Download Section 8.2: Complex Numbers in Trigonometric Form 1. Complex

Trigonometry, Chapter 8
Section 8.2: Complex Numbers in Trigonometric Form
1. Complex numbers are numbers with two components which we represent algebraically as the sum of two
unlike terms: a + bi
where a and b are any real numbers and i is defined as
i=
−1
or
i2 =
−1
We call the first term the ‘real’ component and the second term the ‘imaginary’ component.
2. Graphically these numbers can be represented as ordered pairs.
imaginary axis
●
2
Note that the ‘i’ here is different
from the basis vector i = <1, 0>
from the last chapter. The
context and the italic vs. bold
print should make it clear which
is being referred to.
7 + 2i
real axis
7
3. The trigonometric form of a complex number involves r, the absolute value or ‘modulus’, and θ , the
‘argument’, which is the smallest positive angle counterclockwise from the positive real axis.
Terminology and Formulas: let z represent some complex number, z = a + bi.
im
Given z = a + bi
Modulus of z:
Argument of z:
a + bi
●
r = |z| =
r
a 2 + b2
θ
b
re
a
θ = Arg z
Given r = |z| and θ
Re z: a = r cos θ
Im z: b = r sin θ
b
has reference angle θˆ = tan −1
a
Notation: =
z
a + bi
= r cos θ + (r sin θ )i
z=
r (cos θ + i sin θ )
Standard form
or
z=
r cis θ
Trig form
Example. Find r and θ for z = –4 + 4i. Write z in trigonometric form.
Trigonometry, Chapter 8
Section 8.3: Operations of Complex Numbers
1. Addition/Subtraction: These operations are easier using the standard form (the rectangular components),
and is done component-wise as with vectors.
Add : (a + bi ) + (c + di ) = (a + c) + (b + d )i
Sub : (a + bi ) − (c + di ) = (a − c) + (b − d )i
2. Multiplication/Division: These operations are easier to define using trig form. These were not defined
for vectors, the dot product was not vector multiplication. We can give a meaningful and logically
consistent definition for multiplication and division of ordered pairs as follows.
Product Rule:
z1=
⋅ z2 r1 (cos θ1 + i sin θ1 ) ⋅ r2 (cos θ 2 + i sin θ 2 )
foil
r1 ⋅ r2 (cos θ1 + i sin θ1 )(cos θ 2 + i sin θ 2 )
=
=
z1 r1=
cis θ1 , z2 r2 cis θ 2
r1 ⋅ r2 [cos θ1 cos θ 2 + i cos θ1 sin θ 2 + i sin θ1 cos θ 2 + i 2 sin θ1 sin θ 2 ]
=
r1 ⋅ r2 [cos θ1 cos θ 2 + i (cos θ1 sin θ 2 + sin θ1 cos θ 2 ) − sin θ1 sin θ 2 ]
=
= r1 ⋅ r2 [(cos θ1 cos θ 2 − sin θ1 sin θ 2 ) + i (cos θ1 sin θ 2 + sin θ1 cos θ 2 )]
Multiply the magnitudes.
Add the angles.
Quotient Rule:
r1 ⋅ r2 [
=
cos (θ1 + θ 2 )
+
z1 ⋅ z2 = r1 ⋅ r2 [cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 )]
z1 r1 (cos θ1 + i sin θ1 )
=
⋅
z2 r2 (cos θ 2 + i sin θ 2 )
r (cos θ1 + i sin θ1 ) ⋅ (cos θ 2 − i sin θ 2 )
= 1⋅
r2 (cos θ 2 + i sin θ 2 ) ⋅ (cos θ 2 − i sin θ 2 )
Divide the magnitudes.
Subtract the angles.
Power Rule:
i sin (θ1 + θ 2 )]
z1 r1
=
[cos(θ1 − θ 2 ) + i sin(θ1 − θ 2 )]
z2 r2
for any integer n: z n = r n cis (nθ )
Example: Let z1 =
− 2 3 + 2i and z2 =
5i .
a. Write both in trig form, using degrees.
b. Find z1 ⋅ z2 , z1 / z2 , and z13 using the trig form.
c. Convert your answers back to standard form.
Foil top and
bottom and
simplify as
above. Should
obtain this
expression.
(De Moivre’s Theorem)
Trigonometry, Chapter 8
Section 8.4: Roots of Complex Numbers
1. Recall that in algebra we show the equivalence between roots and rational exponents:
n
1
a = an .
2. When we apply this to complex numbers and De Moivre’s Thm we have to include one additional detail:
There are exactly n complex nth roots of a number (2 square roots, 3 cube roots, 4 fourth roots, etc. )
so we must use θ + 360° ⋅ k or θ + 2π ⋅ k in place of θ as follows:
For z = r cis θ the nth roots are given by the formula:
1
n
1
n
 θ + 360°⋅ k 
z = r cis 

n


where k = 0, 1, 2, ...., n-1
These values of k
give exactly n
roots.
Example 1: Find the two square roots of -4.
Step 1: Trig form: Verify that z = -4 + 0i = 4 cis 180°.
Parameters: n = 2 (square root)
r=4
θ = 180°
k = 0, 1
Step 2:
1
2
 180° + 360°⋅ k 
4 cis 
=

2


2 cis (90° + 180°⋅ k )
Step 3: If directed, convert roots to standard form:
 2 cis 90°
⇒ 
 2 cis 270°
k=
0
k=
1
2cis90° = 0 + 2i = 2i
2cis270° = 0 − 2i = − 2i
Example 2: Find the three cube roots of -8.
4π
4π 

of z 81 cos
Example 3: Find the four 4th roots
=
+ i sin

3
3 

Notice the geometric relationship between the complex roots in each example by graphing them.
Trigonometry, Chapter 8
Section 8.5: Polar Coordinates
1. Application. A microphone is characterized by its sensitivity to sounds coming from different directions,
called its directionality. A two dimensional graph is used to represent this, with the microphone at the
origin and a closed curve showing the area the microphone is designed to capture sound from. For
example, a cardioid microphone has a pattern as follows
Compare the two equations that models this graph
Rectangular Coordinates:
x 4 − 16 x 2 − 8 x 2 y + 2 x 2 y 2 − 8 y 3 + y 4 =
0
Polar Coordinates:=
r 2sin θ − 2
As we will see, in polar coordinates the equations of certain circular graphs are much simpler and
easier to work with.
2. Example. Consider the ordered pair (3, 3) in rectangular coordinates. Just as we did with complex
numbers in standard form we can find the trigonometric components r = 3 2 , θ = 45°.
y
3
●
(3, 3)
y
3 2
3
x
●
45°
x
3. Polar Coordinates (r, θ):
r = distance from the point to a fixed point called the pole
θ = the angle formed from a fixed axis called the polar axis
From the example above we can see that the ordered pair
Rectangular: (3, 3) = Polar: ( 3 2 , 45°)
3 2
When comparing rectangular and polar coordinates:
pole = origin
polar axis = positive x-axis
●
45°
Trigonometry, Chapter 8
4. Conversion formulas.
Rectangular (x , y) to Polar (r, θ)
Polar (r, θ) to Rectangular (x , y)
r = ± x2 + y 2
x = r cos θ
y
θ has reference angle θˆ = tan −1
x
y = r sin θ
Note: Polar coordinates and rectangular coordinates are related in exactly the same way as the
trigonometric form and standard form of complex numbers with one difference:
In polar coordinates r can be negative or positive, and θ can be any angle with the correct
terminal side. When r is negative the point is reflected through the origin.
Example. The ordered pair ( 3 2 , 45°) can be written in several other ways:
y
y
3 2
●
3 2
●
225°
45°
x
y
y
-315°
x
−3 2
3 2
●
−3 2
x
3 2
●
-135°
( 3 2 , 45°)
( 3 2 , -315°)
(- 3 2 , 225°)
(- 3 2 , -135°)
positive r
positive θ
positive r
negative θ
negative r
positive θ
negative r
negative θ
Example. Graph the ordered pair (4, 120°) and find three other ways to write the label the same point in
polar coordinates.
x
Trigonometry, Chapter 8
5. Graphs of polar equations. A polar equation involves r and θ, and the solutions (r, θ) can be graphed.
We will do this in detail in section 8.6. In this section we will focus on converting equations between
polar form and rectangular form.
Example. r 2 = 4 . In this equation r must equal ±2 and can θ be anything. This gives a circle of radius 2.
We can convert to rectangular coordinates by replacing r 2 with x 2 + y 2
2
2
circle: (h, k) = (0, 0), r = 2
2
r = 4 ⇒ x +y = 4
(2,90°)
•
•
(2, 0°)
•(2, 270°)
(-2, 90°)
Example: r = 6sin θ . We will see in the next section how to make a table of values plug in different
values of θ. For example, θ= 90° ⇒ r= 6sin90° = 6 , so we get the point (6, 90°).
Similarly you can obtain other points: (0, 0°), (3, 30°), etc.
We can convert to rectangular form by substituting sin θ =
r =6sin θ
⇒
r =6 ⋅
y
r
⇒
r 2 =6 y
⇒
y
:
r
•
x 2 + y 2 =6 y
– 3
•
Completing the square gives
x 2 + ( y − 3) 2 =
9
Circle with center (0, 3), r = 3
(6,90°)
•
(3,30°)
(0, 0°)
To convert equations between polar and rectangular coordinates use the following substitutions, or
anything equivalent. There can be many ways to do this.
y
= tan θ
x2 + y 2 =
r 2 , x = r cos θ ,
y = r sin θ ,
x
Example: Write the equation in rectangular coordinates: r (cosθ + sin θ ) = 2
Trigonometry, Chapter 8
Section 8.6: Graphing Equations in Polar Coordinates
Example. Sketch the graph of r = 2sin 3θ .
θ
To do this we will start with a table of values.
(= 2sin 3θ )
r
( 2 ,45°)
0°
0
(=
2sin 0°)
15°
30°
2 (=
2sin 45°)
2 (=
2sin 90°)
45°
60°
2 (=
2sin135°)
0 (=
2sin180°)
θ
r
1
r: 0→2
2
r: 2→0
(0, 0°)
(0, 60°)
(= 2sin 3θ )
60°
75°
90°
105°
120°
r:
r:
θ
●
● ● (2, 30°)
● ( 2 ,15°)
3
4
→
→
(= 2sin 3θ )
r
120°
135°
150°
165°
180°
r:
r:
5
6
→
→
Putting this all together we get the following graph.
y = 2sin 3 x
r = 2sin 3θ
5
2
6
1
3
4
Can you see how to
obtain the polar
graph by comparing
to this rectangular
graph?
This can be used
instead of, or in
addition to, a table
of values.
1
2
5
90°
30°
60°
3
6
150°
120°
180°
4