The document provides steps to find roots of complex numbers in trigonometric form. It gives 4 examples: 1) Finding cube roots of 4 4 3z i   2) Finding fourth roots of 81 3) Finding cube roots of 1z i  4) Finding fourth roots of  0 0 81 cos240 sin 240i For each, it calculates the modulus and arguments of the roots based on dividing the original by the root index.

1. 4.3 Roots of Complex Numbers Solutions
1. Find the cube roots of 4 4 3z i   . Write your answer in trigonometric form using radian measures for
the arguments.
First we will put 4 4 3z i   in trigonometric form.
   
222 2
4 4 3 16 48 64 8r z a b         
 1 14 3
tan tan 3
4 3

  
 
       
BUT we can see that our complex number in quadrant II so OUR angle
must have the same reference angle as
3

 . The reference angle is
3

so the argument for our complex number
must be
2
3

.
2 2
4 4 3 8 cos sin
3 3
z i i
  
     
 
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
3 3 9
 


Fourth, to get the remaining arguments of the next 2 cube roots we just add
2
"3"

to the argument of the first cube
root….
2 2 8
9 3 9
  
  and to get the argument of the last cube root we add
2
"3"

to the argument of the second
cube root….
8 2 14
9 3 9
  
  .
So the 3 cube roots of 4 4 3z i   are as follows…
2 2
2 cos sin
9 9
8 8
2 cos sin
9 9
14 14
2 cos sin
9 9
i
i
i
 
 
 
    
    
    
    
    
    
    
    
    
Note: I will leave these complex numbers written in trigonometric form because none of the arguments have
“nice” cosine or sine values.

2. 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.  81 81 cos sini   
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
2 2 3 2 3 2
3 cos sin 3
4 4 2 2 2 2
3 3 2 2 3 2 3 2
3 cos sin 3
4 4 2 2 2 2
5 5 2 2 3 2 3 2
3 cos sin 3
4 4 2 2 2 2
7
3 cos
4
i i i
i i i
i i i
 
 
 

     
                  
     
                    
      
                        
 

 
7 2 2 3 2 3 2
sin 3
4 2 2 2 2
i i 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.  1 11
tan tan 1
1 4

   
   
 
1z i 
3. Find the cube roots of 1z i  . Leave your answer in trigonometric form using radian measures for the
arguments.
1. Rewrite in trigonometric form.    
2 22 2
1 1 1 1 2r z a b       
So 1 2 cos sin
4 4
z i i
     
       
    
2. The modulus for each of the three cube roots will be 3 6
2 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
2 2
"3" 3
 
 to each subsequent argument
so the second cube root will have argument
2 9 3
12 3 12 4
   
   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
are as follows….
6
6
6
2 cos sin
12 12
3 3
2 cos sin
4 4
17 17
2 cos sin
12 12
i
i
i
 
 
 
    
    
    
    
    
    
    
    
    

4. 4. Find the fourth roots of  0 0
81 cos240 sin 240i . 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.
0
0240
60
4

4. The remaining three arguments will be found by continuously adding
0
0360
90
"4"

So the 2nd
fourth root will have argument = 0 0 0
60 90 150  and the 3rd
fourth root will have argument
0 0 0
150 90 240  and finally the 4th
fourth root will have argument 0 0 0
240 90 330  .
5. So the 4 fourth roots of  0 0
81 cos240 sin 240i
are as follows…….
    
    
    
    
0 0
0 0
0 0
0 0
3 1 3 3 1
3 cos 60 sin 60 3
2 2 2 2
1 3 1 3
3 cos 150 sin 150 3
2 2 2 2
1 3 1 3
3 cos 240 sin 240 3
2 2 2 2
3 1 3 3 1
3 cos 330 sin 330 3
2 2 2 2
i i i
i i i
i i i
i i i
 
       
 
  
           
  
  
            
  
 
       
 