Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex ...

Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers

Page 1 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6

Syllabus Objectives: 5.1 – The student will explore methods of vector addition and subtraction. 5.2

– The student will develop strategies for computing a vector’s direction angle and magnitude given

its coordinates. 5.4 – The student will resolve vectors into unit vectors. 5.7 – The student will solve

real-world application problems using vectors in two and three dimensions.

Directed Line Segment: a segment with direction and distance

AB A: Initial Point (start); B: Terminal Point (end)

Coordinates of A: ,x yA A Coordinates of B: ,x yB B

Magnitude (length) of a Directed Line Segment AB : 22

x x y yAB B A B A

Note: This is the distance formula!

Vector (v): the set of all directed line segments that are equivalent to a given directed line segment

Note: Equivalent means same magnitude and direction.

Component Form of a Vector: ,x x y yB A B A

Ex: Graph the vector 3, 2AB and find the magnitude.

One possible graph:

Note: 3, 2AB could be placed anywhere on the coordinate grid. We have

placed it in standard position, which is with the initial point at the origin.

Magnitude: 2 2

3 0 2 0 9 4 13AB

Note: If a vector u is written in component form, 1 2,u uu , then the magnitude of u is

2 2

1 2u uu . This is because the initial point is the origin, 0,0 .

Vector Addition: Let 1 2,u uu and 1 2,v vv . Then 1 1 2 2,u v u vu v .

Scalar Multiplication: Let 1 2,u uu and k be any constant. Then 1 2,k ku kuu .

Note: If 0k , then ku is in the opposite direction.

A

B



Ex: Use the graph of the vectors to complete each example below.

1. Show that u v .

Show that u v . 2 2

6 1 4 1 25 9 34u

2 2

4 1 1 4 25 9 34v

Show that the direction of u is the same as the direction of v.

Use slope: Direction of u = 4 1 3

6 1 5; direction of v =

1 4 3

4 1 5

The direction and magnitude are the same, so u v .

2. Find the component form and the magnitude of and u w .

Component form of u: 1 6,1 4 5, 3u 34u (see above)

Component form of w: 2 4 ,5 3 2,8w 2 22 8 4 64 68 2 17w

3. Find the component form of 2 3u w .

2 3 2 5, 3 3 2,8 10, 6 6, 24 10 6, 6 24 16, 30u w

Unit Vector: a vector with a magnitude of 1

A unit vector in the direction of a vector v can be found by dividing v by the magnitude of v.

Unit Vector in the Direction of v: v

v

Standard Unit Vectors: unit vectors i and j in standard position along the positive x- and y-axes

1,0 & 0,1i j

Any vector can be written in terms of the standard unit vectors.

Ex: Write the vector 2,5v in terms of the standard unit vectors.

2,5 2 1,0 5 0,1 2 5v i j

u

v

w



Ex: Find a unit vector in the direction of the given vector. Verify your answer is a unit vector

and give your answer in component form and standard unit vector form. 2 4i j

Find the magnitude: 22

2 4 2 4 20 2 5i j

Divide the original vector by its magnitude: 2 4 2 4 2 5 2 5

5 52 5 2 5 2 5 5 5

i j i j i ji j (SUV)

Component Form: 5 2 5

,5 5

Verify magnitude of unit vector:

2 2

5 2 5 5 2 5 5 20 25, 1

5 5 5 5 25 25 25☺

Recall: In the unit circle, cos , sinx y . This leads into another way of expressing a vector, in

terms of its direction angle, θ.

Direction Angle: in standard position, the angle the vector makes with the positive x-axis

(counterclockwise)

Resolving a Vector: in terms of its direction angle, θ, a vector can be written as

cos ,sin cos sinu u i u j

Ex: Find the magnitude and direction angle of 2 6v i j .

Magnitude: 2 2

2 6 40 2 10v

Direction angle: cos sinv i v j 2 cosv

1

2 2 10 cos

1cos

10

1cos 108.43

10

OR 6 sinv

1

6 2 10 sin

3sin

10

3sin 71.57

10

but since we know 2 6v i j is in Quadrant II, 180 71.57 108.43



Ex: Find the component form of v given its magnitude and its direction angle. 5, 30v

5 3 5cos sin 5cos30 5sin30

2 2v v i v j v i j v i j

Application: Resultant Force

Ex: Two forces act on an object: 3, 45uu and 4, 30uv . Find the direction and

magnitude of the resultant force.

Write each vector in component form:

3 2 3 2

cos sin 3cos45 3sin 452 2

u uu u i u j u i j u i j

cos sin 4cos 30 4sin 30 2 3 2v vv v i v j v i j v i j

The resultant force is the sum u v : 3 2 3 2

2 3, 22 2

Application: Bearing

Ex: A plane flies due east at 500 km/h and there is a 60 km/h with a bearing of 45°. Find the

ground speed and the actual bearing of the plane.

Sketch a diagram:

Find the vectors p and w: 500 cos ,500sinp

60cos 45 ,60sin 45w Note: The 45° is the direction angle, not the bearing.

Vector v is the sum p + w: 60cos45 500cos ,60sin 45 500sinv

The second component of vector v must equal zero, because the plane is headed due east.

160sin 45 60sin 4560sin 45 500sin 0 sin sin 4.868

500 500

Bearing of the plane: 90 94.868

Ground speed of the plane:

2

260 cos 45 500 cos 0 60 cos 45 500 cos 4.868 540.623km/hrv

500 km/hr

60 km/hr

45°

θ

w v

p



You Try: Find the component form of v given its magnitude and the angle it makes with the positive x-

axis. 2v , direction: 2 3i j

QOD: In the examples in your notes, we used sine or cosine to find the direction angle of a vector.

Explain how you could use tangent to find the direction angle.



Syllabus Objective: 5.3 – The student will explore methods of vector multiplication. 5.5 – The

student will determine if two vectors are parallel or perpendicular (orthogonal). 5.6 – The student

will derive an equation of a line or plane by using vector operations. 5.7 – The student will solve

real-world application problems using vectors in two and three dimensions.

Dot Product: Let 1 2,u uu and 1 2,v vv . The dot product is 1 1 2 2u v u vu v .

Note: The dot product is a scalar.

Ex: Evaluate 5, 2 3,4 .

5, 2 3,4 5 3 2 4 15 8 7

Properties of the Dot Product:

1. u v v u

2. 2

u u = u

3. 0 u 0

4. u v + w = u v u w

5. c c cu v u v u v

Ex: Evaluate the following given 3,6 ; 1,0 ; 5, 2u v w

(a) w w 5 5 2 2 29w w

(b) w 2 2

5 2 25 4 29w

(c) v w u 1 5,0 2 6, 2 3,6 6 3 2 6 30v w u u

(d) v u w u 1 3 0 6 5 3 2 6 3 27 30v u w u

Angle Between Two Vectors: 1cos cos

u v u v

u v u v

Proof: Use the triangle.

Law of Cosines: 2 2 2

2 cosv u u v u v

Property of Dot Product: 2 2

2 cosv - u v - u u v u v

Expand: 2 2

2 cosv v v u u v u u u v u v

Property of Dot Product: 2 2 2 2

2 2 cosv u v u u v u v

Property of Equality: 2 2 cos cosu v

u v u vu v

u

v u − v

θ



Ex: Find u v , where θ is the angle between u and v. 5

6, 8,6

u v

5 3cos cos 48 24 3

6 6 8 2

u v u vu v u v

u v

Application

Ex: Find the interior angles of the triangle with vertices .

4 3,2 0 1,2 ; 5 3,1 0 2,1AB AC 11 2 2 1 4cos cos 36.87

55 5A A A

3 4,0 2 1, 2 ; 5 4,1 2 1, 1BA BC

11 1 2 1 1cos cos 71.565

105 2B B B

180 36.87 71.565 71.565C Angles: 36.87 , 71.565 ,71.565

Orthogonal Vectors: two vectors whose dot product is equal to 0

What is the angle between two non-zero orthogonal vectors?

0

cos cos cos 0 90u v

u v u v

Note: If the angle between the vectors is 90°, we may also say they are perpendicular. The word

orthogonal is used instead for vectors because the zero vector is orthogonal to any other vector, but is not

perpendicular.

What is the dot product of two vectors that are parallel? The angle “between” them would have to be

either 180° or 360°.

cos cos180 1u v u v

u vu v u v

or cos cos360 1u v u v

u vu v u v

Parallel Vectors: two vectors whose dot product is equal to −1 or 1

Ex: Are the vectors orthogonal, parallel, or neither? 3 2 , 3 4v i j w i j

Find v w : 3 3 2 4 1v w The vectors are parallel.

A

B

C



Vector Projection: the projection of u onto v is denoted by: 2

projv

u vu v

v

Ex: Find the projection of v onto w. Then write v as the sum of two orthogonal vectors, with

one the projwv . 1,3 ; 1,1v w

2 22 2

1 1 3 1proj proj 1,1 proj 2 1,1 proj 2, 2

1 1

w w w w

v wv w v v v

w

proj 1,3 2,2 1,1wv v 2,2 1,1v

Application: Force

Ex: Find the force required to keep a 200-lb cart from rolling down a 30° incline.

Draw a diagram and label The force due to gravity: 200g j (gravity acts vertically downward)

Incline vector: 3 1

cos 30 sin 302 2

v i j i j

Force vector required to keep the cart from rolling: 2

projv

g v

vf g v

2 2

3 10, 100 ,

2 2 3 1 3 1, 100 , 50 3, 50

2 2 2 21

g vv

vf

Magnitude of Force: 2 250 3, 50 (50 3) ( 50) 100 poundsf

u

v

u1 = projvu

u2 = u – u1 Note: The projection of u onto v is

the “shadow” formed by vector u

onto v as light comes straight down.

2 23 1 3 1

12 2 4 4

v

g

v

f

20030



Application: Work cos force distanceW

Ex: A person pulls a wagon with a constant force of 15 lbs at a constant angle of 40° for 500 ft.

What is the person’s work?

cos 40 15 lbs 500 ft 5745.33 foot lbsw

You Try: Find the projection of v onto u. Then write v as the sum of two orthogonal vectors, with one

the projuv . 2 3 ;v i j u i j

QOD: If u is a unit vector, what is u u ? Explain why.

40°



Syllabus Objective: 1.10 – The student will solve problems using parametric equations.

Parametric Curve: the set of all points ,x y , where x f t and y g t are continuous functions of t

on an interval I (called the parameter interval)

Parameter: the variable t

Parametric Equations: x f t and y g t

Orientation: the directions that results from plotting the points as the values of t increase

Graphing Parametric Equations

Ex: Graph 22 , 2 1, 1 2x t y t t

Note: Choose appropriate values for t first. Then substitute in and find x and y.

Make a table:

Plot points:

Eliminating the Parameter

1. Solve one of the equations for t. (Or if a trig function, isolate the trig function.)

2. Substitute for t in the other equation. (Or use an identity if a trig function.)

Ex: Write the parametric equation as a function of y in terms of x.

a) 22 , 2 1x t y t

Solve for t in the x-equation (easier to solve for): 22

xx t t

Substitute into the y-equation: 2

2 212 1 2 1 1

2 2

xy t y y x

b) 1

,22

tx y

tt

Solve for t in the x-equation (easier to solve for):

2 2

2 2

1 1 1 12 1 2 2

22x x t x t t

tt x x

t x y

−1 2 1

1 −2 1

0 0 −1

2 −4 7



Substitute into the y-equation:

2

2 22

2 2

1 1 22

1 21 12

2 2

x

t x xy y y y xt

x x

Ex: Write the parametric equation as a function of y and graph.

tan , 2 tan 1, 02

x y

Note: A parametric equation can be written in terms of θ instead of t.

The x-equation is already solved for the trig function: tanx

Substitute into the y-equation: 2 1y x

Graph:

Using a Trig Identity

Ex: Eliminate the parameter. 6cos , 6sin , 0 2x t y t t

Solving for a trig function won’t help, so we need to use the identity 2 2sin cos 1t t .

Square both equations: 2 2 2 236cos , 36sinx t y t

Add the equations: 2 2 2 2 2 2 2 236cos 36sin 36 cos sinx y t t x y t t

Trig identity: 2 2 2 2 2 236 cos sin 36x y t t x y

Note: The graph is a circle. The parameter interval lets us know that it would go around 1 time.

Writing a Parameterization

Ex: Find the parameterization of the line segment through the points 1, 2A and 2, 3B .

Sketch a graph:

OP OA AP ; AP is a scalar multiple of AB , so OP OA t AB

A

B

O

P



, 1, 2 2 1 , 3 2 , 1, 2 3, 1OP OA t AB x y t x y t

, 1 3 , 2x y t t 1 3 , 2x t y t These equations define the LINE.

Find the parameter interval for the line segment: We want 1 2x .

1 3x t : 1 1 3 0t t 2 1 3 1t t So, 0 1t

Solution: 1 3 , 2 , 0 1x t y t t

Simulating Horizontal Motion

Ex: A dog is running on a horizontal path with the coordinates of his position (in meters) given

by 3 20.2 19 100 70s t t t where 0 15t . Use parametric equations and a graphing

calculator to simulate the dog’s motion.

Choose any horizontal line to simulate the motion: We will choose 3y .

Parametric Equations: 2 20.2 19 100 70 , 3, 0 15x t t t y t

Graph (Calculator must be in Parametric mode):

Note: To see the motion, change the type of line to a “bubble”. If you would like the bubble to move

slower, make the Tstep smaller.

Parametric Equations for Projectile Motion

distance: 0 cosx v t height: 20 0

1sin

2y gt v t h

Note: On Earth, 232 ft/secg or 2

9.8 m/secg . 0v is initial velocity; 0

h is initial height.

Ex: A golf ball is hit at 150 ft/sec at a 30° angle to the horizontal.

a) When does it reach its maximum height?

Height: 2 20 0

1sin 16 150sin 30 0

2y gt v t h y t t

( 0 0h because a golf ball is hit from the ground)

Simplify: 2 216 150sin 30 0 16 75y t t y t t

Maximum height is at vertex: 75 75 11

2 sec2 2 16 32 32

bt t

a

b) How far does it go before it hits the ground?

Hits the ground when 0y : 2 1116 75 0 16 75 0 16 75 0 4 sec

16y t t t t t t

Note: We could have doubled the time it took for the ball to reach its highest point!



Distance: 0

11cos 150cos 30 4 608.84 ft

16x v t

c) Does the ball hit a 6 ft tall golfer, standing directly in the path of the ball 580 feet away?

Find the time it takes for the ball to be 580 ft away: 0 cos 580 150cos 30 4.46secx v t t t

Find the height of the ball at this time: 22

16 75 16 4.46 75 4.46 16.23fty t t

No – the ball misses him by about 10 feet.

Application: Ferris Wheel

Ex: Zac is on a Ferris wheel of radius 20 ft that turns counterclockwise at a rate of one

revolution every 24 sec. The lowest point of the Ferris wheel (6 o’clock) is 10 ft above ground

level at the point (0, 10) on a rectangular coordinate system. Find the parametric equations for the

position of Zac as a function of time t (in seconds) if the Ferris wheel starts (t = 0) with Zac at the

point (20, 30).

Time to complete one revolution = 24 sec: 2

24 12

So, in one second, the ferris wheel travels through an angle of 12

.

When 0, 20 & 30t x y : 20cos , 30 20sin , 012 12

x t y t t

You Try: A baseball is hit at 3 ft above the ground with an initial speed of 160 ft/sec at an angle of 17°

with the horizontal. Will the ball clear a 20-ft wall that is 400 ft away?

QOD: How would you write a parametrization for a semicircle?

Remember: cos , sinx r y r



Syllabus Objectives: 3.3 – The student will differentiate between polar and Cartesian (rectangular)

coordinates. 6.2 – The student will transform functions between Cartesian and polar form. 6.4 –

The student will solve real-world application problems using polar coordinates.

Polar Coordinate: ,r ; r: the directed distance from the pole (origin); θ: the directed angle from the

polar axis (x-axis)

Plotting Points on a Polar Graph

Ex: Plot the points 3 5

3, , 8, 240 , & 2,2 6

A B C .

Point A: Start at the polar axis and go counter-clockwise 3

2 (270°). Place the point 3 units from the

pole (origin). Point B: Start at the polar axis and go clockwise 240°. Place the point 8 units from the

pole. (Note: Each radius drawn in the grid is 15°.) Point C: Start by going counter-clockwise 5

3

(300°) from the polar axis. Place a point 2 units from the pole. Because 2r , you must place the point

on the opposite side of the pole.

Writing the Polar Coordinates of a Point

Ex: Find four different polar coordinates of P.

Starting at the polar axis and going counter-clockwise: 6,30

6

4

2

2

4

6

10 5 5 10

P

6

4

2

2

4

6

10 5 5 10

A

B

C



Starting at the polar axis and going clockwise: 6, 330

Going counter-clockwise more than one revolution: 6,390

Using 0r and rotating counter-clockwise: 6,210

Note: There are infinitely many correct answers!

Polar Conversions

Polar to Rectangular: cos

sin

r x

r y

Rectangular to Polar:

1

2 2 2 2 2

tany

x

x y r r x y

Converting from Polar to Rectangular Coordinates

Ex: Convert to rectangular coordinates.

a) 5 2,45

cos

sin

r x

r y

15 2 cos 45 5 2 5

2

15 2 sin 45 5 2 5

2

x

y

5,5

b) 3,3

cos

sin

r x

r y

1 33cos 3

3 2 2

3 3 33sin 3

3 2 2

x

y

3 3 3

,2 2

Converting from Rectangular to Polar Coordinates (Note: Be careful with the quadrant!)

Ex: Convert to polar coordinates.

a) 1, 3

1

2 2 2 2 2

tany

x

x y r r x y

1

tan 3 60

1 3 2r

1, 3 is in Quadrant II, so the polar coordinates are 2,120 .

b) 0, 4

This point is on the negative y-axis, so we know 270 . 4,270

Note: There are other possible answers to these!



Converting from Polar to Rectangular Equations

Ex: Convert the equations and sketch the graph.

a) 2r

2 2 22 4 4r r x y Graph is a circle with center at origin & radius 2.

b) 4

2cos cos

4 2

2sin sin

4 2

x r x r x r

y r y r y r

y x

c) secr

1sec cos 1 1

cosr r r x

Graphing in Polar Coordinates on the Calculator

We will check our graphs above. Calculator must be in Polar mode.

a) c)

Note: We cannot check the graph of b) on the calculator, but the line y x represents the angle

454

for all values of r.



Converting from Rectangular to Polar Equations

Ex: Convert the equations.

a) 4x

cosx r : cos 4 4 secr r

b) 3 6 2 0x y

2 23 6 2 0 3 cos 6 sin 2 0 3cos 6sin 2 or

3cos 6sin 6sin 3cosx y r r r r r

c) 2 2

2 1 5x y

Expand: 2 2 2 24 4 2 1 5 4 2 0x x y y x y x y

Substitute: 24 cos 2 sin 0 4 cos 2sin 0r r r r r

So 0r or 4cos 2sin 0r . But 0r is a single point. So 4 cos 2 sinr

Application: Finding Distance

Ex: The location of two ships from the shore patrol station, given in polar coordinates, are

2 mi, 150 & 3mi, 80 . Find the distance between the ships.

Sketch a diagram:

Note: The angle between the ships (from the patrol station) is 150 80 70 .

Using the Law of Cosines: 2 2 22 3 2 2 3 cos70 13 12cos70 2.983mid d

You Try: Convert the coordinates. Polar: 2, ; Rectangular: 7,7

QOD: How could you write an expression for all of the possible polar coordinates of a point?



Syllabus Objectives: 6.3 – The student will sketch the graph of a polar function and analyze it.

Tests for Symmetry of Polar Curves

1. Symmetry about x-axis: ,r is equivalent to ,r

2. Symmetry about y-axis: ,r is equivalent to ,r

3. Symmetry about origin: ,r is equivalent to ,r

Graphing Polar Curves

Ex: Graph and find the domain, range, symmetry, and maximum r-value.

a) 2 4cosr

θ 0

3

2

2

3

r 6 4 2 0 −2

Domain: All real numbers Range: 2 6r Max r-value: 6r

Symmetry: Substitute . 2 4 cos 2 4 cosr Symmetric about x-axis

This curve is called a limaçon.

b) 6sin 3r

θ 0

18

6

5

18

3

7

18

2

r 0 3 6 3 0 −3 −6


Symmetry: Substitute &r . 6sin 3 6sin 3 6sin 3r r r Symmetric about y-axis

This curve is called a rose.

c) 24 cos 2r

θ 0

6

4

r 2 2 0


Symmetry:

Substitute . 2 24 cos 2 4 cos 2r r Symmetric about x-axis



Substitute r . 2 2

4cos 2 4cos 2r r Symmetric about origin

Substitute &r . 2 2

4cos 2 4cos 2r r Symmetric about y-axis

This curve is called a lemniscate.

Classifications of Polar Curves

Limaçon Curves: sinr a b and cosr a b

Rose Curves: cosr a n and sinr a n

Petals: odd = n and even = 2n

Lemniscate Curves: 2 2cos 2r a and 2 2

sin 2r a

You Try: Use your graphing calculator to explore variations of sinr a n . Describe the effects of

changing the window, the θ-step, a, n, and changing sin to cos .

QOD: Are all polar curves bounded? Explain.



Syllabus Objectives: 7.1 – The student will graph a complex number on the complex/Argand plane.

7.2 – The student will represent a complex number in trigonometric (polar) form. 7.3 – The student

will simplify expressions involving complex numbers in trigonometric (polar) form. 7.4 – The

student will compute the powers of complex numbers using DeMoivre’s Theorem and find the nth

roots of a complex number.

Complex Number Plane (Argand Plane): horizontal axis: real axis; vertical axis: imaginary axis

Plotting Points in the Complex Plane

Ex: Plot the points 3 4 , 1 3 , & 2A i B i i in the complex plane.

Absolute Value (Modulus) of a Complex Number: the distance a complex number is from the origin on

the complex plane 2 2a bi a b (This can be shown using the Pythagorean Theorem.)

Ex: Evaluate 3 i . 22

3 3 1 10i

Recall: Trigonometric form of a vector: cos ,sinu

Trigonometric Form of a Complex Number z = a + bi: cos sinz r i

Note: This can also be written as cisz r .

2 2cos , sin ,a r b r r a b tan

b

a

r = modulus; θ = argument

Writing a Complex Number in Trig Form

Ex: Find the trigonometric form of 3 i .

Find r: 2 22 2

3 1 4 2r a b

Find θ: 1 11

tan tan6 63

b

a

11 11 113 2 cos sin or 2cis

6 6 6i i

A

B

C



Writing a Complex Number in Standard Form (a + bi)

Ex: Write 9cis in standard form.

Expand: 9cis 9 cos sin 9 1 0 9 0 9i i i

Multiplying and Dividing Complex Numbers

Let 1 1 1 1cos sinz r i and 2 2 2 2cos sinz r i .

Multiplication: 1 2 1 2 1 2 1 2cos sinz z r r i

Division: 1 11 2 1 2

2 2

cos sinz r

iz r

Ex: Express the product of 1z and 2z in standard form.

1 24 cos sin , 2 cos sin4 4 6 6

z i z i

1 2

5 54 2 cos sin 4 2 cos sin 1.464 5.464

4 6 4 6 12 12z z i i i

Powers of a Complex Number: De Moivre’s Theorem cos sin cos sinnn n

z r i r n i n

Ex: Evaluate 5

2 2i .

Rewrite in trig form: 1 2tan 45 135

2

2 2

2 2 4 2r

55

2cis135 2 cis 5 135 32cis 675 32cis 315

nth Roots of a Complex Number:

2 2 2cos sin cisn nn k k k

z r i rn n n n n

, 0,1, 2, ... 1k n

Note: Every complex number has a total of “n” nth roots.

Ex: Find the cube roots of 8i.

Write in trig form: 8, 90r 8cis90

Evaluate the roots: 3 3 90 3608cis90 8cis 2cis 30 120

3

kk

0 : 2cis 30 0 2cis 30 2 cos30 sin 30 3k i i

1: 2cis 30 120 2cis 150 2 cos150 sin150 3k i i

2 : 2cis 30 240 2cis 270 2 cos270 sin 270 2k i i

Roots of Unity: the nth roots of 1



Ex: Express the fifth roots of unity in standard form and graph them in the complex plane.

5th Roots of Unity: 5 1 0i 1, 0r 1cis0

55 0 2 21cis0 1cis cis

5 5

k k

0 : cis0 1k

21: cis 0.31 0.95

5k i

42 : cis 0.81 0.59

5k i

63 : cis 0.81 0.59

5k i

84 : cis 0.31 0.95

5k i

You Try:

1. Write each complex number in trigonometric form. Then find the product and the quotient.

1 3 , 2 2 3i i

2. Solve the equation 41 0x . (You should have 4 solutions!)

QOD: Is the trigonometric form of a complex number unique? Explain.

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