The document discusses representing complex numbers geometrically using vectors on an Argand diagram. It explains that complex numbers can be represented as vectors, with addition of complex numbers corresponding to placing the vectors head to tail and subtraction corresponding to head to head placement. It also discusses properties like the triangle inequality for complex number operations and representing multiplication as scaling of the vector.
3. Geometrical Representation of
Complex Numbers
Complex numbers can be represented on the Argand Diagram as vectors.
y
x
4. Geometrical Representation of
Complex Numbers
Complex numbers can be represented on the Argand Diagram as vectors.
y
z x iy
x
5. Geometrical Representation of
Complex Numbers
Complex numbers can be represented on the Argand Diagram as vectors.
y
z x iy
x
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. 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. 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. 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
14. Addition / Subtraction
y
z2
z1
x
To add two complex numbers, place the vectors “head to tail”
15. Addition / Subtraction
z1 z2
y
z2
z1
x
To add two complex numbers, place the vectors “head to tail”
16. Addition / Subtraction
z1 z2
y
z2
z1
x
To add two complex numbers, place the vectors “head to tail”
17. Addition / Subtraction
z1 z2
y
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. Addition / Subtraction
z1 z2
y
z2
z1
x
z1 z2
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. Addition / Subtraction
z1 z2
y
z2
z1
x
z1 z2
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. Addition / Subtraction
z1 z2
y
z2
NOTE :
z1 the parallelogram formed by adding
vectors has two diagonals;
x
z1 z2 and z1 z2
z1 z2
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. Addition / Subtraction
z1 z2
y
z2
NOTE :
z1 the parallelogram formed by adding
vectors has two diagonals;
x
z1 z2 and z1 z2
z1 z2
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. Addition / Subtraction
z1 z2
y
z2
NOTE :
z1 the parallelogram formed by adding
vectors has two diagonals;
x
z1 z2 and z1 z2
z1 z2
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. Addition / Subtraction
z1 z2
y
z2
NOTE :
z1 the parallelogram formed by adding
vectors has two diagonals;
x
z1 z2 and z1 z2
z1 z2
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. Addition / Subtraction
z1 z2
y
z2
NOTE :
z1 the parallelogram formed by adding
vectors has two diagonals;
x
z1 z2 and z1 z2
z1 z2
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. Addition / Subtraction
z1 z2
y
z2
NOTE :
z1 the parallelogram formed by adding
vectors has two diagonals;
x
z1 z2 and z1 z2
z1 z2
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 z2 z1 z2
26. Addition
If a point A represents z1 and point B
represents z 2 then point C representing
z1 z2 is such that the points OACB
form a parallelogram.
27. Addition
If a point A represents z1 and point B
represents z 2 then point C representing
z1 z2 is such that the points OACB
form a parallelogram.
Subtraction
If a point D represents z1
and point E represents z2 z1
then the points ODEB form a
parallelogram.
Note: AB z2 z1
arg z2 z1
28. e.g.1995 y
Q
P
O x
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. e.g.1995 y
Q
P
O x
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. e.g.1995 y
Q
P
O x
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. e.g.1995 y
Q
P
O x
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. e.g.1995 y
Q
P
O x
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. e.g.1995 y
Q
P
O x
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 triangula r inequality on OPQ
The length of OQ is w
zw z w
The length of PQ is z w
34. e.g.1995 y
Q
P
O x
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 triangula r inequality on OPQ
The length of OQ is w
zw 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. e.g.1995 R y
Q
P
O x
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 triangula r inequality on OPQ
The length of OQ is w
zw 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. e.g.1995 R y
Q
P
O x
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 triangula r inequality on OPQ
The length of OQ is w
zw 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. w
iii If z w z w , what can be said about ?
z
38. w
iii If z w z w , what can be said about ?
z
zw zw
39. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
40. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
41. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
arg w arg z
2
42. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
arg w arg z
2
w
arg
z 2
43. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
arg w arg z
2
w w
is purely imaginary
arg
z 2 z
44. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
arg w arg z
2
w w
is purely imaginary
arg
z 2 z
Multiplication
45. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
arg w arg z
2
w w
is purely imaginary
arg
z 2 z
Multiplication
z1 z2 z1 z2
46. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
arg w arg z
2
w w
is purely imaginary
arg
z 2 z
Multiplication
z1 z2 z1 z2 arg z1 z2 arg z1 arg z2
47. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
arg w arg z
2
w w
is purely imaginary
arg
z 2 z
Multiplication
z1 z2 z1 z2 arg z1 z2 arg z1 arg z2
r1cis1 r2cis 2 r1r2cis1 2
48. w
iii If z w z w , what can be said about ?
z
z w z w i.e. diagonals in OPRQ are =
OPRQ is a rectangle
arg w arg z
2
w w
is purely imaginary
arg
z 2 z
Multiplication
z1 z2 z1 z2 arg z1 z2 arg z1 arg z2
r1cis1 r2cis 2 r1r2cis1 2
i.e. if we multiply z1 by z2 , the vector z1 is rotated anticlockwise by 2
and its length is multiplied by r2
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. 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. 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
Multiplica tion by i is a rotation anticlockwise by
2
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
Multiplica tion by i is a rotation anticlockwise by
2
REMEMBER: A vector is HEAD minus TAIL
55. y DC DA i
B
C 1 1 i
A
C
O D(1) x
56. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
57. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
B A DC
58. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
B A DC
B C 1
59. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
B A DC
B C 1
B 1 i
i 1 i
60. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
B A DC
B C 1
B 1 i
i 1 i
OR
61. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
B A DC
B C 1
B 1 i
i 1 i
OR
B C DA
62. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
B A DC
B C 1
B 1 i
i 1 i
OR
B C DA
B 1 i i ( 1)
63. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
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
64. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
OR
B A DC
B C 1
B 1 i
i 1 i
OR
B C DA
B 1 i i ( 1)
i 1 i
65. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
OR
B A DC DB 2cis DA
4
B C 1
B 1 i
i 1 i
OR
B C DA
B 1 i i ( 1)
i 1 i
66. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
OR
B A DC DB 2cis DA
4
B C 1
B 1 2 cos i sin ( 1)
B 1 i 4 4
i 1 i
OR
B C DA
B 1 i i ( 1)
i 1 i
67. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
OR
B A DC DB 2cis DA
4
B C 1
B 1 2 cos i sin ( 1)
B 1 i 4 4
i 1 i B 1 i ( 1) 1
OR
B C DA
B 1 i i ( 1)
i 1 i
68. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
OR
B A DC DB 2cis DA
4
B C 1
B 1 2 cos i sin ( 1)
B 1 i 4 4
i 1 i B 1 i ( 1) 1
OR 1 i i 1
B C DA
B 1 i i ( 1)
i 1 i
69. y DC DA i
B
C 1 1 i
A
C C 1 1 i
O x 1 i i
D(1)
OR
B A DC DB 2cis DA
4
B C 1
B 1 2 cos i sin ( 1)
B 1 i 4 4
i 1 i B 1 i ( 1) 1
OR 1 i i 1
i 1 i
B C DA
B 1 i i ( 1)
i 1 i
70. e.g.2000 B
y
C
A
O x
In the Argand Diagram, OABC is a rectangle, where OC = 2OA.
The vertex A corresponds to the complex number
71. e.g.2000 B
y
C
A
O x
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. e.g.2000 B
y
C
A
O x
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. e.g.2000 B
y
C
A
O x
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. e.g.2000 B
y
C
A
O x
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. e.g.2000 B
y
C
A
O x
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. e.g.2000 B
y
C
A
O x
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. e.g.2000 B
y
C
A
O x
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
AC
D
2
78. e.g.2000 B
y
C
A
O x
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
diagonals bisect in a rectangle D
2
D midpoint of AC
AC
D
2
79. e.g.2000 B
y
C
A
O x
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
diagonals bisect in a rectangle D
2
D midpoint of AC
AC 1
D D i
2 2
83. Examples
OB OA cis
3
B 1cis
3
1 3
B i
2 2
84. Examples
OB OA cis
3
B 1cis
3
1 3
B i
2 2
OD OB i
85. Examples
OB OA cis
3
B 1cis
3
1 3
B i
2 2
OD OB i
D iB
86. Examples
OB OA cis
3
B 1cis
3
1 3
B i
2 2
OD OB i
D iB
3 1
D i
2 2
87. Examples
OB OA cis
3
B 1cis
3
1 3
B i
2 2
C B OD
OD OB i
D iB
3 1
D i
2 2
88. Examples
OB OA cis
3
B 1cis
3
1 3
B i
2 2
C B OD
OD OB i
1 3 3 1
D iB C i i
D
3 1
i 2 2 2 2
2 2
89. Examples
OB OA cis
3
B 1cis
3
1 3
B i
2 2
C B OD
OD OB i
1 3 3 1
D iB C i i
D
3 1
i 2 2 2 2
2 2 1 3 1 3
C i
2 2
93. (i) AP AO i
P A i O A
P z1 i0 z1
94. (i) AP AO i
P A i O A
P z1 i0 z1
P z1 iz1
95. (i) AP AO i
P A i O A
P z1 i0 z1
P z1 iz1
P 1 i z1
96. (i) AP AO i
(ii) QB BO i
P A i O A
P z1 i0 z1
P z1 iz1
P 1 i z1
97. (i) AP AO i
(ii) QB BO i
P A i O A Q B i O B
P z1 i0 z1
P z1 iz1
P 1 i z1
98. (i) AP AO i
(ii) QB BO i
P A i O A Q B i O B
P z1 i0 z1 Q B iB
P z1 iz1
P 1 i z1
99. (i) AP AO i
(ii) QB BO i
P A i O A Q B i O B
P z1 i0 z1 Q B iB
P z1 iz1 Q 1 i z2
P 1 i z1
100. (i) AP AO i
(ii) QB BO i
P A i O A Q B i O B
P z1 i0 z1 Q B iB
P z1 iz1 Q 1 i z2 M
PQ
P 1 i z1 2
101. (i) AP AO i
(ii) QB BO i
P A i O A Q B i O B
P z1 i0 z1 Q B iB
P z1 iz1 Q 1 i z2 M
PQ
P 1 i z1 2
1 i z1 1 i z2
M
2
102. (i) AP AO i
(ii) QB BO i
P A i O A Q B i O B
P z1 i0 z1 Q B iB
P z1 iz1 Q 1 i z2 M
PQ
P 1 i z1 2
1 i z1 1 i z2
M
2
z1 z2 z1 z2 i
M
2
103. Exercise 4K; 1 to 8, 11, 12, 13
Exercise 4L; odd
Terry Lee: Exercise 2.6