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
36. d) Solve z 5 z 4 z 3 z 2 z 1 0
Now z 6 1 z 1z 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 1z 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 1z 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 1z 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 1z 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 1z 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 1z 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
2k
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
2k
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
2k
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
2k
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
2k
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
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