Download Slide 10-2 - Cloudfront.net

7.6 Powers and Roots of Complex
Numbers
• Powers of Complex numbers
[r (cos   i sin  )]2  [r (cos   i sin  )r (cos   i sin  )]
 r  r[cos(   )  i sin(   )]
 r 2 (cos 2  i sin 2 )
In the same way,
[r (cos   i sin  )]3  r 3 (cos 3  i sin 3 )
Copyright © 2007 Pearson Education, Inc.
Slide 10-1
7.6 De Moivre’s Theorem
De Moivre’s Theorem
If r(cos  + i sin ) is a complex number, and n is any
real number, then
[r (cos   i sin  )]n  r n (cos n  i sin n ).
In compact form, this is written
[r cis  ]n  r n (cis n ).
Copyright © 2007 Pearson Education, Inc.
Slide 10-2
7.6 Finding a Power of a Complex
Number
Example Find (1  i 3) and express the result in
rectangular form.
8
Solution
(1  i 3)8  [2(cos 60  i sin 60 )]8
 2 [cos(8  60 )  i sin(8  60 )]
8

Convert to
trigonometric form.

 256(cos 480  i sin 480 )
 256(cos 120  i sin 120 )
3
 1
 256   i 
2 
 2
 128  128i 3
Copyright © 2007 Pearson Education, Inc.
480º and 120º are
coterminal.
cos120º = -1/2;
sin120º = 3 / 2
Rectangular form
Slide 10-3
7.6 Roots of Complex Numbers
nth Root
For a positive integer n, the complex number a+bi is
the nth of the complex number x + yi if
(a + bi)n = x + yi.
• To find three complex cube roots of
8(cos 135º + i sin 135º), for example, look for a
complex number, say r(cos  + sin ), that will
satisfy
[r (cos   i sin  )]3  8(cos135  i sin 135 ).
Copyright © 2007 Pearson Education, Inc.
Slide 10-4
7.6 Roots of Complex Numbers
By De Moivre’s Theorem,
[r (cos   i sin  )]3  8(cos135  i sin 135 ).
becomes
r 3 (cos 3  i sin 3 )  8(cos135  i sin 135 ).
Therefore, we must have r3 = 8, or r = 2, and
3  135  360  k , k any integer
135  360  k

, k any integer.
3
Copyright © 2007 Pearson Education, Inc.
Slide 10-5
7.6 Roots of Complex Numbers
Let k take on integer values 0, 1, and 2.
135  0

k  0,  
 45
3
135  360
k  1,  
 165
3
135  720

k  2,  
 285
3
It can be shown that for integers k = 3, 4, and 5, these
values have repeating solutions. Therefore, all of the cube roots
(three of them) can be found by letting k = 0, 1, and 2.
Copyright © 2007 Pearson Education, Inc.
Slide 10-6
7.6 Roots of Complex Numbers
When k = 0, the root is
When k = 1, the root is
When k = 2, the root is
2(cos 45º + i sin 45º).
2(cos 165º + i sin 165º).
2(cos 285º + i sin 285º).
nth Root Theorem
If n is any positive integer, r is a positive real number, and  is
in degrees, then the nonzero complex number r(cos  + i sin  )
has exactly n distinct nth roots, given by
n
r (cos   i sin  ),
where

  360 k
n
Copyright © 2007 Pearson Education, Inc.

360 k
or   
, k  0,1, 2, , n  1
n
n
Slide 10-7
7.6 Finding Complex Roots
Example Find the two square roots of 4i. Write the
roots in rectangular form, and check your results
directly with a calculator.
Solution First write 4i in trigonometric form as



4i  4 cos  i sin .
2
2

Here, r = 4 and  = /2. The square roots have modulus
4  2 and arguments as follows.
2 k 
 
 k
2
2
4

Copyright © 2007 Pearson Education, Inc.
2
Slide 10-8
7.6 Finding Complex Roots
Since there are two roots, let k = 0 and 1.
If k = 0, then
If k = 1, then



 0  .
4
4

5
    1  .
4
4
Using these values for , the
square roots are 2 cis 4 and
2 cis 54 , which can be written
in rectangular form as
2  i 2 and  2  i 2.
Copyright © 2007 Pearson Education, Inc.
Slide 10-9
7.6 Finding Complex Roots
Example Find all fourth roots of  8  8i 3. Write the roots in
rectangular form.
Solution
 8  8i 3  16 cis 120

r  16 and   120

Modulus 4 16  2
120 360 k


Arguments  

 30  90  k
4
4
If k = 0, then
If k = 1, then
If k = 2, then
If k = 3, then
Copyright © 2007 Pearson Education, Inc.
 = 30º + 90º·0 = 30º.
 = 30º + 90º·1 = 120º.
 = 30º + 90º·2 = 210º.
 = 30º + 90º·3 = 300º.
Slide 10-10
7.6 Finding Complex Roots
Using these angles on the previous slide, the fourth
roots are
2 cis 30º, 2 cis 120º, 2 cis 210º, and 2 cis 300º.
These four roots can be written in rectangular form as
3  i,  1  i 3,  3  i, 1  i 3.
Copyright © 2007 Pearson Education, Inc.
Slide 10-11
7.6 Solving an Equation by Finding
Complex Roots
Example Find all complex number solutions of
x5 – 1 = 0. Graph them as vectors in the complex plane.
Solution Write the equation as
x5  1  0
x  1.
To find the five complex number solutions, write 1 in
polar form as
1  1  0i  1(cos 0  i sin 0 ).
5
The modulus of the fifth roots is 5 1  1.
Copyright © 2007 Pearson Education, Inc.
Slide 10-12
7.6 Solving an Equation by Finding
Complex Roots
The arguments are given by
0  72  k , k  0, 1, 2, 3, or 4.
Using these arguments, the fifth roots are
1(cos 0º + i sin 0º),
1(cos 72º + i sin 72º),
1(cos 144º + i sin 144º),
1(cos 216º + i sin 216º),
1(cos 288º + i sin 288º),
Copyright © 2007 Pearson Education, Inc.
k=0
k=1
k=2
k=3
k=4
Slide 10-13