10.6 Roots of Complex Numbers Notice these numerical statements. 32  2  125   5 4 16   2 These are true!

Download Report

Transcript 10.6 Roots of Complex Numbers Notice these numerical statements. 32  2  125   5 4 16   2 These are true!

Slide 1

10.6 Roots of Complex Numbers


Slide 2

Notice these numerical statements.
5

32  2

3

 125   5

1
4

16   2

These are true! But I would like to write them a bit differently.

32 = 25

–125 = (–5)3

16 = 24

Keeping this “change in appearance” in mind, let’s extend this to the
complex plane.
3i is a fourth root of 81 because (3i)4 = 81
In general …
For complex numbers r and z and for any positive integer n,
r is an nth root of z iff rn = z.
We will utilize DeMoivre’s Theorem to verify.


Slide 3

Ex 1) Find the four fourth roots of 81 cis

2
3

Let w = r cis θ represent a fourth root. Then w 4  81 cis

2
3

 r cis    r 4 cis 4  81 cis
4

r  81
4

4 

2
3

but there are lots of angles that terminate at

we must consider multiples  4 

r 3
4 

2
3

4 

2
3

 

2
12

4 

 



 

Four 4th roots: 3 cis

6


6

2
3

3 cis

4 

2
3

8
3

4 

14 
3

2
3

 



2
3

6
3

3 cis

7
6
7
6



12 
3

 2 k

2
3

4 

2
3

4 

20 
3

 
3 cis

2
3

5
3
5
3



18 
3


Slide 4

Complex Roots Theorem
For any positive integer n and any complex number z = r cis θ,
the n distinct nth roots of z are the complex numbers
n

r cis

  2k
n

for k = 0, 1, 2, …, n – 1

Now, what does this mean?? Explain what “to do” in plain words!
Ex 2) Find the cube roots of 1000i
r 

0  1000  1000
2


think!
 
(graph in head) 2
3


1000 cis 



2

 2 k 

3


 0 + 1000i


,

 4 2


6

6



5

6

,

,

6

10 cis

63

,

 8 4


6

63

3
2


6

, 10 cis

5
6

, 10 cis

3
2


Slide 5

Ex 3) Graph the five fifth roots of 32.
2 cis

1
5

5

32  2

 

0  2 k

2 cis

4
5

5
0,

2
5

,

2

4
5

,

6
5

,

8
5
2 cis 0

2 cis

6
5

2 cis

8
5


Slide 6

The various nth roots of 1 are called the roots of unity.
1 in polar is: 1 cis 0
so nth roots of unity are of the form
1 cis

2 k

for k = 0, 1, 2, …, n – 1
n is nth root

n

Ex 4) Find the three cube roots of unity and locate them on complex
plane.
r = 1 and they are spaced 23 rad apart
1 cis 0

1 cis

2

1 cis

3


1

1
2



3

4
3



i

2

1
2



3

i

2

Ex 5) You can use these answers to find the cube root of 8
*multiply #4 answers by 3 8  2
2

1 

3i

1

3i


Slide 7

The complex roots theorem provides a connection between the roots of a
complex number and the zeros of a polynomial.
Ex 6) Find all solutions of the equation
x3 + 2 = 2i
x3 = –2 + 2i
(aka find 3 roots of –2 + 2i)
polar r  (  2) 2  (2) 2  8
 

3

8 cis

4
8 8

3

1 1

2 3

3
4

1
6

8 

6

1 3

 2
 4






3 4
4
4
3
4
3

8

3



12
6

8 cis


4

,

6

8 cis

11
12

,

6

8 cis

8

3

12

12



16 
12

19 

11

19 

12

12

12


Slide 8

Homework
#1007

Pg 526 #1, 2, 7, 8, 12, 13, 16, 17, 19, 23,
27, 42, 43, 44