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Chapter 7
Applications of
Trigonometric
Functions
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All rights reserved
© 2010
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SECTION 7.8 Polar Form of Complex Numbers; DeMoivre’s Theorem
OBJECTIVES
1
2
3
4
5
6
Represent complex numbers geometrically.
Find the absolute value of a complex number.
Write a complex number in polar form.
Find products and quotients of complex numbers in
polar form.
Use DeMoivre’s Theorem to find powers of a
complex number.
Use DeMoivre’s Theorem to find the nth roots of a
complex number.
GEOMETRIC REPRESENTATION OF
COMPLEX NUMBERS
The geometric representation of the complex
number a + bi is the point P(a, b) in a rectangular
coordinate system.
When a rectangular coordinate system is used to
represent complex numbers, the plane is called the
complex plane.
The x-axis is also called the real axis, because the
real part of a complex number is plotted along the
x-axis. Similarly, the y-axis is also called the
imaginary axis.
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GEOMETRIC REPRESENTATION OF
COMPLEX NUMBERS
A complex number a + bi may be viewed as a
position vector with initial point (0, 0) and
terminal point (a, b).
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EXAMPLE 1
Plotting Complex Numbers
Plot each number in the complex plane.
1 + 3i, –2 + 2i, –3, –2i, 3 – i
Solution
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ABSOLUTE VALUE OF
A COMPLEX NUMBER
The absolute value (or magnitude or modulus)
of a complex number z = a + bi is
z  a  bi  a  b .
2
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6
POLAR FORM OF A COMPLEX NUMBER
The complex number z = a + bi can be written in
polar form
z  r  cos  i sin   ,
where a = r cos θ, b = r sin θ, r  a 2  b 2 , and
b
tan   .
a
When a nonzero complex number is written in polar
form, the positive number r is the modulus or
absolute value of z; the angle θ is called the
argument of z (written θ = arg z).
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EXAMPLE 4
Writing a Complex Number in Rectangular
Form
Write the complex number
in rectangular form.
Solution
The rectangular form of z is
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PRODUCT AND QUOTIENT RULES FOR
TWO COMPLEX NUMBERS IN POLAR FORM
Let z1 = r1(cos 1 + isin 1) and
z2 = r2(cos θ2 + isin θ2) be two complex numbers in
polar form. Then
z1 z2  r1r2 cos1  2   i sin 1  2 
and
z1 r1
 cos1   2   i sin 1   2 , z2  0.
z 2 r2
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EXAMPLE 5
Finding the Product and Quotient of Two
Complex Numbers
Let z1  3cos 65º i sin 65º  and
z1
z2  4 cos15º i sin15º . Find z1 z2 and .
z2
Leave the answers in polar form.
Solution
z1 z2  3cos 65  i sin 65  4cos15  i sin 15
 3  4cos65  15  i sin 65  15
 12cos 80  i sin 80
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EXAMPLE 5
Finding the Product and Quotient of Two
Complex Numbers
Solution continued
z1 3cos 65  i sin 65

z2 4cos15  i sin 15
3
 cos65  15  i sin 65  15
4
3
 cos 50  i sin 50
4
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DEMOIVRE’S THEOREM
Let z = r(cos + isin) be a complex number
in polar form. Then for any integer n,
z n  r n  cos n  i sin n .
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EXAMPLE 7
Finding the Power of a Complex Number
Let z = 1 + i. Use DeMoivre’s Theorem to find
each power of z. Write answers in rectangular
form.
a. z16
b. z 10
Solution
Convert z to polar form. Find r and .
r  a 2  b 2  12  12  2
b 1

tan     1, so  
a 1
4



z  1  i  2  cos  i sin 

4
4
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EXAMPLE 7
Finding the Power of a Complex Number
Solution continued
16
a. z



z  2  cos  i sin 
4
4

16
 

 
z   2  cos  i sin 
4
4 
 
16 

 


16
z  2 cos16    i sin 16  
4
4 

 
16
 
 28 cos4   i sin 4 
 2561  i  0  256
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EXAMPLE 7
Finding the Power of a Complex Number
Solution continued
b. z 10



z  2  cos  i sin 
4
4

10
z
10
z 10
 

 
  2  cos  i sin 
4
4 
 
10 

 


 2 cos  10    i sin   10  
4
4 

 
 
1   5

cos 

32   2

 5

i
sin



 2
1
1
 0  i  1   i
32
32
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


15
DEMOIVRE’S nth ROOTS THEOREM
Let z and w be two complex numbers and let n
be a positive integer. The complex number z is
called an nth root of w if zn = w.
The nth roots of a complex number
w = r(cos  + isin ), where r > 0 and  is in
degrees, are given by
    360º k 
   360º k  
zk  r  cos 
 i sin 
,






n
n
 
for k = 0, 1, 2, …, n – 1.
1n
If  is in radians, replace 360º with 2π in zk.
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EXAMPLE 8
Finding the Roots of a Complex Number
Find the three cube roots of 1 + i in polar form,
with the argument in degrees.
Solution
In the previous example, we showed that



1  i  2  cos  i sin  and

4
4
1  i  2 cos 45º i sin 45º .
Use DeMoivre’s Theorem with n = 3.
zk 
 2
1/ 3
  45  360k 
 45  360k 
  i sin 
, k  0, 1, 2.
cos
3
3



 
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EXAMPLE 8
Finding the Roots of a Complex Number
Solution continued
Substitute k = 0, 1, and 2 in the expression for zk
and simplify to find the three cube roots.
  45  360  0 
 45  360  0 
z0  2 cos
  i sin 

3
3



 
1/ 6
 21/ 6 cos15  i sin 15
  45  360 1 
 45  360 1 
z1  2 cos
  i sin 

3
3



 
 21/ 6 cos135  i sin 135
1/ 6
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EXAMPLE 8
Finding the Roots of a Complex Number
Solution continued
  45  360  2 
 45  360  2 
z2  2 cos
  i sin 

3
3



 
 21/ 6 cos 255  i sin 255
1/ 6
The three cube roots of 1 + i are as follows:
z0  2
cos15º i sin15º 
z1  2
cos135º i sin135º 
1/6
1/6
z2  21/6  cos 255º i sin 255º 
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