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# X2 t01 06 geometrical representation (2013)

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### X2 t01 06 geometrical representation (2013)

1. 1. Geometrical Representation of Complex Numbers
2. 2. Geometrical Representation of Complex Numbers Complex numbers can be represented on the Argand Diagram as vectors.
3. 3. Geometrical Representation of Complex Numbers Complex numbers can be represented on the Argand Diagram as vectors. y x
4. 4. Geometrical Representation of Complex Numbers Complex numbers can be represented on the Argand Diagram as vectors. y z  x  iy x
5. 5. Geometrical Representation of Complex Numbers Complex numbers can be represented on the Argand Diagram as vectors. y z  x  iy x
6. 6. Geometrical Representation of Complex Numbers Complex numbers can be represented on the Argand Diagram as vectors. y z  x  iy x The advantage of using vectors is that they can be moved around the Argand Diagram
7. 7. Geometrical Representation of Complex Numbers Complex numbers can be represented on the Argand Diagram as vectors. y z  x  iy x The advantage of using vectors is that they can be moved around the Argand Diagram
8. 8. Geometrical Representation of Complex Numbers Complex numbers can be represented on the Argand Diagram as vectors. y z  x  iy x The advantage of using vectors is that they can be moved around the Argand Diagram No matter where the vector is placed its length (modulus) and the angle made with the x axis (argument) is constant
9. 9. Geometrical Representation of Complex Numbers Complex numbers can be represented on the Argand Diagram as vectors. y z  x  iy A vector always x represents HEAD minus TAIL The advantage of using vectors is that they can be moved around the Argand Diagram No matter where the vector is placed its length (modulus) and the angle made with the x axis (argument) is constant
11. 11. Addition / Subtraction y x
12. 12. Addition / Subtraction y z1 x
13. 13. Addition / Subtraction y z2 z1 x
14. 14. Addition / Subtraction y z2 z1 x To add two complex numbers, place the vectors “head to tail”
15. 15. Addition / Subtraction y z1  z 2 z2 z1 x To add two complex numbers, place the vectors “head to tail”
16. 16. Addition / Subtraction y z1  z 2 z2 z1 x To add two complex numbers, place the vectors “head to tail”
17. 17. Addition / Subtraction y z1  z 2 z2 z1 x To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector)
18. 18. Addition / Subtraction y z1  z 2 z2 z1 x z1  z 2 To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector)
19. 19. Addition / Subtraction y z1  z 2 z2 z1 x z1  z 2 To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector)
20. 20. Addition / Subtraction y z1  z 2 z2 NOTE : the parallelogram formed by adding vectors has two diagonals; z1 x z1  z 2 z1  z 2 and z1  z 2 To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector)
21. 21. Addition / Subtraction y z1  z 2 z2 NOTE : the parallelogram formed by adding vectors has two diagonals; z1 x z1  z 2 z1  z 2 and z1  z 2 To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector)
22. 22. Addition / Subtraction y z1  z 2 z2 NOTE : the parallelogram formed by adding vectors has two diagonals; z1 x z1  z 2 z1  z 2 and z1  z 2 To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector) Trianglar Inequality In any triangle a side will be shorter than the sum of the other two sides
23. 23. Addition / Subtraction y z1  z 2 z2 NOTE : the parallelogram formed by adding vectors has two diagonals; z1 x z1  z 2 z1  z 2 and z1  z 2 To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector) Trianglar Inequality In any triangle a side will be shorter than the sum of the other two sides In ABC ; AC  AB  BC
24. 24. Addition / Subtraction y z1  z 2 z2 NOTE : the parallelogram formed by adding vectors has two diagonals; z1 x z1  z 2 z1  z 2 and z1  z 2 To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector) Trianglar Inequality In any triangle a side will be shorter than the sum of the other two sides In ABC ; AC  AB  BC (equality occurs when AC is a straight line)
25. 25. Addition / Subtraction y z1  z 2 z2 NOTE : the parallelogram formed by adding vectors has two diagonals; z1 x z1  z 2 z1  z 2 and z1  z 2 To add two complex numbers, place the vectors “head to tail” To subtract two complex numbers, place the vectors “head to head” (or add the negative vector) Trianglar Inequality In any triangle a side will be shorter than the sum of the other two sides In ABC ; AC  AB  BC (equality occurs when AC is a straight line) z1  z 2  z1  z 2
26. 26. Addition If a point A represents z1 and point B represents z 2 then point C representing z1  z 2 is such that the points OACB form a parallelogram.
27. 27. Addition If a point A represents z1 and point B represents z 2 then point C representing z1  z 2 is such that the points OACB form a parallelogram. Subtraction If a point D represents  z1 and point E represents z 2  z1 then the points ODEB form a parallelogram. Note: AB  z 2  z1 arg z 2  z1   
28. 28. e.g .1995 y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w
29. 29. e.g .1995 y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w
30. 30. e.g .1995 y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w The length of OP is z
31. 31. e.g .1995 y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w The length of OP is z The length of OQ is w
32. 32. e.g .1995 y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w The length of OP is z The length of OQ is w The length of PQ is z  w
33. 33. e.g .1995 y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w The length of OP is z Using the triangular inequality on OPQ The length of OQ is w zw  z  w The length of PQ is z  w
34. 34. e.g .1995 y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w The length of OP is z Using the triangular inequality on OPQ The length of OQ is w zw  z  w The length of PQ is z  w (ii) Construct the point R representing z + w, What can be said about the quadrilateral OPRQ?
35. 35. e.g .1995 R y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w The length of OP is z Using the triangular inequality on OPQ The length of OQ is w zw  z  w The length of PQ is z  w (ii) Construct the point R representing z + w, What can be said about the quadrilateral OPRQ?
36. 36. e.g .1995 R y Q P x O The diagram shows a complex plane with origin O. The points P and Q represent the complex numbers z and w respectively. Thus the length of PQ is z  w i  Show that z  w  z  w The length of OP is z Using the triangular inequality on OPQ The length of OQ is w zw  z  w The length of PQ is z  w (ii) Construct the point R representing z + w, What can be said about the quadrilateral OPRQ? OPRQ is a parallelogram
37. 37. w z iii  If z  w  z  w , what can be said about ?
38. 38. w z iii  If z  w  z  w , what can be said about ? zw  zw
39. 39. w z i.e. diagonals in OPRQ are = iii  If z  w  z  w , what can be said about ? zw  zw
40. 40. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ?
41. 41. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ? arg w  arg z   2
42. 42. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ? arg w  arg z   2 w  arg  z 2
43. 43. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ? arg w  arg z   2 w  arg  z 2 w  is purely imaginary z
44. 44. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ? arg w  arg z   2 w  arg  z 2 Multiplication w  is purely imaginary z
45. 45. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ? arg w  arg z   2 w  arg  z 2 Multiplication z1 z 2  z1 z 2 w  is purely imaginary z
46. 46. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ? arg w  arg z   2 w  arg  z 2 Multiplication z1 z 2  z1 z 2 w  is purely imaginary z arg z1 z 2  arg z1  arg z 2
47. 47. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ? arg w  arg z   2 w  arg  z 2 Multiplication z1 z 2  z1 z 2 w  is purely imaginary z arg z1 z 2  arg z1  arg z 2 r1cis1  r2 cis 2  r1r2 cis1   2 
48. 48. w z z  w  z  w i.e. diagonals in OPRQ are =  OPRQ is a rectangle iii  If z  w  z  w , what can be said about ? arg w  arg z   2 w  arg  z 2 Multiplication z1 z 2  z1 z 2 w  is purely imaginary z arg z1 z 2  arg z1  arg z 2 r1cis1  r2 cis 2  r1r2 cis1   2  i.e. if we multiply z1 by z 2 , the vector z1 is rotated anticlockwise by  2 and its length is multiplied by r2
49. 49. If we multiply z1 by cis the vector OA will rotate by an angle of  in an anti-clockwise direction. If we multiply by rcis it will also multiply the length of OA by a factor of r
50. 50. If we multiply z1 by cis the vector OA will rotate by an angle of  in an anti-clockwise direction. If we multiply by rcis it will also multiply the length of OA by a factor of r   Note: cos  i sin  i  iz1 will rotate OA anticlockwise 90 degrees. 2 2
51. 51. If we multiply z1 by cis the vector OA will rotate by an angle of  in an anti-clockwise direction. If we multiply by rcis it will also multiply the length of OA by a factor of r   Note: cos  i sin  i  iz1 will rotate OA anticlockwise 90 degrees. 2 2 Multiplication by i is a rotation anticlockwise by  2
52. 52. If we multiply z1 by cis the vector OA will rotate by an angle of  in an anti-clockwise direction. If we multiply by rcis it will also multiply the length of OA by a factor of r   Note: cos  i sin  i  iz1 will rotate OA anticlockwise 90 degrees. 2 2 Multiplication by i is a rotation anticlockwise by REMEMBER:  2 A vector is HEAD minus TAIL
53. 53. y B A  C O D(1) x
54. 54. y    DC  DA  i B A  C O D(1) x
55. 55. y B A  C O D(1) x    DC  DA  i C  1    1 i
56. 56. y B A  C O D(1) x    DC  DA  i C  1    1 i C  1    1 i  1  i   i
57. 57. y B A  C O  B  A  DC D(1) x    DC  DA  i C  1    1 i C  1    1 i  1  i   i
58. 58. y B A  C O  B  A  DC B    C 1 D(1) x    DC  DA  i C  1    1 i C  1    1 i  1  i   i
59. 59. y B A  C O  B  A  DC B    C 1 B      1 i  i  1  i   D(1) x    DC  DA  i C  1    1 i C  1    1 i  1  i   i
60. 60. y B A  C O  B  A  DC B    C 1 B      1 i  i  1  i   OR D(1) x    DC  DA  i C  1    1 i C  1    1 i  1  i   i
61. 61. y B A  C O  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA D(1) x    DC  DA  i C  1    1 i C  1    1 i  1  i   i
62. 62. y B A  C O D(1)  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA B  1  i   i  (  1) x    DC  DA  i C  1    1 i C  1    1 i  1  i   i
63. 63. y B A  C O D(1)  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA B  1  i   i  (  1)  i  1  i   x    DC  DA  i C  1    1 i C  1    1 i  1  i   i
64. 64. y    DC  DA  i C  1    1 i B A  C O D(1)  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA B  1  i   i  (  1)  i  1  i   C  1    1 i  1  i   i x OR
65. 65. y    DC  DA  i C  1    1 i B A  C O D(1)  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA B  1  i   i  (  1)  i  1  i   C  1    1 i  1  i   i x OR      DB  2cis  DA 4
66. 66. y    DC  DA  i C  1    1 i B A  C O D(1)  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA B  1  i   i  (  1)  i  1  i   C  1    1 i  1  i   i x OR      DB  2cis  DA 4    B  1  2  cos  i sin  (  1) 4 4 
67. 67. y    DC  DA  i C  1    1 i B A  C O D(1)  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA B  1  i   i  (  1)  i  1  i   C  1    1 i  1  i   i x OR      DB  2cis  DA 4    B  1  2  cos  i sin  (  1) 4 4  B  1  i  (  1)  1
68. 68. y    DC  DA  i C  1    1 i B A  C O D(1)  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA B  1  i   i  (  1)  i  1  i   C  1    1 i  1  i   i x OR      DB  2cis  DA 4    B  1  2  cos  i sin  (  1) 4 4  B  1  i  (  1)  1    1  i  i  1
69. 69. y    DC  DA  i C  1    1 i B A  C O D(1)  B  A  DC B    C 1 B      1 i  i  1  i   OR   B  C  DA B  1  i   i  (  1)  i  1  i   C  1    1 i  1  i   i x OR      DB  2cis  DA 4    B  1  2  cos  i sin  (  1) 4 4  B  1  i  (  1)  1    1  i  i  1  i  1  i  
70. 70. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number 
71. 71. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?
72. 72. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?     OC  OA  2i
73. 73. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?     OC  OA  2i C  2i
74. 74. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?     OC  OA  2i C  2i (ii) What complex number corresponds to the point of intersection D of the diagonals OB and AC?
75. 75. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?     OC  OA  2i C  2i (ii) What complex number corresponds to the point of intersection D of the diagonals OB and AC? diagonals bisect in a rectangle
76. 76. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?     OC  OA  2i C  2i (ii) What complex number corresponds to the point of intersection D of the diagonals OB and AC? diagonals bisect in a rectangle  D  midpoint of AC
77. 77. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?     OC  OA  2i C  2i (ii) What complex number corresponds to the point of intersection D of the diagonals OB and AC? diagonals bisect in a rectangle  D  midpoint of AC AC D 2
78. 78. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?     OC  OA  2i C  2i (ii) What complex number corresponds to the point of intersection D of the diagonals OB and AC?   2i D diagonals bisect in a rectangle 2  D  midpoint of AC AC D 2
79. 79. e.g.2000  B C y A  x O In the Argand Diagram, OABC is a rectangle, where OC = 2OA. The vertex A corresponds to the complex number  i  What complex number corresponds to C?     OC  OA  2i C  2i (ii) What complex number corresponds to the point of intersection D of the diagonals OB and AC?   2i D diagonals bisect in a rectangle 2  D  midpoint of AC 1  AC  D    i  D 2  2
80. 80. Examples
81. 81. Examples     OB  OA cis 3
82. 82. Examples     OB  OA cis 3  B  1cis 3
83. 83. Examples     OB  OA cis 3  B  1cis 3 1 3 B  i 2 2
84. 84. Examples     OB  OA cis 3  B  1cis 3 1 3 B  i 2 2     OD  OB  i
85. 85. Examples     OB  OA cis 3  B  1cis 3 1 3 B  i 2 2     OD  OB  i D  iB
86. 86. Examples     OB  OA cis 3  B  1cis 3 1 3 B  i 2 2     OD  OB  i D  iB D 3 1  i 2 2
87. 87. Examples     OB  OA cis 3  B  1cis 3 1 3 B  i 2 2     OD  OB  i D  iB D 3 1  i 2 2   C  B  OD
88. 88. Examples     OB  OA cis 3  B  1cis 3 1 3 B  i 2 2     OD  OB  i D  iB D 3 1  i 2 2   C  B  OD 1 3   3 1  C   i  i 2 2   2 2 
89. 89. Examples     OB  OA cis 3  B  1cis 3 1 3 B  i 2 2     OD  OB  i D  iB D 3 1  i 2 2   C  B  OD 1 3   3 1  C   i  i 2 2   2 2  1  3  1  3  C  i   2   2 
90. 90. (i )     AP  AO  i
91. 91. (i )     AP  AO  i P  A  i  O  A 
92. 92. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1 
93. 93. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1
94. 94. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1 P  1  i z1
95. 95. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1 P  1  i z1 (ii )     QB  BO  i
96. 96. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1 P  1  i z1     (ii ) QB  BO  i Q  B  i O  B 
97. 97. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1 P  1  i z1     (ii ) QB  BO  i Q  B  i O  B  Q  B  iB
98. 98. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1 P  1  i z1     (ii ) QB  BO  i Q  B  i O  B  Q  B  iB Q  1  i z 2
99. 99. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1 P  1  i z1     (ii ) QB  BO  i Q  B  i O  B  Q  B  iB Q  1  i z 2 M PQ 2
100. 100. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1 P  1  i z1     (ii ) QB  BO  i Q  B  i O  B  Q  B  iB Q  1  i z 2 PQ 2 1  i z1  1  i z2 M 2 M
101. 101. (i )     AP  AO  i P  A  i  O  A  P  z1  i0  z1  P  z1  iz1 P  1  i z1     (ii ) QB  BO  i Q  B  i O  B  Q  B  iB Q  1  i z 2 PQ 2 1  i z1  1  i z2 M 2  z1  z2    z1  z2 i M 2 M
102. 102. Cambridge Ex 1E; 2 to 8, 10 to 15, 18, 19, 21, 22 Terry Lee: Exercise 2.6