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6.5 Trig. Form of a Complex Number
z  a  bi  a  b
2
Ex.
Z = -2 + 5i
2
 2,5
z  (2) 2  52
z  29
-2
5
Trigonometric Form of a Complex Number
z  r cos  i sin  
r = the hypotenuse and theta = the angle.
Write in trig. form
Ex.
z  2  2 3i
2
2 3
Find r (the hyp) & theta
r=4
opp. 2 3
tan  


adj.
2
3
1
  60 0 ref. angle
4
 
or 240
3
4
4
z  4(cos
 i sin
)
3
3
Write the complex number in standard form a + bi.
Ex.
  
  
z  8 cos    i sin   
 3 
  3
1
3 
z  2 2 
i
2 2 
z  2  6i
Multiplication of Complex Numbers
z1 z2  r1r2 cos1   2   i sin 1   2 
Find the Product if
11
11 

2
2 

z1  2 cos
 i sin
 & z2  8 cos 6  i sin 6 


3
3 

  2 11 
 2 11 
z1 z2  16cos    i sin   
6 
6 
 3
  3
5
5 

 16cos
 i sin
 16 0  i(1)  16i

2 
2


2

Dividing Complex Numbers
z1 r1
 cos1   2   i sin 1   2 
z 2 r2

z1
z2
Divide
z1  24 cos 300  i sin 300
o
o
z2  8cos 75  i sin 75 
o
o


z1 24
o
o
o
o

cos(300  75 )  i sin( 300  75 )
z2
8

 3 cos 225  i sin 225
o

2 
2 
  i 

 3 
  2 
2

 

o

3 2 3 2


i
2
2

DeMoivre’s Theorem and nth Roots
z  rcos  isin  
n
n
 r cos n  i sin n 
n
Ex.
Find
(1  3i )
12
Imagine how much fun it would be to multiply this
example out 12 times. It would take forever. Using
DeMovre’s Theorem, however, makes it short and simple.
First, convert to trigonometric form.
12
 
2
2 
(1  3i )  2 cos  i sin 
3
3 
 
2
2 
12 
 2 cos12
 i sin 12   4096cos 8  i sin 8 
3
3 

12
 40961  0i   4096
nth Roots of a Complex Number
For a positive integer n, the complex number
z = r(cos x + i sin x)
has exactly n distinct nth roots given by
n
  2k
  2k 

r  cos
 i sin

n
n 

where k = 0, 1, 2, . . . , n - 1
WATCH THE EXAMPLES
CAREFULLY!!!
Ex.
Find all the sixth roots of 1.
First, write 1 + 0i in trig. form.
1  0i  6
1 + 0i is over 1 and up 0. Therefore, 1 is the
hypotenuse and theta is 0o.
1cos0
o
 isin 0
o

1
6
0  2k
0  2k 

 1 cos
 i sin

6
6 

6
Now, we plug in 0, 1, 2, 3, 4, and 5 for k to find
our six roots.
1
1cos 0  i sin 0  1

 1
3

1 cos  i sin   
i
3
3 2 2

2
2 
1
3

1 cos
 i sin
i
 
3
3 
2 2

1cos   i sin    1
4
4 
1
3

1 cos
 i sin
i
 
3
3 
2 2

5
5  1
3

1 cos
 i sin
i
 
3
3  2 2

Find the three cube roots of z = -2+2i
Again, first convert to trig form.
2  2i
1
3

 8 cos135
o
 isin135
o

1
1  3 

135

360k
135

360k

8 2  cos
 isin

  

3
3
1
For k = 0, 1, and 2, the roots are:

2 cos 45  i sin 45
o

2 cos 285
o

 i sin 285 
2 cos165o  i sin 165o
o

o
3