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Chapter 6: Complex Numbers and Vectors
Chapter 6
Complex Numbers and Vectors
Part I: Complex Numbers
If you were asked to solve the quadratic
equation x
2
 6 x  34 , you would start be applying the
 b  b 2  4ac
x
quadratic formula
2a
with a  1 , b  6 and c  34
Hence,
6   100
6  36  136
x
x
2
2
.
But, how do we find  100 ?  100 cannot be
represented by a number, as there is no real number
whose square is negative so therefore we define
 100  j10 .
Example: Find
Solution:
 49  j 7 .
Example: Find
Solution:
TRY: Find
 49 .
 90.25 .
 90.25  j 9.5 .
 169 .
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Chapter 6: Complex Numbers and Vectors
Powers of j
j  1
j 2  1
 
j 3  j 2 j   1 j   j
j  j
4
   1
2 2
Example: Find
Solution:
j
16
Solution:
TRY:
j
.
2 3
j   13 j   j
16
.
 
 j
Example: Find
25
j
1
7
 
j  j
7
Example: Find
Solution:
j
2
j
4 4
 14  1
25
.
 
 j
Find
j 34 .
Find
j 81 .
4 6
j  16 j  j
Find
j 48 .
Find
j 123 .
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Chapter 6: Complex Numbers and Vectors
Complex Numbers
Lets refer back to our original problem of solving
x 2  6 x  34 using the quadratic formula. We found
that x 
6   100
and we said that  100  j10 .
2
6  j10
Thus x 
2 and therefore:
6  j10
6  j10
 3  j5
x
 3  j 5 and x 
2
2
The result x  3  j 5 consists of 2 separate terms that
cannot be combined any further. In the expression
x  3  j5
 3 is called the real part of x
 5 is called the imaginary part of x
 the two combined form a complex number.
Hence, a complex number = real part + j (imaginary
part)
Addition and Subtraction of Complex Numbers
In order to add and subtract complex number, we match
like with like and then add and subtract as normal.
Example:
Find 4  j 2  5  j 6
Solution: Rewrite as 4  j 2  5  j 6
4  5  j 2  j6
4  5  j2  6
 9  j8
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Chapter 6: Complex Numbers and Vectors
Example:
Solution:
Find
6  j 2  7  j3
Find
3  j 2  2  j6
6  j 2  7  j3
6  7  j 2  j3
6  7  j2  3
 13  j
Example:
Solution: 3  j 2  2  j 6
3  2  j 2  j6
3  2  j2  6
 1 8 j
TRY:
4  j7  8  j 2 .
Find 4  j5  3  j 2 .
Find 3  j5  5  j 4   2  j3 .
Find
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Chapter 6: Complex Numbers and Vectors
Multiplication of complex numbers
Example: Find 3  j 42  j5
Solution:
3  j 42  j5  (3  2)  (3  j5)  ( j 4  2)  ( j 4  j5)  6  j15  j8  ( j 4  j5)
 6  j 23  j 2 20
2
We know that j  1 ,
therefore
6  j 23  j 2 20  6  j 23   120  6  j 23  20  14  j 23
Example: Find 4  j53  j 2
Solution:
4  j53  j 2  (4  3)  (4  j 2)  ( j5  3)  ( j5  j 2)  12  j8  j15  ( j5  j 2)
 12  j 7  j 2 10
We know that j  1 ,
therefore
12  j 7  j 2 10  12  j 7   110  12  j 7  10  22  j 7
.
2
Example: Find 5  j 27  j36  j 4
Solution:
5  j 27  j36  j 4  5  7  5   j3   j 2  7   j 2   j36  j 4
 35  j15  j14  j 2 66  j 4  35  j 29  66  j 4  29  j 296  j 4
Now
29  j 296  j 4  29  6  29  j 4   j 29  6   j 29  j 4


 174  j116  j174   j 2 116  174  j58  116  290  j58.
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Chapter 6: Complex Numbers and Vectors
TRY: Find 6  j 710  j3 Find 3  j96  j 2
Find 12  j314  j5 Find 312  j 26  j 
Find 2  j34  j 6 Find 7  j 2  j32  j 
Example: Find
Solution:
4  j 24  j 2
4  j 24  j 2  (4  4)  (4   j 2)  ( j 2  4)  ( j 2   j 2)  16  j8  j8  ( j 2  j 2)
 16  j 2 4  16  (1)2  16  4  20
In this example, our result contains no imaginary part,
only a real number. This is an exceptional case. Look at
the 2 complex numbers that we have multiplied together;
they are the same except for the middle sign. A pair of
complex numbers like this are known are conjugate
complex numbers and when two conjugate complex
numbers are multiplied together, you always get only a
real number.
TRY:
Find 2  j 72  j 7
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Chapter 6: Complex Numbers and Vectors
Graphical Representation of a Complex Number
Although we cannot evaluate a complex number as a real
number, we can represent it diagrammatically and a
diagram representing a complex number is known as an
Argand Diagram. On an Argand diagram,
 the x-axis represents the real part of the complex
number
 the y-axis represents the imaginary part of the
complex number.
Therefore we can graphically represent the complex
number 6 
j7 .
Argand diagram for
Y (Imaginary)
8
6+j7
a
6
4
b
2
1
2
3
4
5
6
7
X (Real)
TRY: Draw an Argand diagram for 6  j 7 .
Draw an Argand diagram for 2  j 4 .
Find and draw an Argand diagram for 2  j3  6  j 2 .
Find and draw an Argand diagram for 3  j9  12  j3 .
7/17
Chapter 6: Complex Numbers and Vectors
Polar Form of a Complex Number
It is sometimes necessary, or more convenient, to express
a complex number in Polar form.
Argand diagram for
Y (Imaginary)
8
6+j7
a
6
4
r
b
2
1
θ2
3
4
5
6
7
X (Real)
(With respect to the Argand diagram).
We can see that,
r 2  a2  b2 ,
b
a
a  r cos and b  r sin 
z  a  jb , this can be rewritten as
z  r cos   jr sin 
z  r cos  j sin  
tan  
As
r  a2  b2
b
  tan 1
a
8/17
Chapter 6: Complex Numbers and Vectors
This is known as the polar form of the complex number
b
1


tan
a  jb where r  a  b and
a
2
2
 r is known as the modulus of the complex number z
and can be written as mod z
  is known as the argument of the complex number
and can be written as arg z
Example: Express z  4  j 3 in polar form
Solution: We know that for
a  jb
we have
b
r
a.
Therefore for 4  j 3 we have a  4, b  3 so we
1
a  b and   tan
2
2
2
2
calculate: r  4  3  16  9  25  5
1


tan
and
So
3
θ  37degrees
4 So
z  5cos37  jsin37 
Example: Given
arg z .
z  2  j 5 , find mod z and
Solution: For 2  j 5 we have a  2, b  5 so we
2
2
calculate: mod z  r  2  5  4  25  29
1 5
arg
z



tan
and
2 . (So θ  68degrees)
9/17
Chapter 6: Complex Numbers and Vectors
z  4  j 7 , find mod z and arg z .
Given z  2  j 4 , find mod z and arg z .
TRY: Given
Given z  9  j 4 , find mod z and arg z .
10/17
Chapter 6: Complex Numbers and Vectors
Part II: Vectors
Physical quantities can be divided into 2 main groups:

A scalar quantity is one that is defined completely by a
single number with appropriate units e.g. length, area,
volume, mass, time etc. When the units are stated the
quantity is denoted entirely by its size or magnitude
e.g. speed of 10 km/hour is a scalar quantity.

A vector quantity is defined completely when we
know not only its magnitude (units) but also the
direction in which it operates e.g. force, velocity,
acceleration. A vector quantity involves direction as
well as magnitude e.g. a velocity of 10 km/hour due
north is a vector quantity.
Representing a vector.
A vector quantity can be represented graphically by a
line drawn so that its length represents the magnitude of
the vector and the direction of the line is the direction of
the vector. A vector quantity AB is referred to as AB or
bold type a. The magnitude of the vector quantity is
defined as AB or a , or simply AB or a. Note that BA
11/17
Chapter 6: Complex Numbers and Vectors
would represent a vector quantity of the same magnitude
but the opposite direction to AB .
Introduction to 3D vectors.
The axes of reference Ox, Oy and Oz are chosen so that
they form a right handed set. Vector OP is defined by its
components: A along Ox, B along Oy, C along Oz.
The symbols i, j, k denote unit vectors.
Let i = unit vector in Ox direction, j = unit vector in Oy
direction, k = unit vector in Oz direction.
Then OP  ai  bj  ck .
Magnitude of a vector
For the vector quantity OP  r  ai  bj  ck , then the
magnitude of the vector in terms of the unit vectors
2
2
2
OP

r

a

b

c
is
.
Example: Given PQ  4 i  3 j  2k find PQ .
Solution: If PQ  4i  3 j  2k ,
then PQ 
TRY:
4 2  3 2  2 2  16  9  4  29 .
Given OA  2i  4 j-3k find OA .
Given OB  1i-2 j  3k find OB .
12/17
Chapter 6: Complex Numbers and Vectors
Addition and Subtraction.
Example: Let a  2i  3 j  5k and b  4i  1j  6k ,
Find a  b .
Solution:
a  b  2i  3 j  5k  4i  1j  6k  6 i  4 j  11k .
Example: Let a  2i  3 j  5k and b  4 i  1j  6k ,
Find a  b .
a  b  2i  3 j  5k   4i  1j  6k 
Solution:  2i  3 j  5k - 4i  1j  6k
.
 2i  2 j  1k
Scalar products (and dot products).
If a and b are 2 vectors, the scalar product of a and b is
defined as the scalar (number) abcosθ where a and b
are the magnitudes of the vectors a and b and θ is the
angle between them.
The scalar product is denoted by a.b (often called the dot
product): a.b  abcosθ = a
=b


projection of b on a
projection of a on b
In both cases the result is a scalar quantity.
13/17
Chapter 6: Complex Numbers and Vectors
For a  a1i  a2 j  a3 k , b  b1i  b2 j  b3 k
then
a.b  a1i  a 2 j  a3 k b1i  b2 j  b3 k 
 a1ib1i  a1ib2 j  a1ib3 k  a 2 jb1i  a 2 jb2 j  a 2 jb3 k  a3 kb1i  a3 kb2 j  a3 kb3 k
 a1b1ii  a1b2 ij  a1b3 ik  a 2 b1 ji  a 2 b2 jj  a 2 b3 jk  a3 b1ki  a3 b2 kj  a3 b3 kk.
Note: i.i  11cos 0  1 , similarly j.j  1 , k.k  1 .
Also: i.j  11cos 90  0 , similarly j.k  0 , k.i  0 .
Putting these altogether in the formulation above gives:
a.b  a1b1 1  a1b2 0  a1b3 0  a2 b1 0  a2 b2 1  a2 b3 0  a3 b1 0  a3 b2 k 0  a3 b3 1
Therefore to find the scalar product we just sum the
products of the coefficients of the unit vectors along the
corresponding axes:
a.b  a1b1  a2 b2  a3 b3 .
Example: Given a  2i  3 j  5k and b  4i  1j  6k
find the dot product (or scalar product) a.b .
Solution: Comparing a  2i  3 j  5k and b  4i  1j  6k
to a  a1i  a2 j  a3 k and b  b1i  b2 j  b3 k respectively
we can see that a1  2, a 2  3, a3  5, b1  4, b2  1, b3  6.
Therefore we can calculate
a.b  a1b1  a2 b2  a3 b3 .  2  4  3  1  5  6  8  3  30  41
14/17
Chapter 6: Complex Numbers and Vectors
Example: Given a  3i-2 j  1k and b  2i  3 j-4k find
the dot product (or scalar product) a.b .
Solution: Comparing a  3i-2 j  1k and b  2i  3 j-4k
to a  a 1i  a 2 j  a 3k and b  b1i  b 2 j  b 3k respectively
we can see that
a1  3, a 2  2, a3  1, b1  2, b2  3, b3  4.
Therefore we can calculate
a.q  a 1b1  a 2b 2  a 3b 3 .  3  2   2  3  1 4  6  6  4  4
TRY:
Given a  4 i  1j  3k and b  1i  3 j-7 k ,
find a  b , a  b , a  b .
Given
OA  a  i  j  4k
OB  b  8i  2k
OC  c  5i  2 j  11k
Find:
a) b - a
b) c - b
c) the magnitude of the vector OB , i.e. OB
d)
e)
f)
the magnitude of the vector OC , i.e. OC
the scalar product a.b
the scalar product b.c
15/17
Chapter 6: Complex Numbers and Vectors
Vector products (and cross products).
If a and b are 2 vectors, the vector product of a and b
(a×b) is defined as a vector having magnitude absin θ
where a and b are the magnitudes of the vectors a and b
and θ is the angle between them. The product vector
acts in a direction perpendicular to both a and b in the
sence that a, b and a×b form a right-handed set in that
order.
|a×b| = absin θ
Note that |b×a| reverses the direction.
The vector product is denoted by a×b (often called the
cross product):
For a  a1i  a2 j  a3 k , b  b1i  b2 j  b3 k then
a  b  a1i  a2 j  a3 k  b1i  b2 j  b3 k 
 a1b1i  i  a1b2i  j  a1b3i  k  a2b1 j i  a2b2 j j  a2b3 j k  a3b1k  i  a3b2 k  j  a3b3 k  k
Note i×i = j×j= k×k = 0 and i×j=k, j×k=i, k×i=j
Also note i×j = -(j×i), j×k= -(k×j), k×i= -(i×k)
Putting these altogether in the formulation above gives:
a  b  a1b1 0  a1b 2 k   a1b3  j  a 2 b1  k   a 2 b 2 0  a 2 b3 i  a 3 b1  j  a 3 b 2  i  a 3 b3 
Therefore the vector product is given by:
16/17
Chapter 6: Complex Numbers and Vectors
a  b  (a 2 b 3  a 3 b 2 )i  (a 1 b 3  a 3 b1 ) j  (a 1 b 2  a 2 b1 )k
The cross product can be visualised as follows:
a1 a2 a3
a1 a2 a3 a1 a2 a3
b1 b2 b3
+ i
b1 b 2 b3 b1 b 2 b3
- j
+ k
Example: Given a  2i  3 j  5k and b  4i  1j  6k
find the cross product (or vector product) a  b .
Solution: Comparing a  2i  3 j  5k and b  4i  1j  6k
to a  a1i  a2 j  a3 k and b  b1i  b2 j  b3 k
respectively we can see that:
a1  2, a 2  3, a3  5, b1  4, b2  1, b3  6.
Therefore we can calculate
a  b  (a 2 b 3  a 3 b 2 )i  (a 1b 3  a 3 b1 ) j  (a 1b 2  a 2 b1 )k
a  b  (3  6  5  1)i  (2  6  5  4) j  (2  1  3  4)k
a  b  (13)i  (8) j  (10)k
a  b  13i  8 j  10k
TRY:
Given a  4i  1j  3k and b  1i  3 j-7 k , find the
vector product
a b .
17/17