The document discusses nth roots of unity. It states that the solutions to equations of the form zn = ±1 are the nth roots of unity. These solutions form a regular n-sided polygon with vertices on the unit circle when placed on an Argand diagram. As an example, it shows that the solutions to z5 = 1 are the fifth roots of unity located at angles that are integer multiples of 2π/5 around the unit circle. It then proves that if ω is a root of z5 - 1 = 0, then ω2, ω3, ω4 and ω5 are also roots. Finally, it proves that 1 + ω + ω2 + ω3 + ω4 = 0.

1. nth Roots Of Unity  z
n
 1

2. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity

3. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity
When placed on an Argand Diagram, they form a regular n sided
polygon, with vertices on the unit circle.

4. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity
When placed on an Argand Diagram, they form a regular n sided
polygon, with vertices on the unit circle.
e.g . z 5  1

5. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity
When placed on an Argand Diagram, they form a regular n sided
polygon, with vertices on the unit circle.
e.g . z 5  1
 2 k  0 
z  cis 
 k  0, 1, 2
 5 

6. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity
When placed on an Argand Diagram, they form a regular n sided
polygon, with vertices on the unit circle.
e.g . z 5  1
 2 k  0 
z  cis 
 k  0, 1, 2
 5 
2
2
4
4
z  cis 0, cis
, cis 
, cis
, cis 
5
5
5
5

7. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity
When placed on an Argand Diagram, they form a regular n sided
polygon, with vertices on the unit circle.
e.g . z 5  1
 2 k  0 
z  cis 
 k  0, 1, 2
 5 
2
2
4
4
z  cis 0, cis
, cis 
, cis
, cis 
y
5
5
5
5
cis 0
x

8. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity
When placed on an Argand Diagram, they form a regular n sided
polygon, with vertices on the unit circle.
e.g . z 5  1
 2 k  0 
z  cis 
 k  0, 1, 2
 5 
2
2
4
4
z  cis 0, cis
, cis 
, cis
, cis 
y
2
cis
5
5
5
5
5
cis 0
x
2
cis 
5

9. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity
When placed on an Argand Diagram, they form a regular n sided
polygon, with vertices on the unit circle.
e.g . z 5  1
 2 k  0 
z  cis 
 k  0, 1, 2
 5 
2
2
4
4
z  cis 0, cis
, cis 
, cis
, cis 
y
2
4
cis
5
5
5
5
cis
5
5
cis 0
x
4
cis 
5
2
cis 
5

10. nth Roots Of Unity  z
n
 1
The solutions of equations of the form, z n  1, are the nth roots of unity
When placed on an Argand Diagram, they form a regular n sided
polygon, with vertices on the unit circle.
e.g . z 5  1
 2 k  0 
z  cis 
 k  0, 1, 2
 5 
2
2
4
4
z  cis 0, cis
, cis 
, cis
, cis 
y
2
4
cis
5
5
5
5
cis
5
5
cis 0
x
4
cis 
5
2
cis 
5

11. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.

12. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1
If  is a solution then  5  1

13. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1
If  is a solution then  5  1
5 5
    15
 5 is a solution
1

14. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1
If  is a solution then  5  1
5 5
    15
 5 is a solution
1
    
2 5
5 2
 12
1
 2 is a solution

15. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1
If  is a solution then  5  1
5 5
    15
 5 is a solution
1
    
2 5
5 2
 12
1
 2 is a solution
    
3 5
5 3
 13
1
 3 is a solution

16. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1
If  is a solution then  5  1
5 5
    15
 5 is a solution
1
    
2 5
5 2
 12
1
 2 is a solution
    
3 5
5 3
 13
1
 3 is a solution
    
4 5
5 4
 14
1
 4 is a solution

17. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1
If  is a solution then  5  1
5 5
    15
 5 is a solution
1
    
2 5
5 2
 12
1
 2 is a solution
    
3 5
5 3
 13
1
 3 is a solution
    
4 5
5 4
 14
1
 4 is a solution
Thus if  is a root then  2 ,  3 ,  4 and  5 are also roots

18. ii  Prove that 1     2   3   4  0

19. ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2
3
4
5
sum of the roots 

5
 1

20. ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2
3
4
5
5
OR   1  0
  11     2   3   4   0
1     2   3   4  0   1
sum of the roots 

 1
 NOTE :

 n

 1  0


  11     2     n1   0


5

21. ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2
3
4
sum of the roots 
5

5
OR   1  0
  11     2   3   4   0
1     2   3   4  0   1
OR
1   2  3  4 

a  r n  1
r 1
1 5  1
 1
 1
 NOTE :

 n

 1  0


  11     2     n1   0


5
GP: a  1, r   , n  5
0

5
1  0

22. ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2
3
4
5
sum of the roots 

5
OR   1  0
  11     2   3   4   0
1     2   3   4  0   1
OR
1   2  3  4 
a  r n  1
r 1
1 5  1
 1
 NOTE :

 n

 1  0


  11     2     n1   0


5
GP: a  1, r   , n  5

1  0

0
 1
iii  Find the quadratic equation whose roots are    4 and  2   3
5

23. ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2
3
4
5
sum of the roots 

5
OR   1  0
  11     2   3   4   0
1     2   3   4  0   1
OR
1   2  3  4 
a  r n  1
r 1
1 5  1
 1
 NOTE :

 n

 1  0


  11     2     n1   0


5
GP: a  1, r   , n  5

1  0

0
 1
iii  Find the quadratic equation whose roots are    4 and  2   3
4
2
3
     4  2   3 
       
 3  4  6  7
 1
 3  4    2
 1
5

24. ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2
3
4
5
sum of the roots 

5
OR   1  0
  11     2   3   4   0
1     2   3   4  0   1
OR
1   2  3  4 
a  r n  1
r 1
1 5  1
 1
 NOTE :

 n

 1  0


  11     2     n1   0


5
GP: a  1, r   , n  5

1  0

0
 1
iii  Find the quadratic equation whose roots are    4 and  2   3
4
2
3
     4  2   3 
       
 3  4  6  7
 1
 3  4    2
 equation is x 2  x  1  0
 1
5

25. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
(i ) Evaluate 1  

2 3

26. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
(i ) Evaluate 1  
   

2 3
3

27. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
(i ) Evaluate 1  
   
3
  3
 1

2 3

28. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1   
1  1 2
3
   
  3
 1

29. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1   
1  1 2
3
   
1 1
 2 
3
 
 
 1

1  
2

30. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1   
1  1 2
3
   
1 1
 2 
3
 
 
 1

1  
2
2
 2

1

31. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1   
1  1 2
3
   
1 1
 2 
3
 
 
 1
(iii) Form the cubic equation with roots 1, 1   , 1   2

1  
2
2
 2

1

32. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1   
1  1 2
3
   
1 1
 2 
3
 
 
 1
(iii) Form the cubic equation with roots 1, 1   , 1   2
 z  1z 2  2     2 z  1     2   3   0

1  
2
2
 2

1

33. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1   
1  1 2
3
   
1 1
 2 
3
 
 
 1
(iii) Form the cubic equation with roots 1, 1   , 1   2
 z  1z 2  2     2 z  1     2   3   0
 z  1z 2  z  1  0

1  
2
2
 2

1

34. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1   
1  1 2
3
   
1 1
 2 
3
 
 
 1
(iii) Form the cubic equation with roots 1, 1   , 1   2
 z  1z 2  2     2 z  1     2   3   0
 z  1z 2  z  1  0
z  z  z  z  z 1  0
z3  2z 2  2z 1  0
3
2
2

1  
2
2
 2

1

35. d) Solve z 5  z 4  z 3  z 2  z  1  0

36. d) Solve z 5  z 4  z 3  z 2  z  1  0
Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1

37. d) Solve z 5  z 4  z 3  z 2  z  1  0
Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
And z 6  1  0 has solutions;
 2 k 
z  cis 
 k  0, 1, 2,3
 6 

38. d) Solve z 5  z 4  z 3  z 2  z  1  0
Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
And z 6  1  0 has solutions;
 2 k 
z  cis 
 k  0, 1, 2,3
 6 
z5  z4  z3  z2 1  0

39. d) Solve z 5  z 4  z 3  z 2  z  1  0
Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
And z 6  1  0 has solutions;
 2 k 
z  cis 
 k  0, 1, 2,3
 6 
z5  z4  z3  z2 1  0
 z  1z 5  z 4  z 3  z 2  1
0
 z  1
z 6  1  0, z  1

40. d) Solve z 5  z 4  z 3  z 2  z  1  0
Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
And z 6  1  0 has solutions;
 2 k 
z  cis 
 k  0, 1, 2,3
 6 
z5  z4  z3  z2 1  0
 z  1z 5  z 4  z 3  z 2  1
0
 z  1
z 6  1  0, z  1


2
2
z  cis , cis  , cis
, cis 
, cis
3
3
3
3
1
3 1
3
1
3
1
3
z 
i, 
i,  
i,  
i, 1
2 2 2 2
2 2
2 2

41. e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer

42. e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1

43. e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1
 2k 
z  cis 

 9 
k
2 
z  cis 

9 

k  0,1,2,3,4,5,6,7,8

44. e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1
 2k 
z  cis 

 9 
k
2 
z  cis 

9 

z  k
k  0,1,2,3,4,5,6,7,8

45. e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1
 2k 
z  cis 
k  0,1,2,3,4,5,6,7,8

 9 
k
2 
z  cis 

9 

z  k
(ii) Prove that    2   3   4   5   6   7   8  1
z9 1  0

46. e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1
 2k 
z  cis 
k  0,1,2,3,4,5,6,7,8

 9 
k
2 
z  cis 

9 

z  k
(ii) Prove that    2   3   4   5   6   7   8  1
z9 1  0
1     2   3   4   5   6   7   8  0 sum of roots 

47. e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1
 2k 
z  cis 
k  0,1,2,3,4,5,6,7,8

 9 
k
2 
z  cis 

9 

z  k
(ii) Prove that    2   3   4   5   6   7   8  1
z9 1  0
1     2   3   4   5   6   7   8  0 sum of roots 
   2   3   4   5   6   7   8  1

48. 
2
4 1
(iii) Hence show that cos cos
cos

9
9
9 8

49. 
2
4 1
(iii) Hence show that cos cos
cos

9
9
9 8
9
z 1
  z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 

50. 
2
4 1
(iii) Hence show that cos cos
cos

9
9
9 8
9
z 1
  z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
2
4
 2
 2

z  1 z  2cos
z  1
  z  1  z  2cos
9
9



6
8
 2


z  2cos
z  1 z 2  2cos
z  1

9
9



2
4
 2
 2

z  1 z  2cos
z  1
  z  1  z  2cos
9
9



8
 2

2
 z  z  1  z  2cos 9 z  1



51. 
2
4 1
(iii) Hence show that cos cos
cos

9
9
9 8
9
z 1
  z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
2
4
 2
 2

z  1 z  2cos
z  1
  z  1  z  2cos
9
9



6
8
 2


z  2cos
z  1 z 2  2cos
z  1

9
9



2
4
 2
 2

z  1 z  2cos
z  1
  z  1  z  2cos
9
9



8
 2

2
 z  z  1  z  2cos 9 z  1


Let z  i
2

i  1   i  1  2cos
9

9
4

i  2cos
9

8
 
i   i   2cos
9
 

i


52. 2

i  1   i  1  2cos
9

9
4

i  2cos
9

8
 
i   i   2cos
9
 

i


53. 2

i  1   i  1  2cos
9

9
4

i  2cos
9

8
 
i   i   2cos
9
 
2 
4 
8 

i  1  i  i  1  2cos
 2cos
 2cos 
9 
9 
9 

4

i


54. 2

i  1   i  1  2cos
9

9
4

i  2cos
9

8
 
i   i   2cos
9
 
2 
4 
8 

i  1  i  i  1  2cos
 2cos
 2cos 
9 
9 
9 

2
4
8
cos
cos
1  8cos
9
9
9
4

i


55. 2

i  1   i  1  2cos
9

9
4

i  2cos
9

8
 
i   i   2cos
9
 
2 
4 
8 

i  1  i  i  1  2cos
 2cos
 2cos 
9 
9 
9 

2
4
8
cos
cos
1  8cos
9
9
9
2
4 

cos
1  8cos
  cos 
9
9 
9
4

i


56. 2

i  1   i  1  2cos
9

9
4

i  2cos
9

8
 
i   i   2cos
9
 
2 
4 
8 

i  1  i  i  1  2cos
 2cos
 2cos 
9 
9 
9 

2
4
8
cos
cos
1  8cos
9
9
9
2
4 

cos
1  8cos
  cos 
9
9 
9
4

2
4 1
cos cos
cos

9
9
9 8

i


57. 2

i  1   i  1  2cos
9

9
4

i  2cos
9

8
 
i   i   2cos
9
 

i

2 
4 
8 

i  1  i  i  1  2cos
 2cos
 2cos 
9 
9 
9 

2
4
8
cos
cos
1  8cos
9
9
9
2
4 

cos
1  8cos
  cos 
9
9 
9
4

2
4 1
cos cos
cos

9
9
9 8
Cambridge: Exercise 7C; 1 to 4, 5ac, 6, 7, 8, 9, 11
Patel: Exercise 4I; 1, 3, 5 (need 4b), 7
Patel: Exercise 4J; 1 to 4, 7ac