Download 6.5

Section 6-5
Day 1& 2
1) Trig form of a Complex #
2) Multiplying , Dividing,
and powers (DeMoivre’s
Theorem) of Complex #s
Warm-Up
Find the work done by a man pushing a car
with 60 lbs of force at an angle 30 degrees
below horizontal for 1000 feet.
51961.52 ft.lbs
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3
Section 6-5 Day 1
The Trigonometric form of a Complex Number
The standard form of a complex number is z = a+bi
In the complex plane, every complex number corresponds
to a point.
Imaginary axis
Example:
4
Plot the points 3 + 4i and
–2 – 2i in the complex plane.
2
Real
axis
–2
(– 2, – 2)
or
– 2 – 2i
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(3, 4)
or
3 + 4i
2
–2
5
The absolute value of the complex number z = a + bi is
the distance between the origin (0, 0) and the point (a, b).
| a  bi |  a  b
2
2
Example:
Plot z = 3 + 6i and find its absolute value.
Imaginary axis
z  3 6
2
8
6
z = 3 + 6i
 9  36
4
2
–4
–2
3 5 units
4
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2
Real
axis
 45
3 5
6
To write a complex number a + bi in trigonometric
form, let  be the angle from the positive real axis
(measured counter clockwise) to the line segment
connecting the origin to the point (a, b).
Imaginary axis
a = r cos 
b = r sin 
(a, b)
r  a 2  b2
r
b
a  bi  (r cos  )  (r sin  )i

a
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Real
axis
7
The trigonometric form of a complex number
z = a + bi is given by
z = r(cos  + i sin )
where a = r cos , b = r sin , r  a 2  b2 , and tan   b .
a
The number r is the modulus of z, and  is the
argument of z.
Example:

z  1 cos   i sin 
2
3
3
modulus
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
argument
8
How is graphing in trig form different?
In a rectangular system, you go
left or right and up or down.
z  2  2i
In a trigonometric or polar system, you
have a direction to travel and a distance
to travel in that direction.
z  2  cos45  i sin 45
Polar form (2,45)
Example:
Write the complex number z = –7 + 4i in trigonometric
form.
z  r  7  4i
Imaginary axis
 (7) 2  42  49  16  65
z = –7 + 4i
z  65
tan   b   4
a
7
  150.26
150.26°
Real
axis
z  r  cos θ  i sin θ 
 65(cos150.26  i sin150.26)
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10
You try:
Write the complex number in trigonometric form.
z  2  2 3i
z  r  cos θ  i sin θ 

 4 cos 4   i sin 4 
3
3
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
11
You try:
Write the complex number in trigonometric form.
z  3  i
z  r  cos θ  i sin θ 
 10  cos161.56  i sin161.56 
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12
Example:
Write the complex number 3.75 cos 3  i sin 3
4
4
in standard (rectangular) form a + bi.


cos 3   2
4
2
sin 3  2
4
2


2
2
z  3.75  

i
2 
 2
 15 2  15 2 i Standard form
8
8
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13
You Try:
Write the complex number 3  cos 330  i sin 330 
2
in standard form a + bi.
cos 330  3
2
sin 330   1
2
 3 1 
3
z 
 i
2 2 2 
3
3

 3 i Standard form
4
4
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14
You Try:
Write the complex number 6  cos 23030  i sin 23030
in standard form a + bi.
 3.8165  4.6297i
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15
• Homework Day 1:
• Pg. 440, 1-47 odd
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16
Section 6-5 Day 2
Multiplying , Dividing, and Powers (DeMoivre’s
Theorem) of Complex #s
Homework Quiz
Find the trigonometric form of the complex number:
z  2  6i
 2 10  cos108.43  i sin108.43 
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18
Multiplication of Complex Numbers
Multiply the 2 complex numbers.
 
 
z1  2 cos 2  i sin 2 

3
3 
 
 
z2  8 cos 11  i sin 11 

6
6 
Hint: Write the numbers in standard form and multiply algebraically.
Solution : 16i
There is an easier way
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19
Multiplication of Complex Numbers
z  r  cos θ  i sin θ 
To multiply 2 complex numbers, you multiply the
moduli and add the arguments.
Example: Find the product z 1z2 of the complex
numbers and write it in standard form:
 
 
z1  2 cos 2  i sin 2 

3
3 
 
 
z2  8 cos 11  i sin 11 

6
6 

Solution : 16 cos   i sin 
2
2
 16  0  i 
 16 i
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
20
You Try:
Find the product of the two complex numbers. Find
both the trigonometric form and standard form of the
product.
1
z1   cos115  i sin115 
2
4
z2   cos 300  i sin 300 
5
2
2
 cos 415  i sin 415  or  cos 55  i sin 55 
5
5
.229  .328i
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21
Division of Complex Numbers
z  r  cos θ  i sin θ 
To divide 2 complex numbers, you divide the
moduli and subtract the arguments.
Example: Find the quotient z 1 z2 of the complex
numbers and write it in standard form:
z1  24 cos  300   i sin  300  
z2  8 cos  75   i sin  75 
3
2
3
2



i
z1 / z2  3  cos 225  i sin 225
2
2
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22
Division of Complex Numbers
To divide 2 complex numbers, you divide the moduli and subtract the arguments.
You Try: Find the quotient z 1 z2 of the complex
numbers and write it in standard form:
z1  15 cos 7  i sin 7 
4
4 


3 cos 3  i sin 3
4
4
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z2  5 cos   i sin  

or
3 2  3i 2
2
2
23
Powers of Complex Numbers
Example:
Write (1 + i)2 in standard form a + bi.
Solution : 2i
Write (1 + i)5 in standard form a + bi.
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24
To raise a complex number to a power, you can use it’s
trigonometric form:
z  r  cos θ  i sin θ 
Find z 2
z 2  r  cos θ  i sin θ  r  cos θ  i sin θ 
 r 2  cos 2θ  i sin 2θ 
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25
Raising complex numbers by powers creates a pattern:
z  r  cos θ  i sin θ 
z 2  r  cos θ  i sin θ  r  cos θ  i sin θ   r 2  cos 2θ  i sin 2θ 
z3  r 2  cos 2θ  i sin 2θ  r  cos θ  i sin θ   r 3  cos 3θ  i sin 3θ 
z 4  r 4  cos 4θ  i sin 4θ 
z 5  r 5  cos 5θ  i sin 5θ 
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27
DeMoivre’s Theorem
If z = r(cos  + i sin ) is a complex number and n is
a positive integer, then
zn = [r(cos  + i sin )]n
= rn (cos n + i sin n)].
Example:
[2(cos 20  i sin 20)]  2 [cos(3  20)  i sin(3  20)]
3
3
 8(cos 60  i sin 60)
1

3
 8 
i   4  4 3i
2 2 
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28
Example:
Use DeMoivre’s Theorem to write (1 + i)5 in standard
form a + bi.
1 i
Convert the complex
number into
trigonometric form.
1  i  
1  i 
5

r  1  i  12  12  2
  tan 1 1  
1
4
2 cos   i sin 
4
4




  2 cos   i sin  

4
4 
5
Example continues.
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29
Example continued:
1  i 
5


  2 cos   i sin  

4
4 

 
5
 
 
2 cos 5    i sin 5   

4
4 
5

 4 2 cos 5  i sin 5
4
4

 1  1  
 4 2 

 i
2 
 2 
 4  4i
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30
You Try:
Write (3 + 4i)3 in standard form a + bi.
r  3  4  25  5
2
 3  4i 
3
2
3
  tan 1 4  53.13
 5  cos 53.13  i sin 53.13 
3
 53 cos(3  53.13)  i sin(3  53.13)
 125cos(159.39)  i sin(159.39)
 1250.936  i0.352)
 117  44i
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31
• Homework Day 2:
• Pg. 441, 51-89 odd
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32
Warm-Up
Find the indicated power of the complex
number by rewriting the number in trig
form and using DeMoivre’s Theorem.
Write the result in standard form.
 2  2i 
4
64
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33
Section 6-5 Day 3
Roots of Complex Numbers
Solve : x  1  0
6
How many solutions?
Difference of squares?
Sum and difference of cubes?
1  i 3 1  i 3
x   1,
,
2
2
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35
Roots of Complex Numbers
• There will be as many answers as the index of the
root you are looking for.
– Square root = 2 answers
– Cube root = 3 answers, etc.
• Answers will be spaced symmetrically around the
circle
– You divide a full circle by the number of answers to
find out how far apart they are
General Process
1. The complex number must be in trig. form.
2. Take the nth root of r. All answers have the
same value for r.
3. Divide theta by n to find the first angle.
4. Divide a full circle by n to find out how
much you add to theta to get to each
subsequent answer.
5. Write your numbers, increasing by theta/n
each time.
The formula
z  r  cos   i sin  
  360k
  360k 

z  r  cos
 i sin
 or
n
n


  2 k
  2 k 

n
r  cos
 i sin

n
n


n
n
k starts at 0 and goes up to n-1
This is easier than it looks.
Solve : x 1  0,
6
1  1  0i
x  1,
6
x 1
6
First, write as a root and write
the radicand in trig. form.
1 + 0i is over 1 and up 0. Therefore, 1 is the hypotenuse
and theta is 0o. Or use arctan b/a.
1 cos 0o  i sin 0o 
First angle?
0  2k
0  2k 

 1 cos
 i sin

6
6 

6
Divide theta by n to find the first angle.
How far apart will the evenly spaced angles be?
The first
angle is 0.
2 

6
3
1 cos 0  i sin 0   1
1 cos   i sin    1

 1
3

1 cos  i sin   
i
3
3 2 2

4
4

1 cos
 i sin
3
3

2
2

1 cos
 i sin
3
3

1
3




i

2 2

1
3

i
 
2 2

5
5

1 cos
 i sin
3
3

3
 1
i
 
 2 2
The sixth roots of 1.
1
3
 
i
2
2
1  0i
1
3
 
i
2
2
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1
3

i
2
2
1  0i
1
3

i
2
2
41
Example:
Find the 4th roots of: z  81 cos80  i sin80
1. Find the 4th root of 81
r  4 81  3
2. Divide theta by 4 to get the first angle.
80
   20
4
3. Divide 360 by 4 to determine the spacing
between angles
360
 90 between angles
4
4. List the 4 answers.
• The only thing that changes is the angle.
• The number of answers equals the number of roots.
z1  3  cos 20  i sin 20  2.819  1.026i
z2  3  cos110  i sin110  1.026  2.819i
z3  3 cos 200  i sin 200  2.819 1.026i
z4  3 cos 290  i sin 290  1.026  2.819i
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43
You Try:
Find the square roots of
z  88 3i
Write them in standard form.
Remember to convert to trig form first.
2 3  2i
2 3  2i
You Try: Find the cube roots of z = -2+2i
Again, first convert to trig form.
3
2  2i 
3
8
3
3
8  cos135  i sin135
o
o

8 2
2  cos 45o  i sin 45o 
2  cos165o  i sin165o 
2  cos 285o  i sin 285o 
Homework Day 3:
Pg. 442, 93-115 odd, don’t need to graph
them.
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46