4.3 Roots of Complex Numbers Solutions i . Write your answer in trigonometric form using radian measures for 1. Find the cube roots of z the arguments.
First we will put z i in trigonometric form. r
 z
 a 2  b 2   
2 
2

16 48

64

8
  tan

1


4 3

4

  tan

1
 
 

3
BUT we can see that our complex number in quadrant II so OUR angle must have the same reference angle as


3
2
 must be .
3 z
   i

8 cos
2

3
 i sin
2

3
. The reference angle is

3
so the argument for our complex number
Second, we can now tell that the “modulus” for each of the 3 cube roots will be the cube root of the original complex number modulus. 3 8

2
Third we will calculate the argument for the first of the 3 cube roots by dividing the argument of the original complex number by “3” (because we are finding cube roots).
2



2
9

2

Fourth, to get the remaining arguments of the next 2 cube roots we just add root….
2


2


8

9 3 9 cube root….
and to get the argument of the last cube root we add
8


2

9 3

14
9

.
"3"
2

"3" to the argument of the first cube to the argument of the second
So the 3 cube roots of z i are as follows…

2 cos

2
 

9

 i sin

2
 

9




2 cos

8
 

9

 i sin

8
 

9




2 cos

14

9

 i sin

14

9



Note: I will leave these complex numbers written in trigonometric form because none of the arguments have
“nice” cosine or sine values.

2. Find the fourth roots of

81 . Write your answer in trigonometric form using radian measures for the arguments. Also write your answers in rectangular form.
1.
   i sin
 
2. The modulus for each of the 4 fourth roots will be 4 81

3
3. The first argument will be the original argument divided by “4” since we are looking for the fourth roots.

4
4. The remaining three arguments will be found by adding
2



"4" 2 to each subsequent argument so the second fourth root will have argument
 
3

4 2 4
and the third fourth root will have argument
3
 
5

4 2 4
and the final fourth root will have argument
5
 
7
4 2 4

So the 4 fourth roots of

81 are as follows

3 cos
 
 i sin
 



3


 2
2
2
2

 
3 2

3 2
2 2 i

3 cos

3
 

4

 i

3
  sin

4




3



2
2
 

3 cos

5
 
 4 
 i sin

5
 
 4 



3

 
2
2


2
2

  
3 2

3 2
2 2 i
2
2



  
3 2

3 2
2 2 i

3 cos 
7

4

 i sin

7
 
 4 



3


2
2


2
2



 
3 2

3 2
2 2 i
Note: Unlike problem #1 I chose to rewrite my complex roots in rectangular form because the arguments had
“nice” cosine and sine values.

3. Find the cube roots of z
  i . Leave your answer in trigonometric form using radian measures for the arguments.
1. Rewrite in trigonometric form. r
 z
 a
2  b
2     
2 
1 1 2
  tan

1
1
 
 tan

1
  

4
So z 1

2 cos

 
 i sin

 


2. The modulus for each of the three cube roots will be
3
2

6 2 (multiply the indices of the nested roots)
3. The first argument will be the original argument divided by “3” since we are looking for the cube roots.




3 4 12
4. The remaining two arguments will be found by adding so the second cube root will have argument
2


2

"3" 3


2


9


3

12 3 12 4 to each subsequent argument
and the third and final cube root will have argument equal to
3


2


17

4 3 12
5. So the three cube roots of z i
6

2 cos



12


 i sin



12



 are as follows…. 6

2 cos

3
 

4

 i sin

3
 

4



6

2 cos

17

12

 i sin

17

12




4. Find the fourth roots of

0  i sin 240
0

. Write your answer in rectangular form using no decimals.
1. Since this complex number is already written in trigonometric form we obviously don’t need to do this step.
2. The modulus for each of the 4 fourth roots will be 4 81

3
3. The first argument will be the original argument divided by “4” since we are looking for the fourth roots.
240
0
4

60
0
4. The remaining three arguments will be found by continuously adding
360
0
"4"

90
0
So the 2 nd
fourth root will have argument = 60
0 
90
0 
150
0 and the 3 rd
fourth root will have argument
150
0 
90
0 
240
0
and finally the 4 th
fourth root will have argument 240
0 
90
0 
330
0
.
5. So the 4 fourth roots of

0  i sin 240 0
 are as follows…….

  
 i
  

0

 i

0
 

3


2
3
 
1
2




3 3
2

1
2 i

3


1
2 i


2
3




1 3 i
2 2
 
0

 i

0
 

3


1
2 i


2
3




1 3 i
2 2
 
0

 i

0
 

3


2
3
 
1
2




3 3
2

1
2 i