MTH 111–112 Supplement

MTH 111–112 Supplement

Mathematics Faculty
Portland Community College

Greta Swanson, Editor

Alex Jordan, Editor

December 15, 2023

Contents

1 MTH 111 Supplement 1

1.1 Graph Transformations. . . . . . . . . . . . . . . . 1
1.2 Inverse Functions . . . . . . . . . . . . . . . . . . 3
1.3 Exponential Functions . . . . . . . . . . . . . . . . 3
1.4 Logarithmic Functions . . . . . . . . . . . . . . . . 3

2 MTH 112 Supplement 6

2.1 Angles . . . . . . . . . . . . . . . . . . . . . . 6
2.2 Generalized Definitions of Trigonometric Functions . . . . . 9
2.3 Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift . 14
2.4 Complex Numbers and Polar Coordinates . . . . . . . . . 17

Appendices

A Answers and Solutions to Exercises 22

iv
Chapter 1

MTH 111 Supplement

1.1 Graph Transformations

Example 1.1.1 The table below defines the functions f , g, and h. Express
g(x) and h(x) in terms of f .

x −3 −2 −1 0 1 2 3
f (x) 8 6 4 2 0 −1 −2
g(x) −8 −6 −4 −2 0 1 2
h(x) 5 3 1 1 −3 −4 −5

Answer. g(x) = −f (x) and h(x) = f (x) − 3. □

Example 1.1.2
(a) If f (x) = x2 and g(x) = 2x2 + 5, express g(x) in terms of f .
Answer. g(x) = 2f (x) + 5

(b) If f (x) = x2 and h(x) = (x + 5)2 − 3, express h(x) in terms of f .

Answer. h(x) = f (x + 5) − 3
□

Exercises
One Function in Terms of Another. In Exercises 1–4, the table below
defines the functions f , g, h, k, and l.
x −2 −1 0 1 2
f (x) 0 1 2 3 4
g(x) 4 3 2 1 0
h(x) 0 −1 −2 −3 −4
k(x) 6 7 8 9 10
l(x) 0 3 6 9 12
1. Express g(x) in terms of f and describe how the graph of y = f (x)
can be transformed into the graph of y = g(x).
2. Express h(x) in terms of f and describe how the graph of y = f (x)
can be transformed into the graph of y = h(x).

1
CHAPTER 1. MTH 111 SUPPLEMENT 2

3. Express k(x) in terms of f and describe how the graph of y = f (x)

can be transformed into the graph of y = k(x).
4. Express l(x) in terms of f and describe how the graph of y = f (x) can
be transformed into the graph of y = l(x).
5. The second row in the table below gives values for the function f . Complete
the rest of the table. If you don’t have sufficient information to fill in some
of the cells, leave those cells blank.

x −4 −3 −2 −1 0 1 2 3 4
f (x) −2 −1 0 1 2 3 4 5 6
1
2x
−2f (x)
f (x) + 5
f (x + 2)
f 12 x

f (2x)
f (x − 3)

Find the Transformations. In Exercises 6–9, first write g(x) in terms of f .

Then compose a sequence of transformations that will transform the graph of
y = f (x) into the graph of y = g(x).
√ √
6. f (x) = x g(x) = x−74
1 2
7. f (x) = x g(x) = x +3

2
8. f (x) = x2 g(x) = −4 21 x − 5 + 3
√ √
9. f (x) = 3 x g(x) = 12 3 10x + 30 − 6

Sketch Transformations. In Exercises 10–13, use the provided graph of

y = f (x) to sketch a graph of each given function.
10 y
8
6
4
2
x
−8 −6 −4 −2 2 4 6 8
−2
−4
−6
−8
−10

10. k1 (x) = f (2x) 11. k2 (x) = 2f (−2x) − 1

10 y 10 y
8 8
6 6
4 4
2 2
x x
−8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8
−2 −2
−4 −4
−6 −6
−8 −8
−10 −10
CHAPTER 1. MTH 111 SUPPLEMENT 3

12. k3 (x) = −2f (2x + 4) 1

13. k4 (x) = f 2x +2
10 y 10 y
8 8
6 6
4 4
2 2
x x
−8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8
−2 −2
−4 −4
−6 −6
−8 −8
−10 −10

1.2 Inverse Functions

These exercises examine the invertibility of a function defined using a table.

Exercises
1. The table below defines the function m. Is m an invertible function? Why
or why not? If your answer is yes, construct a table-of-values for m−1 .

x 1 2 3 4 5
m(x) 0 5 10 15 20
2. The table below defines the function p. Is p an invertible function? Why
or why not? If your answer is yes, construct a table-of-values for p−1 .

x 1 2 3 4 5
p(x) 4 0 −2 0 2

1.3 Exponential Functions

These exercises find the formula for an exponential funciton given a pair of
input-output coordinates.

Exercises

Find the Formula. In Exercises 1–6, find an algebraic rule for an exponential
function f that passes through the given two points.
2. (0, 4) and 4, 41


1. (0, 50) and (3, 400)
−1, 23 and (2, 18) −2, 125



3. 4. 8 and (1, 8)
1 27 4




5. (−2, 125) and 3, 25 6. −3, 16 and 3, 27

1.4 Logarithmic Functions

Example 1.4.1 The graph of f (x) = loga (x) is given in the graph below. Find
the value of a. Note, the points (1, 0) and (9, 2) are on the graph of f .
CHAPTER 1. MTH 111 SUPPLEMENT 4

y
3

1
x
−1 1 2 3 4 5 6 7 8 9 10 11 12 13
−1

−2

−3

Solution. Since the function has the form f (x) = loga (x) and (9, 2) is on the
graph, we know that f (9) = 2. Thus,

f (9) = 2 =⇒ loga (9) = 2 (since f (9) = loga (9))

=⇒ a2 = 9 (translate to an exponential statement)
=⇒ a = 3 (positive square root because bases are positive)

Notice that we didn’t attempt to use (1, 0), the other obvious point on the
graph of f (x) = loga (x), to find the value of a. Why not? The point (1, 0) is
on the graph of all functions of the form f (x) = loga (x), so it doesn’t provide
information that will help us find the paerticular function graphed here. □

Exercises
1. The graph of f (x) = loga (x) is given below. Find the value of a. Note,
the points (1, 0) and (25, 4) are on the graph of f .
y

x
5 10 15 20 25

−2

−4

Find the Base. In Exercises 2–3, each table represents a table-of-values for
a function f (x) = loga (x). Find the value of a.
2.
√
x 0.000125 0.05 1 2 5 400
f (x) −3 −1 0 0.5 2
3.
CHAPTER 1. MTH 111 SUPPLEMENT 5

1
x 9 1 3 81 243
f (x) −4 0 2 8 10
Chapter 2

MTH 112 Supplement

2.1 Angles
2.1.1 Coterminal Angles
Definition 2.1.1 Two angles are coterminal if they have the same terminal
side when in standard position. ♢
◦
Since 360 represents a complete revolution, if we add integer-multiples of
360◦ to an angle measured in degrees, we’ll obtain a coterminal angle. Similarly,
since 2π represents a complete revolution in radians, if we add integer-multiples
of 2π to an angle measured in radians, we’ll obtain a coterminal angle. We can
summarize this information as follows
If θ is measured in degrees, θ and θ + 360◦ · k, where k ∈ Z, are coterminal.
If θ is measured in radians, θ and θ + 2π · k, where k ∈ Z, are coterminal.
Example 2.1.2 The angles 45◦ , 405◦ , and −315◦ are coterminal as illustrated
in Figure 2.1.3.
y
−315◦

45◦
x

405◦

Figure 2.1.3 Coterminal angles

□

2.1.2 Reference Angles

Definition 2.1.4 The reference angle for an angle in standard position is
the positive acute angle formed by the x-axis and the terminal side of the angle.
♢
Depending on the location of the angle’s terminal side, we’ll have to use a
different calculation to determine the angle’s reference angle.

6
CHAPTER 2. MTH 112 SUPPLEMENT 7

Example 2.1.5 The angles π3 and 30◦ are their own reference angles since they
are acute angles; seen in Figure 2.1.6 and Figure 2.1.7.
y y

π
3
x 30◦ x

Figure 2.1.6 Figure 2.1.7

□
2π 2π π
Example 2.1.8 The reference angle for 3 is π − 3 = 3
(see Figure 2.1.9),
while the reference angle for 150◦ is 180◦ − 150◦ = 30 (see Figure 2.1.10).
◦

y y
2π
3
150◦
π
3
x 30◦ x

Figure 2.1.9 Figure 2.1.10

□
4π 4π π
Example 2.1.11 The reference angle for 3 is − π = (see Figure 2.1.12),
3 3
while the reference angle for 210◦ is 210◦ − 180 = 30◦ (see Figure 2.1.13).
◦

y y
4π
3 210◦
x x
◦
30
π
3

Figure 2.1.12 Figure 2.1.13

□
5π 5π π
Example 2.1.14 The reference angle for 3 is 2π − 3 = 3
(see Figure 2.1.15),
while the reference angle for 330◦ is 360◦ − 330◦ = 30 (see Figure 2.1.16).
◦
CHAPTER 2. MTH 112 SUPPLEMENT 8

y y

5π
3
x 330◦ x
◦
π 30
3

Figure 2.1.15 Figure 2.1.16

□
Example 2.1.17 The reference angle for 7.5 radians is 7.5 − 2π ≈ 1.2 radians
(see Figure 2.1.18), and the reference angle for −137◦ is 180◦ + ( −137◦ ) = 43◦
(see Figure 2.1.19).
y y

7.5 − 2π
x x
◦
43
7.5
−137◦

Figure 2.1.18 Figure 2.1.19

□

2.1.3 Degrees, Minutes, and Seconds

When measuring angles in degrees, fractions of a degree can be represented in
minutes and seconds.
1
Definition 2.1.20 One minute is 60 of a degree, so 60 minutes is equal to
′ ◦
one degree (written 60 = 1 ).
1 1
One second is 60 of a minute, i.e., 3600 of a degree, so 60 seconds is equal to
′′ ′
one minute (written 60 = 1 ) and 3600 seconds is equal to one degree (written
3600′′ = 1◦ ). ♢
◦ ′ ′′
When we write an angle’s measure in the form D M S where (D, M , and
S are real numbers), then the angle’s measure is D degrees plus M minutes
plus S seconds.
Example 2.1.21 Convert 34◦ 15′ 27′′ into decimal form.
Solution.
1◦ 1◦
   
34◦ 15′ 27′′ = 34◦ + 15′ + 27′′
60′ 3600′′
= 34◦ + 0.25◦ + 0.0075◦
= 34.2575◦

□
◦ ◦ ′ ′′
Example 2.1.22 Convert 61.72 into D M S form.
Solution.

61.72◦ = 61◦ + 0.72◦

CHAPTER 2. MTH 112 SUPPLEMENT 9

60′
 
= 61◦ + 0.72◦ ·
1◦
= 61◦ + 43.2′
= 61◦ 43′ + 0.2′
60′′
 
= 61◦ 43′ + 0.2′ ·
1′
= 61◦ 43′ 12′′

2.1.4 Exercises

Coterminal Angles. In Exercises 1–3, find both a positive and negative

angle that is coterminal angle with the following angles.
1. 63◦ 2. π9 3. 13π
8

Reference Angles. In Exercises 4–12, find the reference angle for the
following angles.
4. 120◦ 5. 5π 6. 400◦
4
7. 13π 8. 2 9. 10π
8 11
10. 2000◦ 11. − 9π 12. −100◦
5

Convert to Decimal Form. In Exercises 13–16, convert the angle measure to

decimal form (round your answers to the nearest thousandth when necessary).
13. 243◦ 10′ 14. 3◦ 25′
15. −23◦ 3′ 16. 75◦ 32′ 17′′

Convert to D◦ M ′ S ′′ Form. In Exercises 17–20, convert the angle measure

to D◦ M ′ S ′′ form.
17. 12.4◦ 18. 1.53◦
◦
19. −144.9 20. 0.416◦

2.2 Generalized Definitions of Trigonometric Func-

tions
We can generalize the definitions of the trigonometric functions so that they
are applicable to circles of any size.
Definition 2.2.1 If the point P = (x, y) is specified by the angle θ on the
circumeference of a circle of radius r, then cos(θ) = xr and sin(θ) = yr .
CHAPTER 2. MTH 112 SUPPLEMENT 10

y
P = (x, y)

r
θ x

Figure 2.2.2
♢
Notice that if we are on a unit circle, where r = 1, then these definitions for
cos(θ) and sin(θ) simplifly accordingly:
x x
cos(θ) = = =x
r 1
y y
sin(θ) = = = y
r 1
We can use Definition 2.2.1 to express the other four trigonometric functions
in terms of x, y, and r.

sin(θ) 1
tan(θ) = csc(θ) =
cos(θ) sin(θ)
y/r 1
= =
x/r y/r
y r
= =
x y

1 1
sec(θ) = cot(θ) =
cos(θ) tan(θ)
1 1
= =
x/r y/x
r x
= =
x y
We summarize this in the following definition.
Definition 2.2.3 If the point P = (x, y) is specified by the angle θ on the
circumeference of a circle of radius r, then we can define the six trigonemtric
functions as follows.
x y y
cos(θ) = sin(θ) = tan(θ) =
r r x
r r x
sec(θ) = csc(θ) = cot(θ) =
x y y
CHAPTER 2. MTH 112 SUPPLEMENT 11

y
P = (x, y)

r
θ x

Figure 2.2.4
♢
Example 2.2.5 Find the exact value of each of the six trigonometric functions
of an angle θ if (−1, 2) is a point on its terminal side.
y

(−1, 2)

r θ
x

Figure 2.2.6
Solution. We need the values of x, y, and r to determine the exact value of
each of the trigonemtric function. We are given x and y, but will need to find
the value of r.
We can think of r as the hypotenuse of a right triangle whose
horiztonal leg has a length of | − 1| = 1 unit and vertical leg
has a length of 2 units. We can use the Pythagorean Theorem
to solve for r: r
2
r2 = 12 + 22
r2 = 5 1
√
r= 5
Now that we have values for x, y, and r, we can use Definition 2.2.3 to state
the exact values of each trigonmetric function.
CHAPTER 2. MTH 112 SUPPLEMENT 12

x y
cos(θ) = sin(θ) =
r r
−1 2
=√ =√
5 5
√ √
−1 5 2 5
=√ ·√ =√ ·√
5 5 5 5
√ √
5 2 5
=− =
5 5

y r
tan(θ) = sec(θ) =
x x
√
2 5
= =
−1 −1
= −2 √
=− 5

r x
csc(θ) = cot(θ) =
y y
√
5 −1
= =
2 2
1
=−
2
□
We can use Definition 2.2.1 to express the coordinates of point P in Fig-
ure 2.2.2. We do this by solving the equations cos(θ) = xr and sin(θ) = yr for x
and y, respectively:
x
=⇒ x = r cos(θ)
cos(θ) =
r
y
sin(θ) = =⇒ y = r sin(θ)
r
We summarize this below.
Definition 2.2.7 If the point P = (x, y) is specified by the angle θ on the
circumeference of a circle of radius r, then x = r cos(θ) and y = r sin(θ).
y
P = (r cos(θ), r sin(θ))

r
θ x

Figure 2.2.8
♢
Example 2.2.9 Find the coordinates of the point on a circle with a radius of 8
units corresponding to an angle of 7π
4 .
CHAPTER 2. MTH 112 SUPPLEMENT 13

7π
4 x

Figure 2.2.10
Solution. We are given the values of r and θ, so we can use Definition 2.2.7
to determine the coordinate values.
Find x:

x = r cos(θ)
 
7π
= 8 cos
4
√
2
=8·
√ 2
=4 2

Find y:

y = r sin(θ)
 
7π
= 8 sin
4
 √ 
2
=8 −
2
√
= −4 2

So the coordinates of the point

√ on a √circle

with a radius of 8 units corre-
sponding to an angle of 7π
4 are 4 2, −4 2 . □

Exercises

Six Trigonemtric Function Values. In Exercises 1–2, find the exact value
of each of the six trigonometric functions of an angle θ if the given point is on
its terminal side.
1. (3, 4) 2. (−2, −6)

Find the coordinates. In Exercises 3–4, find the coordinates of the point
on a circle with the given radius corresponding to the given angle.
3. r = 3, θ = 7π 6
4. r = 10, θ = π3
CHAPTER 2. MTH 112 SUPPLEMENT 14

2.3 Graphing Sinusoidal Functions: Phase Shift

vs. Horizontal Shift
Let’s consider the function g(x) = sin 2x − 2π


3 . Using what we study in
MTH 111 about graph
transformations, it should be apparent that the graph of
g(x) = sin 2x − 2π 3 can be obtained by transforming the graph of g(x) = sin(x).

that g(x) can be expressed in2πterms of f (x) = sin(x),
(To confirm this, notice
as g(x) = f 2x − 2π 3 .) Since the constants “2” and “ 3 ” are multiplied by and
subtracted from the input variable, x, what we study in MTH 111 tells us that
these constants represent a horizontal stretch/compression and a horizontal
shift, respectively.
It is often recommended in MTH 111 that we factor-out the horizontal
stretching/compressing factor before transforming the
graph, i.e., it’s often πrec-
ommended that we first re-write g(x) = sin 2x − 2π



3 as g(x) = sin 2 x − 3 .
After writing g in this format, we can draw its graph by performing the
following sequence of transformations of the “base function” f (x) = sin(x):
1. Compress horizontally by a factor of 21 .
π
2. Shift 3 units to the right.

The advantage of this method is that the y-intercept of f (x) = sin(x), (0, 0),
ends-up exactly where the horizontal shift suggests: when we compress the
graph by a factor of 12 , the the y-interceot of the graph doesn’t move since
1
2 · 0 = 0; then, when we shift the graph π3 units to the right, the point (0, 0)
ends up at 3 , 0 ; so the y-intercept ends up moving π3 units to the right,
π

exactly how far we shifted.

Compare this with the alternative method: we can leave g(x) = sin 2x − 2π


3
as-is and skip factoring-out the horizontal stretching/compressing factor, but
then we need the following sequence to transform f (x) = sin(x) into the graph
of g:
2π
1. Shift 3 units to the right.

2. Compress horizontally by a factor of 12 .

The disadvantage of this method is that the y-intercept of f (x) = sin(x)
doesn’t end-up where the horizontal shift suggests: when we shift the graph
2π
of f (x)
= sin(x) to the right by 3 units, the y-intercept 1moves from (0, π0) to
2π


3 , 0 ; then, when we compress the graph by a factor of 2 , it moves to 3 , 0 ,
so the y-intercept doesn’t end up shifted 2π3 units to the right.
Figure 2.3.1 shows the graphs of y = f (x) and y = g(x). Notice that the
behavior of y = g(x) at x = π3 is like the behavior of y = f (x) at x = 0, i.e.,
y = g(x) appears to have been shifted π3 units to the

right. For this

reason, 3
π
2π π
is called the horzontal shift of g(x) = sin 2x − 3 = sin 2 x − 3 .
CHAPTER 2. MTH 112 SUPPLEMENT 15

y
y = f (x)
y = g(x)
1

x
− 2π − π3 π 2π π 4π 5π 2π 7π
3 3 3 3 3 3

−1

Figure 2.3.1 y = g(x) with f (x) = sin(x)

The constant 2π 3 is given a different name, phase shift, since it can be used
to determine how far “out-of-phase” a sinusoidal function is in comparison
with y = sin(x) or y = cos(x). To determine how far out-of-phase a sinusoidal
function is, we can determine the ratio of the phase shift and 2π. (We use 2π
because it’s the period of y = sin(x) and y = cos(x).) Since 2π 3 is the phase
shift for g(x) = sin 2x − 3 , the graph of y = g(x) is out-of-phase 2π/3
2π 1


2π = 3
of a period. (Since this number is positive, it represents a horizontal shift to
the right 13 of a period.)
Definition 2.3.2 Given a sinusoidal function of the form y = A sin(wx − C) + k
or y = A cos(wx − C) + k, the phase shift is C and |C| 2π represents the fraction
of a period that the graph has been shifted (shift to the right if C is positive or
to the left if C is negative). ♢
Definition 2.3.3 If we re-write the function as y = A sin w x − C



w + k or
y = A cos w x − C C



w + k, we can see that the horizontal shift is w units
(shift to the right if C
w is positive or to the left if C
w is negative). ♢
Example
2.3.4 Identify the phase shift and horizontal shift of g(x) =
cos 3x − π4 .
Solution. The phase shift of g(x) = cos 3x − π4 is π4 . This tells us that the

graph of y = g(x) is out of phase |π/4| 1

2π = 8 of a period, i.e., compared with
π
y = cos(x), the graph of g(x) = cos(3x − 4 ) has been shifted one-eighth of a
period to the right.
To find the horizontal shift, we need to factor-out 3 from 3x − π4 .
 π
g(x) = cos 3x −
  4 π 
= cos 3 x −
3 · 
4
  π
= cos 3 x −
12
π
So the horizontal shift is 12 . This tells us, that compared with y = cos(x),
the graph of g(x) = cos(3x − π4 ) has been shifted 12 π
to the right.
π
Notice that the period of g(x) = cos(3x − 4 ) is 2π · 13 = 2π 3 , and one-eighth
π
of 2π
3 is 2π 1
3 · 8 = 12 , so a shift of one-eighth of a period is the same as a shift of
π
12 units! □
Example 2.3.5 Draw a graph q(t) = 2 sin(4t + π) + 1. First, find it’s amplitude,
period, midline, phase shift, and horiontal shift.
CHAPTER 2. MTH 112 SUPPLEMENT 16

Solution. Amplitude: |A| = |2| = 2

1 π
Period: P = 2π · |w| = 2π
4 = 2
Midline: y = 1
|−π| 1
Phase shift: −π (this tells is that the graph is out-of-phase 2π = 2 of a
period)
Horizontal shift: π4 units to the left since:

q(t) = 2 sin(4t + π) + 1
  π 
= 2 sin 4 t + +1
  4 π 
= 2 sin 4 t − − +1
4
Now we can draw a graph of q(t) = 2 sin(4t + π) + 1 by drawing a sinusoidal
function with the necessary features; see Figure 2.3.6.

4 y

1
t
− π4 π π 3π π
4 2 4
−1

−2

Figure 2.3.6 y = q(t)

□

Exercises

Sketch Sinusoidal Graphs. In Exercises 1–4, draw a graph of each of

the following functions. List the amplitude, midline, period, phase shift, and
horizontal shift.
1. f (x) = 3 sin 3x − π2


2. g(t) = cos(4t + π) + 3
3. m(θ) = 2 cos(2πθ − π) + 4 4. n(x) = −4 sin πx + π4 − 2

Find the Formula. In Exercises 5–6, find two algebraic rules (one involving
sine and one involving cosine) for each of the functions graphed below.
5. y = p(t) 6. y = q(x)
2 y y
2
t
π x
− π2 − π4 π
4
π
2
3π 5π
4 4
−2 −2 −1 1 2 3

−2
−4

−6 −4
CHAPTER 2. MTH 112 SUPPLEMENT 17

2.4 Complex Numbers and Polar Coordinates

2.4.1 Forms of Complex Numbers
√
Recall that a complex number has the form a + bi where a, b ∈ R and i = −1.
Complex numbers have two parts: a real part and an imaginary part. For
the number a + bi, the real part is a and the imaginary part is b. Because
they have two parts, we can use the two dimensional rectangular coordinate
plane to represent complex numbers. We use the horizontal axis to represent
the real part and the vertical axis to represent the complex part. Thus, the
complex number a + bi can be represented by the point (a, b) on the rectangular
coordinate plane; see Figure 2.4.1.

a + bi
b

x
a

Figure 2.4.1
As we’ve studied in this course, the rectangular ordered pair (a, b) can be
represented in polar coordinates (r, θ) where r represents the distance the point
is from the origin and θ represents the angle between the positive x-axis and
the segment connecting the origin and the point; see Figure 2.4.2.

a + bi
b
r
θ x
a

Figure 2.4.2
We know that if the rectangular pair (a, b) represents the same point as the
polar pair (r, θ), then a = r cos(θ) and b = r sin(θ). Thus,
a + bi = r cos(θ) + r sin(θ) · i
= r(cos(θ) + i · sin(θ))
i.e., we can express a complex number using the "polar information" r and θ.
The expression “r(cos(θ) + i · sin(θ))” is what a textbook might decribe as
the “polar form of a complex number.” But a more appropriate expression
to label as “the polar form of a complex number” involves Euler’s Formula.
Euler’s Formula is an identity that establishes a surprising connection between
the exponential function ex and complex numbers.
CHAPTER 2. MTH 112 SUPPLEMENT 18

This is Euler’s Formula. For reference purposes, we state this in a theorem.

Theorem 2.4.3 Euler’s Formula.

eiθ = cos(θ) + i · sin(θ)

Notice that if we multiply both sides of Euler’s formula by r, we obtain a
formula that allows us to write any complex number in polar form:
eiθ = cos(θ) + i · sin(θ)
=⇒ r · eiθ = r · (cos(θ) + i · sin(θ))
=⇒ r(eiθ ) = r cos(θ) + r sin(θ) · i
Definition 2.4.4 The polar form of the complex number z = r cos(θ) +
r sin(θ) · i is z = reiθ . ♢
Let’s review what we’ve established: First, we observed that we can write a
complex number of the form “a + bi” in the form “r · (cos(θ) + i · sin(θ))”. Then
we noticed that we can write an expression of the form “r · (cos(θ) + i · sin(θ))”
in the form “reiθ ”. Finally, we realized that we can write a complex number
“a + bi” in the form “reiθ ” so we defined “reiθ ” as being the polar form of the
complex number a + bi.
Example 2.4.5 Express in "rectangular form" (i.e. in the form z = a + bi) the
5π
complex number z = 6e 6 ·i , given in polar form.
Solution.
5π
z = 6e 6 ·i
   
5π 5π
= 6 cos + 6 sin ·i
6 6
 √   
3 1
=6 − +6 ·i
2 2
√
= −3 3 + 3i
5π
6 ·i
√ complex number z = 6e
Thus, the can be expressed in "rectangular form"
as z = −3 3 + 3i. □
Example 2.4.6 Express in polar form (i.e. in the form z = reiθ ) the complex
number z = 3 − 3i, given in "rectangular form."
Solution. We can associate the complex number z = 3 − 3i with the rect-
angular ordered pair (3, −3), and then translate this ordered pair into polar
coordinates (r, θ), and finally use the polar ordered pair to obtain the polar
form z = reiθ . First, let’s find r:
p
r = (3)2 + (−3)2
√
= 9+9
√
= 3 2.
Now, let’s find θ:
−3
tan(θ) =
3
=⇒ θ = tan−1 (−1)
π
=⇒ θ = −
4
Thus,
√ the complex number z = 3 − 3i can be expressed in polar form as
π
z = 3 2e− 4 ·i . □
CHAPTER 2. MTH 112 SUPPLEMENT 19

2.4.2 Using the Polar Form to Find Complex Roots

√
Example√ 2.4.7 Find the two square roots of −1 + 3i using the polar form of
−1 + 3i.
Solution. Recall that there are two distinct square roots of any positive real
number (e.g., the two square roots of 4 are 2 and −2). The same is true for any
complex number. We can find two different square roots of a complex number
by using two difference polar forms
√ of the number.
To find polar forms of −1√+
3i, we first associate the number with the
rectangular ordered pair −1, 3 , and then translate it into polar coordinates
(r, θ). First let’s find r:
r √ 2
2
r = (−1) + 3
√
= 1+3
= 2.
√
Now, let’s find θ. tan(θ) = − 3 with θ in Quadrant II.
Both θ = 2π 4π
3 and θ = − 3 satisfy the condition, so we’ll use these two
√
angles to obtain two polar forms of −1 + 3i:
√ 2π √ 4π
−1 + 3i = 2e 3 ·i and − 1 + 3i = 2e− 3 ·i

Therefore,
 √ 1/2 2π
−1 + 3i = (2e 3 ·i )1/2
1 2π 1
= 22 e 3 ·2i
√ π
= 2e 3 ·i
√  π  π 
= 2 cos + i · sin
3√ 3
√ 1
 
3
= 2 + i
2 2
√ √
2 6
= + i
2 2
and
 √ 1/2  − 4π ·i 1/2
−1 + 3i = 2e 3
1 4π 1
= 2 2 e− 3 · 2 i
√ 2π
= 2e− 3 ·i
√
    
2π 2π
= 2 cos − + i · sin −
3 3
√
√
 
1 3
= 2 − − i
2 2
√ √
2 6
=− − i.
2 2
√ √ √ √ √
So both 22 + 26 i and − 22 − 26 i are square roots of −1 + 3i. But just as
2, not −2, is called the principal square root of 4, only one√of thes two square
roots that we found is the principal square root of −1 + 3i. The principal
CHAPTER 2. MTH 112 SUPPLEMENT 20

square root (or principal nth root) of a complex number is the root with the
greatest real component. And if there is a tie between two roots for having
the greatest real component, the one with positive imaginary component is the
principal root. So the first root we found (i.e., the one we found using θ = 2π
3 )
√ √
is the principal square root of −1 + 3i, because its real part is 22 which
√
is greater than the other root’s real part, − 22 . The princial root is the one
represented by the radical symbol, so we can write
√ √
√
q
2 6
−1 + 3i = + i.
2 2
□
p3
√ √ √ √
Example 2.4.8 Find −4 2 + 4 2i using the polar form of −4 2 + 4 2i.
√ √
Solution. To find polar forms of −4√ 2 +√4
2i, we first associate the number
with the rectangular ordered pair −4 2, 4 2 , and then translate it into polar
coordinates (r, θ). First let’s find r:
r
√ 2  √ 2
r= −4 2 + 4 2
p
= 42 · 2 + 4 2 · 2
√
=4 2+2
=8

Now, let’s find θ:

√
4 2
tan(θ) = √
−4 2
 √ √ 
=⇒ θ = tan−1 (−1) + π (add π since −4 2, 4 2 is in Quadrant II)
π
=⇒ θ = − + π
4
3π
=⇒ θ =
4
√ √

Note: we add π since −4 2, 4 2 is in Quadrant II, outside the range of
arctangent. √ √ 3π
So the polar form of −4 2 + 4 2i is z = 8e 4 ·i . Therefore,
√ √
q  3π 1/3
3
−4 2 + 4 2i = 8e 4 ·i
1 3π 1
= 83 e 4 ·3i

π
= 2e 4 ·i
 π  π 
= 2 cos + i · sin
4 √ 4
√ 
2 2
=2 + i
2 2
√ √
= 2 + 2i

□
CHAPTER 2. MTH 112 SUPPLEMENT 21

2.4.3 Exercises

Polar Form. In Exercises 1–3, find the polar form z = reiθ of the following
complex numbers given in rectangular form.
√ √
1. z = 6 + 6 3 · i 2. z = −2 3 + 2i
√ √
3. z = 5 2 − 5 2 · i

Rectangular Form. In Exercises 4–6, find the rectangular form z = a + bi

of the following complex numbers given in polar form.
π
4. z = 8e 6 ·i 5. z = 4eπ·i
4π
6. z = 5e 3 ·i

Principal Roots. In Exercises 7–10, find the following principal roots by

first converting to the polar form of each complex number.
p √ √3
7. 18 − 18 3 · i 8. −16 + 16i
√ p √
9. −i 10.
5
−16 3 − 16i
11. Find all three cube roots of 27i.
√
12. Find both of the square roots of − 12 + 3
2 i.
13. Find all three solutions to the equation of z 3 + 1 = 0.
Appendix A

Answers and Solutions to Ex-

ercises

1 · MTH 111 Supplement

1.1 · Graph Transformations
· Exercises

One Function in Terms of Another.

1.1.1. Answer. g(x) = f (−x). So, we can reflect the graph of y = f (x)
across the y-axis to obtain y = g(x).
1.1.2. Answer. h(x) = −f (x). So, we can reflect the graph of y = f (x)
across the x-axis to obtain y = h(x).
1.1.3. Answer. k(x) = f (x) + 6. So, we can shift the graph of y = f (x)
up 6 units to obtain y = k(x).
1.1.4. Answer. l(x) = 3f (x). So, we can stretch the graph of y = f (x)
vertically by a factor of 3 to obtain y = l(x).
1.1.5. Answer.
x −4 −3 −2 −1 0 1 2 3 4
f (x) −2 −1 0 1 2 3 4 5 6
1
2x −1 − 12 0 1
2 1 3
2 2 5
2 3
−2f (x) 4 2 0 −2 −4 −6 −8 −10 −12
f (x) + 5 3 4 5 6 7 8 9 10 11
f (x + 2) 0 1 2 3 4 5 6
f 12 x


0 1 2 3 4
f (2x) −2 0 2 4 6
f (x − 3) −2 −1 0 1 2 3

Find the Transformations.

1.1.6. Solution.
√
x−7
g(x) =
4
1√
= x−7
4
1
= f (x − 7)
4

22
APPENDIX A. ANSWERS AND SOLUTIONS TO EXERCISES 23

So we can transform y = f (x) into y = g(x) by first shifting right 7

units and then compressing vertically by a factor of 14 . (There are other
correct answers.)
1.1.7. Solution.
2
g(x) = +3
x
1
=2· +3
x
= 2f (x) + 3

So we can transform y = f (x) into y = g(x) by first stretching vertically

by a factor of 2 and then shifting up 3 units. (There are other correct
answers.)
1.1.8. Solution.
 2
1
g(x) = −4 x−5 +3
2
 
1
= −4f x−5 +3
2
 
1
= −4f (x − 10) + 3
2

So we can transform y = f (x) into y = g(x) by first stretching horizon-

tally by a factor of 2 and then shifting right 10 units. Then, stretching
vertically by a factor of 4 and reflecting across the x-axis, and finally shifting
up 3 units. (There are other correct answers.)
1.1.9. Solution.
1√3
g(x) = 10x + 30 − 6
2
1
= f (10x + 30) − 6
2
1
= f (10(x + 3)) − 6
2
So we can transform y = f (x) into y = g(x) by first compressing hori-
1
zontally by a factor of 10 and then shifting left 3 units. Then, compressing
vertically by a factor of 12 and finally shifting down 6 units. (There are
other correct answers.)

Sketch Transformations.
1.1.10. Answer. 1.1.11. Answer.
10 y 10 y
8 8
6 6
4 4
2 2
x x
−8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8
−2 −2
−4 −4
−6 −6
−8 −8
−10 −10
APPENDIX A. ANSWERS AND SOLUTIONS TO EXERCISES 24

1.1.12. Answer. 1.1.13. Answer.

10 y 10 y
8 8
6 6
4 4
2 2
x x
−8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8
−2 −2
−4 −4
−6 −6
−8 −8
−10 −10

1.2 · Inverse Functions

· Exercises
1.2.1. Solution. m is an invertible function since it is one-to-one, i.e., each
output corresponds to exactly one input. Here is a table-of-values for m−1 .

x 0 5 10 15 20
m−1 (x) 1 2 3 4 5
1.2.2. Solution. p isn’t an invertible function since it isn’t one-to-one. Notice
how the output 0 corresponds to two distinct output values.

1.3 · Exponential Functions

· Exercises

Find the Formula.

1.3.1. Answer. f (x) = 50 · 2x 1 x


1.3.2. Answer. f (x) = 4 · 2
1.3.3. Answer. f (x) = 2 · 3x 4 x


1.3.4. Answer. f (x) = 10 · 5
1 x 1 2 x



1.3.5. Answer. f (x) = 5 · 5 1.3.6. Answer. f (x) = 2 · 3

1.4 · Logarithmic Functions

· Exercises
√
1.4.1. Answer. a = 5

Find the Base.

1.4.2. Answer. a = 20
√
1.4.3. Answer. a = 3

2 · MTH 112 Supplement

2.1 · Angles
2.1.4 · Exercises
Coterminal Angles.
2.1.4.1. 2.1.4.2. 2.1.4.3.
Answer. 423◦ and Answer. 19π 9 and Answer. 29π 8 and
−297◦ are coterminal − 17π
9 are coterminal − 3π
8 are coterminal
with 63◦ . with π9 . with 13π
8 .
APPENDIX A. ANSWERS AND SOLUTIONS TO EXERCISES 25

Reference Angles.
2.1.4.4. 2.1.4.5. Answer. π4 2.1.4.6.
◦
Answer. 60 Answer. 40◦
2.1.4.7. 2.1.4.8. 2.1.4.9.
π
Answer. 3π 8 Answer. π − 2 ≈ Answer. 11
1.14
2.1.4.10. 2.1.4.11. 2.1.4.12.
Answer. 20◦ Answer. π5 Answer. 80◦

Convert to Decimal Form.

2.1.4.13. 2.1.4.14.
Answer. 243◦ 10′ ≈ 243.167◦ Answer. 3◦ 25′ ≈ 3.417◦
2.1.4.15. 2.1.4.16.
Answer. −23◦ 3′ = −23.05◦ Answer. 75◦ 32′ 17′′ ≈ 75.538◦

Convert to D◦ M ′ S ′′ Form.
2.1.4.17. 2.1.4.18.
Answer. 12.4◦ = 12◦ 24′ Answer. 1.53◦ = 1◦ 31′ 48′′
2.1.4.19. 2.1.4.20.
Answer. −144.9◦ = −144◦ 54′ Answer. 0.416◦ = 0◦ 24′ 57.6′′

2.2 · Generalized Definitions of Trigonometric

Functions
· Exercises

Six Trigonemtric Function Values.

2.2.1. Answer. 2.2.2. Answer.
√
3 10
cos(θ) = sin(θ) = −
5 10
4 √
sin(θ) = 3 10
5 cos(θ) = −
4 10
tan(θ) = tan(θ) = 3
3 √
5 sec(θ) = − 10
sec(θ) = √
3 10
5 csc(θ) = −
csc(θ) = 3
4 1
3 cot(θ) =
cot(θ) = 3
4

Find the coordinates.

 √  √

2.2.3. Answer. − 3 2 3 , − 23 2.2.4. Answer. 5, 5 3

2.3 · Graphing Sinusoidal Functions: Phase Shift

vs. Horizontal Shift
· Exercises
APPENDIX A. ANSWERS AND SOLUTIONS TO EXERCISES 26

Sketch Sinusoidal Graphs.

2.3.1. Answer. 2.3.2. Answer.

Amplitude Amplitude
3 units 1 unit
2π π
Period 3 units
Period 2 units

Midline y = 0 Midline y = 3
Phase shift Phase shift
π −π
2 Horizontal Shift
π
Horizontal Shift 4 units to the left
π
6 units to the right

4 y
y
3 3

2
x 1
− π3 − π6 π π π 2π 5π π t
6 3 2 3 6
− π2 − π4 π π 3π π 5π
4 2 4 4

−3

2.3.3. Answer. 2.3.4. Answer.

Amplitude Amplitude
2 units 4 units
Period 1 unit Period 2 units
Midline y = 4 Midline y = −2
Phase shift Phase shift
π π
−
Horizontal Shift 4
1
2 of a unit to the right Horizontal Shift
1
4 of a unit to the left

6
y
y
4 2
x
2 −1 1 2 3
θ −2

−1 1 2 −4
−2
−6

Find the Formula.

2.3.5. An- 2.3.6. An-
  π   
1

p(x) = 4 sin 2 x − −2 q(x) = 3 sin π x + −1
swer. 4  4
  π swer.
p(x) = 4 cos 2 x − −2
  
2 1
q(x) = 3 cos π x − −1
4
APPENDIX A. ANSWERS AND SOLUTIONS TO EXERCISES 27

2.4 · Complex Numbers and Polar Coordinates

2.4.3 · Exercises
Polar Form.
π
2.4.3.1. Answer. z = 12e 3 ·i
5π
2.4.3.2. Answer. z = 4e 6 ·i
π
2.4.3.3. Answer. z = 10e− 4 ·i

Rectangular Form.
√
2.4.3.4. Answer. z = 4 3 + 4i 2.4.3.5. Answer. z = −4
2.4.3.6. √
Answer. z = − 52 − 5 2 3 · i

Principal Roots.
2.4.3.7. p √ 2.4.3.8. √
Answer.
√ 18 − 18 3 · i = Answer. 3 −16 + 16i = 2 + 2i
3 3 − 3i (and
√ the non-principal (and the
√ non-principal

√ roots

are
root is −3 3 + 3i) −1 − 3 + −1 + 3 i and
√
√

−1 + 3 + −1 − 3 i)
2.4.3.9. √ √ √ 2.4.3.10. p √
5
Answer. −i = 22 − 22 i (and Answer.
√ −16 3 − 16i =
the√ non-principal
√
root is 3 − i (and
the non-principal roots
are 2 cos
7π 7π


2 2 + 2i sin ,
− 2 + 2 i) 30 30
2 cos 19π 19π


30
+ 2i sin 30
,
2 cos 29π 29π
30
− 2i sin 30
, and
17π
2 cos 30 − 2i sin 17π 30 )
√ √
3 3
2.4.3.11. Answer. 2 + 32 i, −3i, and − 3 2 3 + 32 i
√ √
1 3
2.4.3.12. Answer. 2 + 2 i and − 12 − 3
2 i
n √ √ o
2.4.3.13. Answer. −1, 12 + 3 1
2 i, 2 − 3
2 i