1. Factorising Complex
Expressions

2. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.

3. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;

4. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field

5. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field

6. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
NOTE: odd ordered polynomials must have a real root

7. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
NOTE: odd ordered polynomials must have a real root
e.g . i  x 2  2 x  2

8. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
NOTE: odd ordered polynomials must have a real root
e.g . i  x 2  2 x  2   x  12  1

9. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
NOTE: odd ordered polynomials must have a real root
e.g . i  x 2  2 x  2   x  12  1
  x  1  i  x  1  i 

10. ii  z 4  z 2  12  0

11. ii  z 4  z 2  12  0
z 2
 3z 2  4   0

12. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2

13. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers

14. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0

15. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers

16. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers
z   3 or z  2i

17. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers
z   3 or z  2i
iii  Factorise 2 x 3  3x 2  8 x  5

18. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers
z   3 or z  2i
iii  Factorise 2 x 3  3x 2  8 x  5
as it is a cubic it must have a real factor

19. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers
z   3 or z  2i
iii  Factorise 2 x 3  3x 2  8 x  5
as it is a cubic it must have a real factor
3 2
 1  1  1  1
P    2    3    8    5
 2  2  2  2
1 3
   45
4 4
0

20. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers
z   3 or z  2i
iii  Factorise 2 x 3  3x 2  8 x  5
as it is a cubic it must have a real factor
3 2
 1  1  1  1
P    2    3    8    5
 2  2  2  2
1 3
   45
4 4
0
 2 x 3  3 x 2  8 x  5  2 x  1x 2  2 x  5

21. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers
z   3 or z  2i
iii  Factorise 2 x 3  3x 2  8 x  5
as it is a cubic it must have a real factor
3 2
 1  1  1  1
P    2    3    8    5
 2  2  2  2
1 3
   45
4 4
0
 2 x 3  3 x 2  8 x  5  2 x  1x 2  2 x  5

 2 x  1  x  1  4
2


22. ii  z 4  z 2  12  0
z  3z  4  0
2 2
z  3 z  3z  4  0
2
factorised over Real numbers
z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers
z   3 or z  2i
iii  Factorise 2 x 3  3x 2  8 x  5
as it is a cubic it must have a real factor
3 2
 1  1  1  1
P    2    3    8    5
 2  2  2  2
1 3
   45
4 4
0
 2 x 3  3 x 2  8 x  5  2 x  1x 2  2 x  5

 2 x  1  x  1  4
2

 2 x  1 x  1  2i  x  1  2i 

23. iv  z 5  z 4  z 3  z 2  z  1  0

24. iv  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

25. iv  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 

26. iv  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

27. iv  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

28. iv  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

29. iv  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

30. iv  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

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

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

33. v 1996 HSC
2 2
Let   cos  i sin
9 9
a) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1
2k 
z  cis 
 9  k  0,1,2,3,4,5,6,7,8
 

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

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

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

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

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

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

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

41.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8

42.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B 
sin A  sin B  2sin   cos  
 2   2 

43.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)

44.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos   
 2   2 

45.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos
(2 cos half sum
  
 2   2  cos half diff)

46.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos
(2 cos half sum
  
 2   2  cos half diff)
 A B   A B 
cos A  cos B  2sin   sin  
 2   2 

47.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos
(2 cos half sum
  
 2   2  cos half diff)
 A B   A B  (minus 2 sine half sum
cos A  cos B  2sin   sin  
 2   2  sine half diff)

48.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos
(2 cos half sum
  
 2   2  cos half diff)
 A B   A B  (minus 2 sine half sum
cos A  cos B  2sin   sin  
 2   2  sine half diff)
   2   3   4   5   6   7   8  1

49.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos
(2 cos half sum
  
 2   2  cos half diff)
 A B   A B  (minus 2 sine half sum
cos A  cos B  2sin   sin  
 2   2  sine half diff)
   2   3   4   5   6   7   8  1
roots appear in conjugate pairs

50.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos
(2 cos half sum
  
 2   2  cos half diff)
 A B   A B  (minus 2 sine half sum
cos A  cos B  2sin   sin  
 2   2  sine half diff)
   2   3   4   5   6   7   8  1
roots appear in conjugate pairs
2 4 6 8
2cos  2cos  2cos  2cos  1
9 9 9 9

51.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos
(2 cos half sum
  
 2   2  cos half diff)
 A B   A B  (minus 2 sine half sum
cos A  cos B  2sin   sin  
 2   2  sine half diff)
   2   3   4   5   6   7   8  1
roots appear in conjugate pairs
2 4 6 8
2cos  2cos  2cos  2cos  1
9 9 9 9
2 4 6 8 1
cos  cos  cos  cos 
9 9 9 9 2

52.  2 4 1
c) Hence show that cos cos cos 
9 9 9 8
 A B   A B  (2 sine half sum
sin A  sin B  2sin   cos  
 2   2  cos half diff)
 A  B  cos A  B 
cos A  cos B  2 cos
(2 cos half sum
  
 2   2  cos half diff)
 A B   A B  (minus 2 sine half sum
cos A  cos B  2sin   sin  
 2   2  sine half diff)
   2   3   4   5   6   7   8  1
roots appear in conjugate pairs
2 4 6 8
2cos  2cos  2cos  2cos  1
9 9 9 9
2 4 6 8 1
cos  cos  cos  cos 
9 9 9 9 2
3  7  1
2cos cos  2cos cos  
9 9 9 9 2

53. 3  7  1
2cos cos  2cos cos  
9 9 9 9 2

54. 3  7  1
2cos cos  2cos cos  
9 9 9 9 2
 3 7  1
cos  cos  cos 
9 9 9  4

55. 3  7  1
2cos cos  2cos cos  
9 9 9 9 2
 3 7  1
cos  cos  cos 
9 9 9  4
 5 2  1
cos  2cos cos 
9 9 9  4

56. 3  7  1
2cos cos  2cos cos  
9 9 9 9 2
 3 7  1
cos  cos  cos 
9 9 9  4
 5 2  1
cos  2cos cos 
9 9 9  4
 2 5 1
cos cos cos 
9 9 9 8

57. 3  7  1
2cos cos  2cos cos  
9 9 9 9 2
 3 7  1
cos  cos  cos 
9 9 9  4
 5 2  1
cos  2cos cos 
9 9 9  4
 2 5 1
cos cos cos 
9 9 9 8
5 4
But cos   cos
9 9

58. 3  7  1
2cos cos  2cos cos  
9 9 9 9 2
 3 7  1
cos  cos  cos 
9 9 9  4
 5 2  1
cos  2cos cos 
9 9 9  4
 2 5 1
cos cos cos 
9 9 9 8
5 4
But cos   cos
9 9
 2 4 1
 cos cos cos 
9 9 9 8

59. 3  7  1
2cos cos  2cos cos  
9 9 9 9 2
 3 7  1
cos  cos  cos 
9 9 9  4
 5 2  1
cos  2cos cos 
9 9 9  4
 2 5 1
cos cos cos 
9 9 9 8
5 4
But cos   cos
9 9
 2 4 1
 cos cos cos 
9 9 9 8
 2 4 1
cos cos cos 
9 9 9 8

60. OR
z9 1
  z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 

61. OR
z9 1
  z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
 2  4 
  z  1  z 2  2cos z  1 z 2  2cos z  1
 9  9 
 2 6  2 8 
 z  2cos z  1 z  2cos z  1
 9  9 

62. OR
z9 1
  z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
 2  4 
  z  1  z 2  2cos z  1 z 2  2cos z  1
 9  9 
 2 6  2 8 
 z  2cos z  1 z  2cos z  1
 9  9 
 2 2  2 4 
  z  1  z  2cos z  1 z  2cos z  1
 9  9 
 8 
 z 2  z  1  z 2  2cos
 9
z  1


63. OR
z9 1
  z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
 2  4 
  z  1  z 2  2cos z  1 z 2  2cos z  1
 9  9 
 2 6  2 8 
 z  2cos z  1 z  2cos z  1
 9  9 
 2 2  2 4 
  z  1  z  2cos z  1 z  2cos z  1
 9  9 
 8 
 z 2  z  1  z 2  2cos
 9
z  1

Let z  i

64. OR
z9 1
  z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
 2  4 
  z  1  z 2  2cos z  1 z 2  2cos z  1
 9  9 
 2 6  2 8 
 z  2cos z  1 z  2cos z  1
 9  9 
 2 2  2 4 
  z  1  z  2cos z  1 z  2cos z  1
 9  9 
 8 
 z 2  z  1  z 2  2cos
 9
z  1

Let z  i
 2  4   8 
i 9  1   i  1  2cos i  2cos i   i   2cos i
 9  9   9 

65.  2  4   8 
i  1   i  1  2cos
9
i  2cos i   i   2cos i
 9  9   9 

66.  2  4   8 
i  1   i  1  2cos
9
i  2cos i   i   2cos i
 9  9   9 
 2  4  8 
i  1  i  i  1  2cos
4
 2cos  2cos 
 9  9  9 

67.  2  4   8 
i  1   i  1  2cos
9
i  2cos i   i   2cos i
 9  9   9 
 2  4  8 
i  1  i  i  1  2cos
4
 2cos  2cos 
 9  9  9 
2 4 8
1  8cos cos cos
9 9 9

68.  2  4   8 
i  1   i  1  2cos
9
i  2cos i   i   2cos i
 9  9   9 
 2  4  8 
i  1  i  i  1  2cos
4
 2cos  2cos 
 9  9  9 
2 4 8
1  8cos cos cos
9 9 9
2 4  
1  8cos cos   cos 
9 9  9

69.  2  4   8 
i  1   i  1  2cos
9
i  2cos i   i   2cos i
 9  9   9 
 2  4  8 
i  1  i  i  1  2cos
4
 2cos  2cos 
 9  9  9 
2 4 8
1  8cos cos cos
9 9 9
2 4  
1  8cos cos   cos 
9 9  9
 2 4 1
cos cos cos 
9 9 9 8

70.  2  4   8 
i  1   i  1  2cos
9
i  2cos i   i   2cos i
 9  9   9 
 2  4  8 
i  1  i  i  1  2cos
4
 2cos  2cos 
 9  9  9 
2 4 8
1  8cos cos cos
9 9 9
2 4  
1  8cos cos   cos 
9 9  9
 2 4 1
cos cos cos 
9 9 9 8
Exercise 4J; 1 to 4, 7ac