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Chapter 10
Additional Topics in Trigonometry
RECAP
1
Complex Numbers
2
Defining a Complex Number
• We define i as the number that solves x2 = -1
i =  -1
i2 = -1
For and non-negative real k, sqrt(-k) = i sqrt(k)
i is called an imaginary number
A complex number is the sum of an imaginary and a real number
3
Complex Numbers
If x = ½ [ -6 + sqrt(-16)] we write this as
x = ½ [ -6 + 4 i] = -3 + 2 i
All real numbers are complex numbers since we can write
a=a+0i
The form a + b i is called standard form
4
Adding and Subtracting Complex Numbers
• a + b i + c + d i = a + b + (c + d ) i
Think of a + b i as a + bx, where x = i
• Examples:
(2 + 3 i) + (-5 + 2 i) =
(-5 + 4 i) – (-2 – 2i) =
5
Solution
• Examples:
(2 + 3 i) + (-5 + 2 i) = -3 -6i
(-5 + 4 i) – (-2 – 2i) = -3+ (4+ 2) i
6
Multiplying Complex Numbers
• Think of as multiplying (a + bx) (c + dx) where x = i, remembering
that (i)2 = -1
Use FOIL
• Examples:
−4 −9=
(6 – 5 i)(4 + i) =
(2 + 3 i)(2 – 3 i) =
7
• Examples:
−4 −9= -6i
(6 – 5 i)(4 + i) = 24 + 5 + i(6 – 20) = 29 – 14i
(2 + 3 i)(2 – 3 i) =4 + 9 = 13
8
Notes
• It may help to write −𝑎 in terms of i before you start, i 𝑎
• Just as with reals, (a + b i) (a – b i) = a2 – ( i) 2 b2
which, since (i) = -1 is just
a2 + b2
9
Example
Show that -3 + 2 i is as solution to x2 + 6x + 13 = 0
10
Solution
• Show that -3 + 2 i is as solution to x2 + 6x + 13 = 0
• X2 = (9 – 4 + i(-12)) = 5 – 12i
• 6X = -18 + 12i
• X2 + 6x + 13 = 5 – 12 i – 18 + 12 i + 13 = -13 + 13 = 0
11
Example
• Show that 2 – i sqrt(3) is a solution of x2 – 4x = -7
12
Powers of i
(i)3 = (i) 2 (i) = - i
(i)6 = (-1)(-1)(-1)= -1
13
Examples
• i 22 =
• i 28 =
• i 57 =
• i 75 =
14
Division
Think of (3 - i) / (2 + i ) as [3 – 𝑥 ] / [2 + 𝑥 ]
Remember how you were taught to clear the radical out of the
denominator to simplify these expressions using a conjugate?
Here we use the complex conjugate:
The complex conjugate of [a – b i] is a + b i ]
just as the conjugate of [a – sqrt(3)] is [a + sqrt(3)]
15
Examples
2/[5-3 i] =
[3 - i] / [ 2 - i] =
[ 6 + −36 ]/ [ 3 + −9 ] =
16
Solution
2/[5-3 i] =2(5+3i)/ (25 + 9) = (10 + 6i)/ 26 = (5+3i)/13
[3 - i] / [ 2 - i] = (3 – i) (2 + i) / (4 + 1) = (7 + i)/5
[ 6 + −36 ]/ [ 3 + −9 ] = (6 + 6i) /(3 + 3i) = (6+6i)(3-3i)/(18)
(18 + 18 + (18-18)i)/18 =
36/ 18 = 2
17
Note
We can check our work with inverse operations
We found that (3 - i ) / (2 - i ) = (7+i)/5
Check: (2 - i )(7 +i)/5 =[14 + 1 + i(-7 + 2) ] /5 = (15 -5i)/5
= 3-i
18
Complex Numbers in Trigonometric Form
19
Complex Numbers in Trig Form
• Let the y axis be imaginary, the x real
• This is called the complex plane
• Rectangular form is z = a + b i
• Trig form is z = r cos  + r sin  i
r is the magnitude of z (radius of the circle)
 is the argument of z
20
Similar to Unit Circle
• Instead of x and y, we have real and imaginary
• With the unit circle, the radius was 1, here it is r
• On the unit circle x was cos , now it is r cos 
• On the unit circle, y was sin , now it is i r sin 
21
Example
• Convert to trig form:
z = -2 -2i
z = 6 + 2i
22
Alternate Notation
Z = r (cos  + i sin ) = r cis 
23
Example: converting from Trig to Rectangular
• Z = 12 cis(/6)
r =12,  = /6
z = 12 [ cos /6 + i sin /6] = 12/2 [ sqrt(3) + i ]
= 6 [ sqrt(3) + i ]
• Z = 13 cis [arctan(5/12)]
cos  = 12/13, sin = 5/13
z = 13 [ cos  + i sin ] = 12 [ 12/13 + 5/13 i]
= 12 + 5 i
24
Why are we bothering?
• Some of our calculations become much easier
25
Products
• Z1 = 3 + 3 i , Z2 = 0 + 2i
The product, Z1 Z2 = (3 + 3 i )(0 + 2i) = -6 + 6 i
• The magnitude of the product is sqrt (72) = 6 sqrt(2)
and the angle is arctan(-1) = 135 deg
• The angle of Z1 is 45 deg, the length is 3 sqrt(2)
• The angle of Z2 = 90 deg, the length is 2
• Note that the length of the product is the product of the
lengths, and the angle is the sum of the angles of the factors
26
We can do the product in more general form
• Z1 Z2 = r1(cos a + i sin b) r2(cos b + i sin b)
= r1 r2 [(cos a – i sin a ) ( cos b + i sin b)]
= r1 r2 [cos a cos b + i sin a cos b +
i sin b cos a – sin a sin b)
= r1 r2 [(cos a cos b - sin a sin b) +
i ( sin a cos b + sin b cos a)]
= r1 r2 [ cos (a + b) + i sin (a + b)]
• Similarly Z1/Z2 = r1/r2 [ cos (a-b) + i sin ( a-b)]
27
Example
• Z1 = -3 + i sqrt(3), Z2 = sqrt(3) + i
– Write Z1 and Z2 in trig form and compute Z1 Z2
– Compute Z1/Z2
– Verify by using rectangular form
28
Solution
• Z1 = -3 + i sqrt(3), Z2 = sqrt(3) + i
– Write Z1 and Z2 in trig form and compute Z1 Z2
Z1 in in Q2, r = 2sqrt(3),  = 150
Z2 in Q1, r = 2,  = 30
So Z1= 2sqrt(3)(cos 150 + i sin 150)
Z2= 2(cos 30 + i sin 30)
Z1 Z2 = 2 sqrt(3) cos 150 + i sin150) 2 (cos30 + i sin 30)
= 4 sqrt(3)[cos (150 + 30) + i sin (150 + 30)]
= 4 sqrt(3)[ -1 + 0] = -4 sqrt(3)
29
Example Cont
– Compute Z1/Z2
2 sqrt(3)[cos 150 + i sin 150] / [2(cos 30 + i sin 30)]
= sqrt(3) [cos (150 – 30) + i sin (150 – 30)]
= sqrt(3)[cos 120 + i sin 120]
= sqrt(3) [ -1/2 + sqrt(3)/2 i ]
= sqrt(3)/2 + 3/2 i
30
Continued
– Verify by using rectangular form
Z 1 Z 2 = (-3 + sqrt(3) i) (3 + i )
= -3 sqrt(3) – 3 i + 3 i + sqrt(3) i 2
= -4 sqrt(3)
31
Examples
What Quadrant? Write in trig form
7–7i
2 – 2 sqrt(3) i
5 sqrt(7) – 5 sqrt(7)
-6 + 6 sqrt(3) i
32
Review
Rectangular to trig form: x is real, y is imaginary
Calculate the length (r, moduli, |z|, etc.)
Determine the angle {Usually arctan (y/x)]
33
Example
• Graph and write in rectangular form
12 cis (/6)
5 sqrt(3) cis (7 /6)
34
Example
• Find the moduli (length) and angle of the product
Z1 = 1 + sqrt(3) i, Z2 = 3 + sqrt(3) i
35
Review: Products and Quotients in Trig Form
z1 = r1(cos a + i sin a), z2 = r2 ( cos b + i sin b)
z1 z2 = r1 r2 [ cos (a+b) + i sin (a+b)]
z1 /z2 = (r1/r2) [ cos (a-b) + i sin (a-b)]
36
Example
Give the moduli, r, and angles, then compute the product and
verify r1r2 = r and the sum of the angles is the product’s angle
z1 = 1 + sqrt(3) i, z2 = 3 + 3 i
37
Solution
z1 = 1 + sqrt(3) i, z2 = 3 + sqrt(3) i
r1 = sqrt( 1 + 3) = 2; angle = arctan [sqrt(3)] = 60, Q1
r2 = sqrt(9 + 3) = 2 sqrt(3); angle = arctan[sqrt(3)/3] =30, Q1
Rectangular: (1 + sqrt(3) i) [ 3 + sqrt(3)i] =
3 - 3 + i [3 sqrt(3) + sqrt(3)] = i 4 sqrt(3)
angle = 90
r1 r2 = 4 sqrt(3)
Sum of angles 60 +30 = 90
38
Example
Compute z1/z2 in trig form and check by using rectangular
coordinates
z1 = -sqrt(3) + i, z2 = 3
39
Solution
r1 = sqrt(3 + 1) = 2; angle = arctan(-1/sqrt(3) = 150 deg
r2 = 3; angle = 0
z1/z2 = 2/3 [ cos 150 + i sin(150)]
Rectangular: (-sqrt(3) + i)/( 3) = -sqrt(3)/3 + i /3
r = sqrt[3/9 + 1/9} = sqrt(4/9) = 2/3
angle = arctan( -sqrt(3)/3) = 150 deg
40
Chapter 10.3 and 10.4
Vectors in the Plane: Algebraic
41
What is a vector?
• A vector has a direction and a magnitude
• Useful concept in physics – can represent motion, force, etc.
• Our coordinate plane is convenient for illustrating vectors:
– The direction is the direction in the plane
– The length is the magnitude
42
Vector in the Plane
Origin (1, 2)
Terminus (5,7)
Length (1 − 5)2 +(2 − 7)2 = 41
Direction: cos  = 4/41, sin  = 5/41
Each vector has a length and direction
43
Example
• Sketch the position vector is <12, 5> and find its length
44
Solution
• Length is sqrt(12x12 + 5 x 5) = sqrt(144 + 25) = sqrt(169) = 13
45
Magnitude
Vector v = <a, b> has magnitude |v| = sqrt(a2 + b2)
If we are given the magnitude and angle (direction)  of a vector
cos  = a/|v|, or a = |v| cos 
sin  = b/|v|, or b = |v| sin 
Furthermore,  can be determined by
tan  = b/a and knowing the quadrant of v
46
Example
• Find the magnitude and direction of the vectors
v1 = (-2.5, -6), v2 = (3 sqrt(3) ,3)
|v1| = sqrt[ (2.5)2 + (-6) 2] = sqrt[ 6.25 + 36] = sqrt( 42.25) = 6.5
|v2| = sqrt[(3 sqrt 3) 2+ (3) 2] = sqrt[ 27 + 9] = sqrt(36) = 6
1 = arctan(-6/-2.5) = 67.4, 2 = arctan[3/(3 sqrt(3))] = 20
Put in right quadrants: 1 = 247.4, 2 = 30
47
Example
V = <a, b> is in Q3 with a magnitude of 21 and forms an angle of
25 with the negative x axis. Find the horizontal and vertical
components
r = 25 so  = 205
a = |v| cos  = 21 cos 205 = -19
b = |v| sin  = 21 sin 205 = -8.9
Check: sqrt[(-19) 2 + (-8.9)2 ] = 21
48
Operations on Vectors
v = <8, 2>, u = <2, 6>
u + v = <8 + 2, 2 + 6> = <10, 8>, called the resultant vector
To visualize, put the tail of vector v at the head of vector u
u+v
u
v
49
Adding and Subtracting Vectors
If u = <a, b> and v = <c, d> then
u + v = < a + c, b + d >
u – v = <a – c, b – d>
k u = <ka, kb> for k real
if k < 0, the vector points in the opposite direction as u
50
Example
u = <-3, -2>, v = <4, -6>
Multiplying a vector by a number, a scalar:
Find: -2 u, ½ v, and -2 u + ½ v
51
Solution
u = <-3, -2>, v = <4, -6>
Find: -2 u, ½ v, and -2 u + ½ v
• -2u = <-6, -4>
• ½ v = <2, -3>
• -2u + 1/2v = <8, -7>
52
Examples
For a = <2,3>, b= <5,4 >, c= <6, -1>,d= <-2, 0>
Find:
• c+d
• -2b – a
• |4b + c|
53
Solution
For a = <2,3>, b= <5,4 >, c= <6, -1>,d= <-2, 0>
Find:
• c + d =<4, -1>
• -2b – a = <-12, 5>
• |4b + c| =
(𝟐𝟔)𝟐 + +(𝟏𝟓)𝟐 =
54
Overview of Operations
If u = <a, b> and v = <c, d>
• u + v = <a + c, b + d>
•
u – v = <a - c, b - d>
• ku = < ka, kb> for real k
Operations are:
• Commutative, Associative, Distributive (scalar over vector)
• Have multiplicative identity 1 and additive identity 0
• Have inverses
55
Vector Diagrams
• Useful for identifying directions, for example forces
• Two tugboats are trying to pull a barge off a sandbar.
(view from above)
2000 N, 125
1500 N, 35
56
Tug Boats, Cont.
• v1: horiz component 2000 cos (125)= -1157
vert component: 2000 sin(125) = 1638
v1 = -1147 i + 1638 j
• v2: horiz component 1500 cos 35 = 1229
vert component 1500 sin 35 = 860
v2 = 1229 i + 860 j
• v1 + v2 = 82 i + 2498 j
magnitude = sqrt[ 82 x 82 + 2498 x 2498] = 2499 N
angle = arctan(| 2498/83|) = 88
57
Tug Boats
Resultant Vector would be the same as one tug pulling with
2499N of force in a direction of 88
2499 N, 88
2000 N, 125
1500 N, 35
58
Other Applications
• Cross winds
• Currents
• Used in navigation
• Parallel Rules
59
Unit Vectors
• Horizontal Unit Vector <1, 0>, called i (not sqrt(-1)
• Vertical Unit Vector <0, 1>, called j
• Can write any vector as v = ai + bj
• If v = < a, b> = a i + bj
Define u = v/|v| = a / sqrt(a2 + b2) i + b / sqrt(a2 + b2) j
u is a unit vector in the same direction as v
60
Example
• Express <8, -12> in terms of unit vectors
61
• Express <8, -12> in terms of unit vectors
• 8 i -12 j
62
Section 8.3
DeMoivre’s Theorem
63
The Theorem
We know z1 z2 = r1 r2 [cos (a+b) + i sin(a+b) ]
What if z1 and z2 are the same? we get
r2 [ cos 2a + i sin 2a ]
If we multiply it by the vector again, we get
r3 [ cos 3a + i sin 3a ]
So zn = rn [ cos na + i sin na ]
So what???
64
It is easy to compute power of complex numbers!
Find z9 if z = - ½ - ½ i
r = sqrt(2)/2, angle = 5/4
z9= [sqrt(2)/2)] [cos 45/4 + i sin 45/4 ]
= sqrt(2)/32 [cos 5/4 + i sin 5/4 ]
= sqrt(2)/32 [ -sqrt(2)/2 – i (sqrt(2)/2]
= -1/32 – 1/32 i
65
Solutions to Polynomial Equations
Show z = -2 – 2 i is a solution to
z4 - 3z3 – 38 z2 - 128z – 144 = 0
66
Solution
z4 - 3z3 – 38 z2 - 128z – 144 = 0
z = 2 sqrt(2) cis 225
z4 = [2 sqrt(2)] 4 cis (4 x 225) = 64 cis 900 = 64 cis 180 = 64 + i 0
z3 = [2 sqrt(2)] 3 cis (3 x 225) = 16 sqrt(2) cis 315 =
= 16 sqrt(2) [ sqrt(2)/2 – i sqrt(2)/2] = 16 – 16i
z2 = [2 sqrt(2)] 2 cis (2 x 225) = 8 cis (450) = 8[ 0 + i ]
We have 64 – 3(16 – 16 i) – 38(8i) – 128(2 – 2 i) – 144=
64 – 48 + 256 – 144 + i( 48 – 304 + 256) i = 0
67
nths Roots
Similar to De Moivre’s:
The nth root of z (z1/n) =
r1/n [ cos ( /n + 2/n) + i sin ( /n + 2/n) for n = 1 to n-1
68
Polar Coordinates
We learned how to express complex numbers in terms of
r( cos a + i sin a); this is just an extension
A polar coordinate is a point on the x-y plane that is expressed in
terms of an angle and a radius.
Example: if Z = r (cos  + i sin ) that would be the same as
the point at r, , where r is the radius and  is the angle
Note: if r < 0, then the ray points in the opposite direction, add
180 to the angle
Example
P (r, ) = (4, 45) is equivalent (22, 22) in x, y coordinates
If x2 + y2 = 16, what are r and ?
r = 4, x2 + y2 = (r cos ) 2 + (r sin ) 2 = r (1); holds for all 
r = 4 is the equation
y = 1/x
r sin  = 1/ (r cos ) or r 2 sin  cos  = 1
but sin  cos  = ½ sin 2 
r sin 2  = 1 is the equation
Example
“r value analysis”
Sketch r = 4 sin 

0
30
45
60
90
r
0
2
2 2
2 
4
Symmetry
If r = f()
Is symmetric around 90 if  is in terms of sines
Is symmetric around 0 if  is in terms of cosines
To graph polar equations
• Note symmetries
• Determine max and min r
• Determine any intercepts
Examples
• r = 4 sin 
Example
r = 2 – 3 sin 
Parametric Equations
Parametric equations are a way of expressing non-functions
as functions
P (x, y) with x = f(t) and y = g (t); f and g are called parametric
equations with parameter t
Example
x = t2 -3, y = 2 t + 1
t
x
y
-3
6
-5
-2
1
-3
-1
-2
-1
0
-3
1
1
2
-2
1
3
5
3
6
7
Parametric Equations in Rectangular Form
x = t2 - 3, y = 2t + 1
solve for t: t = (y-1)/2
substitute for x: x = (y-1)2 / 4 – 3
x = 2 cos t, y = 4 sin t
x2 = 4 cos2 t, y2 = 16 sin2 t
since cos2 t + sin2 t = 1, we have 1 = x2/4 + y2/16, an ellipse
Section 10.8
DeMoivre’s Theorem
79
The Theorem
We know z1 z2 = r1 r2 [cos (a+b) + i sin(a+b) ]
What if z1 and z2 are the same? we get
r2 [ cos 2a + i sin 2a ]
If we multiply it by the vector again, we get
r3 [ cos 3a + i sin 3a ]
So zn = rn [ cos na + i sin na ]
So what???
80
It is easy to compute power of complex numbers!
Find z9 if z = - ½ - ½ i
r = sqrt(2)/2, angle = 5/4
z9= [sqrt(2)/2)] [cos 45/4 + i sin 45/4 ]
= sqrt(2)/32 [cos 5/4 + i sin 5/4 ]
= sqrt(2)/32 [ -sqrt(2)/2 – i (sqrt(2)/2]
= -1/32 – 1/32 i
81
Solutions to Polynomial Equations
Show z = -2 – 2 i is a solution to
z4 - 3z3 – 38 z2 - 128z – 144 = 0
82
Solution
z4 - 3z3 – 38 z2 - 128z – 144 = 0
z = 2 sqrt(2) cis 225
z4 = [2 sqrt(2)] 4 cis (4 x 225) = 64 cis 900 = 64 cis 180 = 64 + i 0
z3 = [2 sqrt(2)] 3 cis (3 x 225) = 16 sqrt(2) cis 315 =
= 16 sqrt(2) [ sqrt(2)/2 – i sqrt(2)/2] = 16 – 16i
z2 = [2 sqrt(2)] 2 cis (2 x 225) = 8 cis (450) = 8[ 0 + i ]
We have 64 – 3(16 – 16 i) – 38(8i) – 128(2 – 2 i) – 144=
64 – 48 + 256 – 144 + i( 48 – 304 + 256) i = 0
83
nths Roots
Similar to De Moivre’s:
The nth root of z (z1/n) =
r1/n [ cos ( /n + 2/n) + i sin ( /n + 2/n) for n = 1 to n-1
84
Polar Coordinates
We learned how to express complex numbers in terms of
r( cos a + i sin a); this is just an extension
A polar coordinate is a point on the x-y plane that is expressed in
terms of an angle and a radius.
Example: if Z = r (cos  + i sin ) that would be the same as
the point at r, , where r is the radius and  is the angle
Note: if r < 0, then the ray points in the opposite direction, add
180 to the angle
Example
P (r, ) = (4, 45) is equivalent (22, 22) in x, y coordinates
If x2 + y2 = 16, what are r and ?
r = 4, x2 + y2 = (r cos ) 2 + (r sin ) 2 = r (1); holds for all 
r = 4 is the equation
y = 1/x
r sin  = 1/ (r cos ) or r 2 sin  cos  = 1
but sin  cos  = ½ sin 2 
r sin 2  = 1 is the equation
Example
“r value analysis”
Sketch r = 4 sin 

0
30
45
60
90
r
0
2
2 2
2 
4
Symmetry
If r = f()
Is symmetric around 90 if  is in terms of sines
Is symmetric around 0 if  is in terms of cosines
To graph polar equations
• Note symmetries
• Determine max and min r
• Determine any intercepts
Examples
• r = 4 sin 
Example
r = 2 – 3 sin 
Parametric Equations
Parametric equations are a way of expressing non-functions
as functions
P (x, y) with x = f(t) and y = g (t); f and g are called parametric
equations with parameter t
Example
x = t2 -3, y = 2 t + 1
t
x
y
-3
6
-5
-2
1
-3
-1
-2
-1
0
-3
1
1
2
-2
1
3
5
3
6
7
Parametric Equations in Rectangular Form
x = t2 - 3, y = 2t + 1
solve for t: t = (y-1)/2
substitute for x: x = (y-1)2 / 4 – 3
x = 2 cos t, y = 4 sin t
x2 = 4 cos2 t, y2 = 16 sin2 t
since cos2 t + sin2 t = 1, we have 1 = x2/4 + y2/16, an ellipse