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606 • Complex Numbers
respectively. Putting these values of θ, along with r = 2
1/5
, into the equation
z = re
iθ
, we get
z = 2
1/5
e
iπ/6
, 2
1/5
e
i(17π/30)
, 2
1/5
e
i(29π/30)
, 2
1/5
e
i(41π/30)
,
or 2
1/5
e
i(53π/30)
.
Of course, it would be nice to change these to Cartesian form. The first
solution is pretty easy:
2
1/5
e
iπ/6
= 2
1/5
cos
π
6
+ i sin
π
6
= 2
1/5
√
3
2
+ i
1
2
!
= 2
−4/5
(
√
3 + i).
As for the others, they don’t look too nice. For example, the second solution
from the above list works out to be
2
1/5
e
i(17π/30)
= 2
1/5
cos
17π
30
+ i sin
17π
30
,
which can’t easily be simplified. (Do you know what cos(17π/30) is? I don’t
either, and it’s not worth working out.) I leave it to you to write out the other
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
three solutions in (unsimplified) Cartesian form.
Now, let’s see why you only need to let k go from 0 through 4, discarding
all the other possible values of k. Let’s see what happens when k = 5. Using
the equation
θ =
π
6
+
2πk
5
from above, we see that when k = 5, we have
θ =
π
6
+
2π × 5
5
=
π
6
+ 2π.
This is certainly a different value of θ from any of the ones we already listed
above, but it doesn’t lead to a different value of z. Why? Because
2
1/5
e
i(π/6+2π)
= 2
1/5
e
i(π/6)
.
That is, we get the same solution as the case k = 0. Similarly, if you try to
put k = 6, you should get the same value of z as when k = 1. In general,
any time you increase k by 5, you will simply get the same value of z again.
So, the values k = 0, 5, 10, . . . , as well as k = −5, −10, −15, . . ., all lead to
the same solution, z = 2
1/5
e
i(π/6)
. Similarly, the values k = 1, 6, 11, . . . and
k = −4, −9, −14, . . . give the same solution. The same goes for the other
three solutions. While you need to appreciate this fact, in practice it is simple
to apply: unless w = 0, the equation z
n
= w has n different solutions, which
occur when k = 0, 1, . . . , n − 1. Those are the only values of k you need to
use. In our case n = 5, so we only needed k = 0, 1, 2, 3, 4.
It’s interesting to plot the solutions in the complex plane. They all have
modulus 2
1/5
, which means that they lie on the circle centered at the origin
of radius 2
1/5
units. Also, the difference between the arguments (that is,
values of θ) of consecutive solutions is 2π/5, which is one-fifth of a complete
revolution. This means that the solutions are evenly spaced around the circle;
that is, they form a regular pentagon (the solutions are labeled z
0
through
z
4
):
Section 28.4: Solving z
n
= w • 607
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f (
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f (
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f(
t)
F (
x )
y = f(
t)
F (
x + h )
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f (
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r =
1 + 2 cos(θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
In general, there are n solutions to the equation z
n
= w, which when plotted
form the vertices of a regular n-sided polygon. (The exception is if w = 0, in
which case z = 0 is the only solution, but it is of multiplicity n.)
So, let’s outline the main steps in solving z
n
= w:
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f (
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f (
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f(
t)
F (
x )
y = f(
t)
F (
x + h )
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f (
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r =
1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
1. Write z = re
iθ
in polar coordinates. Then z
n
= r
n
e
inθ
.
2. Convert w to polar coordinates. Let’s say that w = Re
iϕ
.
3. Since z
n
= w, we can write the original equation as r
n
e
inθ
= Re
iϕ
.
Here, the values of n, R, and ϕ should be filled in with your values, but
r and θ are always what we need to find (so they appear as variables).
4. Decompose into two equations: r
n
= R and e
inθ
= e
iϕ
.
5. The first is simple to solve: take nth roots to get r = R
1/n
.
6. For the second, use the above triple-boxed principle to get nθ = ϕ+2πk,
where k is an integer.
7. Divide this by n, then write out all the different values for θ when
k = 0, 1, 2, . . . , n − 1.
8. Substitute the value of r and the different values of θ into z = re
iθ
to
get n different values for z, which are the solutions.
9. If necessary, change each and every one of those solutions into Cartesian
coordinates.
Let’s look at one more example: what are the cube roots of i? This
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0, −3)
(−2, 0)
π
2
3π
2
π
r = 3 sin(θ)
3π
2
θ
2π
1
0
−1
−2
−3
0
3
2
−
3
2
0
r = 1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
question is asking us to solve the equation z
3
= i. We start off by writing
z = re
iθ
, so z
3
= r
3
e
i(3θ)
(step 1). Now, we have to convert i into polar
coordinates (step 2), but we have already seen above that i = e
iπ/2
. So,
since z
3
= i, we have r
3
e
i(3θ)
= 1e
iπ/2
(step 3). This leads to the equations
r
3
= 1 and e
i(3θ)
= e
iπ/2
(step 4). Taking cube roots in the first equation
608 • Complex Numbers
gives r = 1 (step 5), and our important principle and the second equation
show that 3θ = π/2 + 2πk, where k is an integer (step 6). This is the same
as θ = π/6 + 2πk/3; since n = 3 in this question, we only need k = 0, 1, 2.
Writing these out, we have
θ =
π
6
,
π
6
+
2π
3
=
5π
6
, or
π
6
+
4π
3
=
3π
2
(step 7). This leads to three possibilities for z, which are
z = e
iπ/6
, e
i(5π/6)
or e
i(3π/2)
(step 8). Finally, we should convert these into Cartesian form (step 9). The
first solution is
z = e
iπ/6
= cos
π
6
+ i sin
π
6
=
√
3
2
+ i
1
2
.
The second solution is
z = e
i(5π/6)
= cos
5π
6
+ i sin
5π
6
= −
√
3
2
+ i
1
2
.
Finally, the third solution is
z = e
i(3π/2)
= cos
3π
2
+ i sin
3π
2
= 0 − i(1) = −i.
Let’s plot these three solutions and check that they do indeed form an equi-
lateral triangle:
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f (
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f (
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(
t)
F (
x )
y = f(
t)
F (
x + h )
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f (
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2π
1
0
−1
−2
−3
0
3
2
−
3
2
0
r = 1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
28.4.1 Some variations
Suppose you want to solve the equation (z − 2)
3
= i. No problem—just let
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3π
2
π
r = 3 sin(θ)
3π
2
θ
2π
1
0
−1
−2
−3
0
3
2
−
3
2
0
r = 1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
Z = z −2, so that the equation is Z
3
= i. Solve this exactly as we just did at
the end of the previous section to find that
Z = z − 2 =
√
3
2
+ i
1
2
, −
√
3
2
+ i
1
2
, or − i.
Section 28.4.1: Some variations • 609
Finally, add 2 to both sides to get
z = 2 +
√
3
2
+ i
1
2
, 2 −
√
3
2
+ i
1
2
, or 2 − i.
There’s not much to that, then. Here’s a tougher one. Let’s try solving the
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
quadratic equation
z
2
+
1
√
2
z −
√
3i
8
= 0.
We can use the quadratic formula to get
z =
−1
√
2
±
q
1
2
+ i
√
3
2
2
.
While this is correct, it’s not in Cartesian form (nor polar form) so we should
try to simplify it. We need to find the square roots of the complex number
1
2
+ i
√
3
2
. How do we do this? By solving the equation Z
2
=
1
2
+ i
√
3
2
.
Following the above steps, we write Z = re
iθ
, and I’ll leave it to you to see
that
1
2
+ i
√
3
2
= e
iπ/3
in polar form. So our equation becomes r
2
e
i2θ
= e
iπ/3
.
This means that r
2
= 1 and 2θ = π/3 + 2πk, where k = 0 or 1. (Remember
the important principle!) So we have r = 1 and θ = π/6 or 7π/6, which means
that Z = e
iπ/6
or e
i7π/6
. Again, you should check that these correspond to
Z =
√
3
2
+
1
2
i or −
√
3
2
−
1
2
i in Cartesian form (respectively). Finally, we can
replace ±
q
1
2
+ i
√
3
2
by ±(
√
3
2
+
1
2
i) in the equation for z above to get
z =
−
1
√
2
±
√
3
2
+
1
2
i
2
.
This simplifies to
z = −
1
2
√
2
+
√
3
4
+
i
4
or −
1
2
√
2
−
√
3
4
−
i
4
.
One more example. How would you factor (z
4
−z
2
+ 1) over the complex
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−3
0
3
2
−
3
2
0
r = 1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
numbers? How about over the real numbers? In the first case, we just need
to find all four complex solutions of the equation z
4
−z
2
+ 1 = 0. To do this,
first we need to realize that this equation is actually a quadratic equation in
z
2
. Let’s set Z = z
2
, so that the equation becomes Z
2
−Z + 1 = 0. This can
be solved using the quadratic formula to get
Z = z
2
=
1 ±
√
−3
2
=
1
2
± i
√
3
2
.
So we need to find the square roots of
1
2
+ i
√
3
2
and
1
2
−i
√
3
2
. We just did the
first one in the previous example, and you can repeat the steps easily enough
to handle the second one. Both of these numbers have two square roots each,
which work out to be
√
3 + i
2
,
−
√
3 − i
2
,
−
√
3 + i
2
, and
√
3 − i
2
.
610 • Complex Numbers
These are the four solutions to z
4
− z
2
+ 1 = 0. It follows that we can factor
z
4
− z
2
+ 1 as follows:
z
4
−z
2
+ 1 =
z −
√
3 + i
2
!
z −
√
3 − i
2
!
z −
−
√
3 + i
2
!
z −
−
√
3 − i
2
!
.
This is the complex factorization. To get the real factorization, we need to
use a nice fact: if w is any complex number, then (z − w)(z − ¯w) has real
coefficients when you multiply it out. Indeed, you get z
2
− (w + ¯w)z + w ¯w,
but it’s easy enough to see that w + ¯w = 2Re(w) (which is real), and we’ve
already seen that w ¯w = |w|
2
, which is also real. Anyway, notice that I have
cunningly grouped the above four factors so that if we multiply out the first
two, we get
z −
√
3 + i
2
!
z −
√
3 − i
2
!
= z
2
−
√
3 + i
2
+
√
3 − i
2
!
z +
√
3 + i
2
!
√
3 − i
2
!
= z
2
−
√
3z + 1.
Similarly, you should check that multiplying out the last two factors gives
z
2
+
√
3z + 1. The conclusion is that
z
4
− z
2
+ 1 = (z
2
−
√
3z + 1)(z
2
+
√
3z + 1).
Notice that there are no complex numbers here, yet working this out without
them would have been pretty darn tricky.
28.5 Solving e
z
= w
Now it’s time to see how to solve equations of the form e
z
= w for given w.
It’d be nice if we could just write z = ln(w), but this isn’t very helpful. For
example, what exactly is ln(−
√
3 + i)? Let’s try to answer this question.
Fortunately, solving e
z
= w isn’t much harder than solving z
n
= w; in fact,
if anything, it’s simpler. Before we see how to do this, we need to understand
e
z
a little better. Let’s see what happens if we write z = x + iy. We get
e
z
= e
x+iy
= e
x
e
iy
.
So what? Well, the main point is that this is already in polar form. The
modulus is e
x
and the argument is y. If you prefer, r = e
x
(remember, e
x
is real and positive) and θ = y. This means that if z is in Cartesian form
x + iy, then e
z
is automatically in polar form: e
z
= e
x
e
iy
. So, the main
difference between solving e
z
= w and z
n
= w is that you don’t need to put
z in polar form in the first case, whereas you do in the second case. A sort of
by-product of this is that there are infinitely many solutions to the equation
e
z
= w (unless w = 0, in which case there are no solutions).
Let’s solve e
z
= −
√
3 + i. We have already converted the right-hand side
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0, 1)
(0, −3)
(−2, 0)
π
2
3π
2
π
r = 3 sin(θ)
3π
2
θ
2π
1
0
−1
−2
−3
0
3
2
−
3
2
0
r = 1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
to polar coordinates as 2e
i(5π/6)
(see page 604). To handle the left-hand side,
Section 28.5: Solving e
z
= w • 611
write z = x + iy in Cartesian coordinates, so e
z
= e
x
e
iy
. So, changing the
original equation to polar form, we get
e
x
e
iy
= 2e
i(5π/6)
.
Now, this separates into two equations:
e
x
= 2 and e
iy
= e
i(5π/6)
.
To solve the first equation, we have to take logarithms to see that x = ln(2).
The second is handled by our important principle to get y = 5π/6 + 2πk,
where k is an integer. Finally, putting these values into z = x + iy, we get
z = ln(2) + i
5π
6
+ 2πk
,
where k is any integer. In this case, we do get a different value of z for each
value of k, so we need to use them all. Let’s plot some of the possible values
of z corresponding to k = −2, −1, 0, 1, and 2 (I’ll use a different scale on the
axes for clarity):
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f (
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f (
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(
t)
F (
x )
y = f(
t)
F (
x + h )
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f (
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
So the solutions are equally spaced on the vertical line x = ln(2). Incidentally,
this means that they form an arithmetic progression of complex numbers.
Although only five solutions are shown in the above picture, you should bear in
mind that there are actually infinitely many solutions to the original equation
e
z
= −
√
3 + i.
Let’s look at one more example. Suppose you want to solve e
2iz+3
= i.
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−1
−2
−3
0
3
2
−
3
2
0
r = 1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
The exponent 2iz + 3 makes this a little more complicated than the previous
example, but it’s not too bad. We’ve already seen that the right-hand side in
polar coordinates is e
iπ/2
, but how about the left-hand side? Once again, we
write z = x +iy, but now we need 2iz +3 = 2i(x+ iy) +3 = (−2y + 3)+ i(2x).
So, the polar form of the left-hand side is given by
e
2iz+3
= e
−2y+3
e
i(2x)
.
612 • Complex Numbers
Notice how the factor of i switched the real and imaginary parts (and also
the sign of y). Anyway, translating our equation e
2iz+3
= i into polar form,
we have
e
−2y+3
e
i(2x)
= 1e
iπ/2
.
This leads to the equations
e
−2y+3
= 1 and e
i(2x)
= e
iπ/2
.
To solve the first equation, take logs to get −2y + 3 = ln(1) = 0, so y =
3
2
.
To solve the second equation, use the boxed principle to get 2x = π/2 + 2πk,
where k is an integer. This means that x = π/4 + πk, so since z = x + iy, we
have
z =
π
4
+ πk +
3
2
i,
where k is an integer. Let’s plot what these solutions look like for k = −2,
−1, 0, 1, and 2, bearing in mind that these are only five of the infinitely many
solutions:
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(
t)
F (
x )
y = f(
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
Once again, the solutions are in arithmetic progression, but this time they lie
on the horizontal line y =
3
2
.
28.6 Some Trigonometric Series
A trigonometric series is a series of the form
∞
X
n=0
(a
n
cos(nθ) + b
n
sin(nθ))
for some coefficients {a
n
} and {b
n
}. In this section, we’ll see that there are a
few such series which can be simplified.
For example, consider the trigonometric series
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(θ)
2π
3
4π
3
0
π
0
pi
−
3
2
3π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
∞
X
n=0
sin(nθ)
n!
,
where θ is real. Note that this is not a power series in θ, since sin(nθ) is not
a power of θ. On the other hand, we can make the whole thing into a power
Section 28.6: Some Trigonometric Series • 613
series by using the complementary series
∞
X
n=0
cos(nθ)
n!
,
in a clever way. In fact, we can find both series at once. The key is Euler’s
identity. Watch carefully, because this is a sneaky trick. Let’s find both series
at once by combining them like this:
∞
X
n=0
cos(nθ)
n!
+ i
∞
X
n=0
sin(nθ)
n!
.
OK, so this is one series plus i times the other. So what? Well, by massaging
the sums
∗
and then using Euler’s identity, this simplifies to
∞
X
n=0
cos(nθ) + i sin(nθ)
n!
=
∞
X
n=0
e
inθ
n!
.
Finally, use the exponential rules to write e
inθ
as (e
iθ
)
n
; the sum becomes
∞
X
n=0
(e
iθ
)
n
n!
.
Now, the last sum looks familiar. In fact, we saw in Section 28.1.1 above that
∞
X
n=0
z
n
n!
= e
z
for all complex numbers z. Now we just have to substitute z = e
iθ
to get
∞
X
n=0
(e
iθ
)
n
n!
= e
e
iθ
.
If you’ve been following this chain of reasoning, you should see that we’ve
proved that
∞
X
n=0
cos(nθ)
n!
+ i
∞
X
n=0
sin(nθ)
n!
= e
e
iθ
.
Now what? Well, we need to convert the right-hand side into Cartesian form.
To do this, write e
iθ
= cos(θ) + i sin(θ), so
e
e
iθ
= e
cos(θ)+i sin(θ)
= e
cos(θ)
e
i sin(θ)
.
This is a good start—this is the polar form of e
e
iθ
. To get the Cartesian
form, we need to convert e
i sin(θ)
into cos(sin(θ)) + i sin(sin(θ)). Putting it all
together, we get
∞
X
n=0
cos(nθ)
n!
+ i
∞
X
n=0
sin(nθ)
n!
= e
cos(θ)
cos(sin(θ)) + ie
cos(θ)
sin(sin(θ)).
∗
This needs some justification. It turns out that everything’s OK because both our
series converge absolutely.
614 • Complex Numbers
Now, if two complex numbers are equal, then their real parts must be equal,
and also their imaginary parts must be equal. This leads to the following two
equations, which are valid for all real θ:
∞
X
n=0
cos(nθ)
n!
= e
cos(θ)
cos(sin(θ)) and
∞
X
n=0
sin(nθ)
n!
= e
cos(θ)
sin(sin(θ)).
Not easy, but this is basically what you have to do. I’ll do one more example,
without all the explanations. Your task is to follow this and explain each step.
The example is to find
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r =
1 + 2 cos(θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r =
1 + cos(θ)
r =
1 +
3
4
cos(
θ)
−
1
4
r =
sin(2θ)
r =
sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
∞
X
n=0
cos(nθ)
3
n
and
∞
X
n=0
sin(nθ)
3
n
.
Following the pattern of the above example, we have
∞
X
n=0
cos(nθ)
3
n
+ i
∞
X
n=0
sin(nθ)
3
n
=
∞
X
n=0
cos(nθ) + i sin(nθ)
3
n
=
∞
X
n=0
e
inθ
3
n
=
∞
X
n=0
(e
iθ
)
n
3
n
=
∞
X
n=0
e
iθ
3
n
.
Now this is a geometric series with ratio e
iθ
/3. This last number is in polar
form with modulus 1/3, which is less than 1; so the geometric series should
converge. By the formula for the sum of a geometric series (see Section 23.1
of Chapter 23), we have
∞
X
n=0
e
iθ
3
n
=
1
1 −
1
3
e
iθ
.
We now have the wretched task of converting this into Cartesian coordinates.
First, try it and see if you can do it. If not, at least try to understand the
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r =
1 + 2 cos(θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−1
−2
−3
0 ≤ θ ≤
2π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2π
r = 1 + cos(θ)
r = 1 +
3
4
cos(θ)
−
1
4
r = sin(2θ)
r = sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4π
r =
2
1 + sin(θ)
−
π
4
≤ θ ≤
5π
4
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
following steps:
1
1 −
1
3
e
iθ
=
1
1 −
1
3
cos(θ) − i
1
3
sin(θ)
=
1
1 −
1
3
cos(θ) − i
1
3
sin(θ)
·
1 −
1
3
cos(θ) + i
1
3
sin(θ)
1 −
1
3
cos(θ) + i
1
3
sin(θ)
=
1 −
1
3
cos(θ) + i
1
3
sin(θ)
1 −
1
3
cos(θ)
2
+
1
3
sin(θ)
2
=
1 −
1
3
cos(θ) + i
1
3
sin(θ)
1 −
2
3
cos(θ) +
1
9
cos
2
(θ) +
1
9
sin
2
(θ)
=
1 −
1
3
cos(θ) + i
1
3
sin(θ)
1 −
2
3
cos(θ) +
1
9
=
9 − 3 cos(θ) + i3 sin(θ)
10 − 6 cos(θ)
=
9 − 3 cos(θ)
10 − 6 cos(θ)
+ i
3 sin(θ)
10 − 6 cos(θ)
.
Section 28.7: Euler’s Identity and Power Series • 615
After all this, we’re ready to write
∞
X
n=0
cos(nθ)
3
n
+ i
∞
X
n=0
sin(nθ)
3
n
=
9 − 3 cos(θ)
10 − 6 cos(θ)
+ i
3 sin(θ)
10 − 6 cos(θ)
;
since the real and imaginary parts must be equal, we conclude that
∞
X
n=0
cos(nθ)
3
n
=
9 − 3 cos(θ)
10 − 6 cos(θ)
and
∞
X
n=0
sin(nθ)
3
n
=
3 sin(θ)
10 − 6 cos(θ)
for any real number θ. As you see, these problems are quite hard!
28.7 Euler’s Identity and Power Series
Let’s finish the chapter with a justification of Euler’s identity
e
iθ
= cos(θ) + i sin(θ)
using power series. By the definition of e
z
from Section 28.1.1 above, with z
replaced by iθ, we see that
e
iθ
= 1 + (iθ) +
(iθ)
2
2!
+
(iθ)
3
3!
+
(iθ)
4
4!
+
(iθ)
5
5!
+
(iθ)
6
6!
+
(iθ)
7
7!
+ ···
= 1 + iθ −
θ
2
2!
− i
θ
3
3!
+
θ
4
4!
+ i
θ
5
5!
−
θ
6
6!
− i
θ
7
7!
+ ··· .
Since the powers of i keep cycling through the values 1, i, −1, −i, we conclude
that the even powers in the above series all have real coefficients, whereas
the odd powers all have imaginary coefficients. Furthermore, every second
even-power term is negative and the others are positive; the same is true for
the odd powers. So, the real part of e
iθ
is
1 −
θ
2
2!
+
θ
4
4!
−
θ
6
6!
+ ··· = cos(θ),
and the imaginary part is
θ −
θ
3
3!
+
θ
5
5!
−
θ
7
7!
+ ··· = sin(θ).
(See Section 26.2 in Chapter 26 to refresh your memory about these Maclaurin
series.) From this last equation, it follows that e
iθ
= cos(θ) + i sin(θ).