝜃 − 2−𝑒 −2𝑗𝜃 ) 2 2 4𝑗 4𝑗

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EULER’S IDENTITY states: 𝑒 𝑗𝜃 = cos 𝜃 + 𝑗 sin 𝜃 𝑒 −𝑗𝜃 = cos 𝜃 − 𝑗 sin 𝜃 Substituting these in to our general formula, we get:

sin2 𝜃 =

1 ((cos 2𝜃 + 𝑗 sin 2𝜃) − 2 − (cos 2𝜃 − 𝑗 sin 2𝜃)) 4𝑗 2

Simplifying the expression, we get:

sin2 𝜃 =

1 (cos 2𝜃 + 𝑗 sin 2𝜃 − 2 − cos 2𝜃 + 𝑗 sin 2𝜃) 4𝑗 2

Cancelling out like terms, we get:

sin2 𝜃 =

1 (2𝑗 sin 2𝜃 − 2) 4𝑗 2

Simplifying the expression again, we get:

sin2 𝜃 =

1 (𝑗 sin 2𝜃 − 1) 2𝑗 2 𝟏

The correct answer choice is C. 𝐬𝐢𝐧𝟐 𝛉 = 𝟐𝒋𝟐 (𝒋 𝒔𝒊𝒏 𝟐𝜽 − 𝟏) Made with

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SOLUTION 7: The FORMULA for the POLAR FORM OF A COMPLEX NUMBER can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The first step is to convert the complex expression in to polar form to match the standard expression that we are accustomed to for the polar form of nonzero complex number. The complex number, in RECTANGULAR FORM, is given as:

1 𝑗√3 𝑧=− + 2 2 The polar form of a nonzero complex number is represented by the expression: 𝑧 = 𝑥 + 𝑗𝑦 = 𝑟(cos 𝜃 + 𝑗 sin 𝜃) = 𝑟𝑒 𝑗𝜃 Where: • 𝑟 is the modulus of the complex number • 𝜃 is the argument of the complex number The radius and angle in polar form of the nonzero complex number can be defined knowing the “𝑥” and “𝑦” coordinates of the complex number in RECTANGULAR FORM using: Made with

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• 𝑟 = |𝑥 + 𝑗𝑦| = √𝑥 2 + 𝑦 2 𝑦

• 𝜃 = arctan (𝑥 ) Where: • 𝑟 is the distance from the origin to the point • 𝑥 is the x-coordinate of the rectangular coordinates • 𝑦 is the y-coordinate of the rectangular coordinates • 𝜃 is the angle between the radius “𝑟” and the positive 𝑥 − 𝑎𝑥𝑖𝑠 Taking these formulas and plugging in what we know, we can solve for the modulus and argument of the complex number in POLAR FORM, such that:

2

1 𝑗√3 1 2 √3 √ 𝑟 = |− + | = (− ) + ( ) = 1 2 2 2 2

And:

1 𝜋 √3 𝜃 = arctan (( ) / (− )) = 2 2 3

The final polar form of the complex number can be re-written as

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𝑧 = 𝑥 + 𝑗𝑦 = 𝑟(cos 𝜃 + 𝑗 sin 𝜃) = 𝑟𝑒 𝑗𝜃

1 𝑗√3 π 𝜋 𝑧=− + = 1 (cos +𝑗 sin ) 2 2 3 3 The FORMULA for DE MOIVRE’S THEOREM can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. De Movire’s Theorem generally states: 𝑧 = (𝑥 + 𝑗𝑦)𝑛 = [𝑟(cos 𝜃 + jsin 𝜃)]𝑛 = 𝑟 𝑛 (cos 𝑛𝜃 + 𝑗 sin 𝑛𝜃) We know the POLAR FORM of our complex expression, which we defined in the previous step, and we also know that n is equal to 10.

Therefore, using De Movire’s Theorem, we can expand the complex expression as: 10

1 𝑗√3 𝑧 = (− + ) 2 2

𝜋 𝜋 10 = [1 (cos + 𝑗 sin )] 3 3

Which gives us:

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𝑧 = 110 [cos

10(𝜋) 10(𝜋) + 𝑗 sin ] 3 3

And the standard form results as:

𝑧 = 1 (cos

10𝜋 10𝜋 1 𝑗√3 + 𝑗 sin )=− + 3 3 2 2 𝟏

The correct answer choice is 𝐁. − 𝟐 +

𝐣√𝟑 𝟐

SOLUTION 8: The FORMULA for the POLAR FORM A COMPLEX NUMBER can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The first step is to convert the complex expression in to polar form, to match the standard expression for polar form of nonzero complex number.

We are given:

𝑧 = −2 − 2𝑗√3 The polar form of a nonzero complex number is represented by the expression:

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𝑧 = 𝑥 + 𝑗𝑦 = 𝑟(cos 𝜃 + 𝑗 sin 𝜃) = 𝑟𝑒 𝑗𝜃 Where: • 𝑟 is the modulus of the complex number • 𝜃 is the argument of the complex number The radius and angle in polar form of the nonzero complex number can be defined knowing the “𝑥” and “𝑦” coordinates of the complex number in RECTANGULAR FORM using: • 𝑟 = |𝑥 + 𝑗𝑦| = √𝑥 2 + 𝑦 2 𝑦

• 𝜃 = arctan (𝑥 ) Where: • 𝑟 is the distance from the origin to the point • 𝑥 is the x-coordinate of the rectangular coordinates • 𝑦 is the y-coordinate of the rectangular coordinates • 𝜃 is the angle between the radius “𝑟” and the positive 𝑥 − 𝑎𝑥𝑖𝑠 Plugging in the values from the given expression, we can solve for modulus and argument of the complex number using these established formulas, which gives us:

2

𝑟 = |−2 + 𝑗(−2√3)| = √(−2)2 + (−2√3 ) = 4

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And:

−2√3 𝜋 𝜃 = arctan ( )= −2 3 The final version of the complex number in polar form can be written as: 𝑧 = 𝑥 + 𝑗𝑦 = 𝑟(cos 𝜃 + 𝑗 sin 𝜃) = 𝑟𝑒 𝑗𝜃 π 𝜋 𝑧 = −2 − 2𝑗√3 = 4 (cos +𝑗 sin ) 3 3 The FORMULA for the ROOTS OF A COMPLEX NUMBER can be referenced under the subject of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The kth root, “𝑤”, of a complex number is found from the equation expressed as: 𝜃 360° 𝜃 360° 𝑘 𝑤 = √𝑟 [cos ( + 𝑛 ) + 𝑗 sin ( + 𝑛 )] 𝑘 𝑘 𝑘 𝑘 The “𝑘” roots of 𝑟(cos 𝜃 + 𝑗 sin 𝜃) can be found by substituting successively 𝑛 = 0,1,2, etc up to (𝑘 − 1) in the stated formula. As 𝑘 = 3, then we would calculate "𝑛" as 𝑛 = 3 − 1 = 2 If “𝑘” is any positive integer, any complex number (other than zero), has “𝑘” distinct roots. Made with

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We will take our complex expression and set it equal to the equation for the roots of a complex number, such that: 𝜋 𝜋 360° 360° 3 √−2 − 2𝑗√3 = √4 [cos ( + 𝑛 ) + 𝑗 sin ( 3 + 𝑛 )] 𝑘 𝑘 𝑘 𝑘

3

3

To solve for the third root, we will plug in a value of 𝑘 = 3 𝑎𝑛𝑑 𝑛 = 2 into the equation above, and re-write 360° as 2𝜋 radians: 𝜋 𝜋 (2𝜋) (2𝜋) 3 𝑧3 = √4 [cos ( 3 + (2) ) + 𝑗 sin ( 3 + (2) )] 3 3 3 3

Simplifying, we find the complex expression for the third root is: 𝜋 4𝜋 𝜋 4𝜋 3 𝑧3 = √4 [cos ( + ) + 𝑗 sin ( + )] 9 3 9 3

Which then simplifies to:

3

𝑧3 = √4 [cos (

13𝜋 13𝜋 ) + 𝑗 sin ( )] 9 9 𝟑

𝟏𝟑𝛑

𝟏𝟑𝛑

𝟗

𝟗

The correct answer choice is D. 𝐳𝟑 = √𝟒 [𝐜𝐨𝐬 (

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) + 𝐣 𝐬𝐢𝐧 (

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)]