4.
COMPLEX NUMBERS
INTRODUCTION
Complex numbers often seem strange when first encountered but it is worth persevering
with them because they provide a powerful mathematical tool for solving several engineering
problems. One of the main applications is to the analysis of Alternating Current (AC) circuits.
Engineers are very interested in these because the mains supply is itself AC and electricity
generation and transportation are dominated by AC voltages and currents. A great deal of
signal analysis and processing uses mathematical models based on complex numbers
because they allow the manipulation of sinusoidal quantities to be undertaken more easily.
Furthermore, the design of filters to be used in communications equipment relies heavily on
their use. One area of particular relevance is control engineering – so much so that control
engineers often prefer to think of control system in terms of a ‘complex plane’ representation
rather than ‘time domain’ representation.
LEARNING OUTCOMES
On completion of this chapter, you will be able to:

Introduce and explain the complex number system.

Represent a complex number in various forms.

Perform algebraic operations with complex numbers in rectangular, polar or
exponential form.

Solve problems involving solutions to complex numbers.
TABLE OF CONTENTS
4.
COMPLEX NUMBERS .................................................................................................. 0
4.1
THE MEANING OF COMPLEX NUMBERS............................................................ 2
4.1.1
The Extended Number System ........................................................................ 3
4.1.2
Graphical Representation of Complex Numbers .............................................. 5
4.1.3
The Rectangular Form of a Complex Number .................................................. 6
4.1.4
The Polar Form of a Complex Number............................................................. 7
4.1.5
The Exponential Form of a Complex Number ................................................ 12
4.2
ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS ................................. 15
4.2.1
Powers of j ..................................................................................................... 16
4.2.2
Equality of Complex Numbers ........................................................................ 17
4.2.3
Algebraic Operations with Complex Numbers in Rectangular Form ............... 18
4.2.4
Arithmetic with Complex Numbers in Polar Form ........................................... 23
4.2.5
Arithmetic with Complex Numbers in Exponential Form ................................. 26
4.3
DE MOIVRE’S THEOREM.................................................................................... 30
4.3.1
Introduction .................................................................................................... 31
4.3.2
Powers of Complex Numbers ........................................................................ 32
4.3.3
Roots of Complex Numbers ........................................................................... 33
4.4
APPLICATIONS OF COMPLEX NUMBERS ........................................................ 36
4.4.1
Theory of Determinants Applied to Solve Complex Numbers ......................... 37
4.4.2
Application of Complex Numbers in Alternating Current Theory ..................... 39
1
4.1
THE MEANING OF COMPLEX NUMBERS
Why it is important to understand: Complex Numbers
“Complex
numbers
are
used
in
many
scientific
fields,
including
engineering,
electromagnetism, quantum physics, and applied mathematics, such as chaos theory. Any
physical motion which is periodic, such as an oscillating beam, string, wire, pendulum,
electronic signal, or electromagnetic wave can be represented by a complex number
function. This can make calculations with the various components simpler than with real
numbers and sines and cosines. In control theory, systems are often transformed from the
time domain to the frequency domain using the Laplace transform. In fluid dynamics,
complex functions are used to describe potential flow in two dimensions. In electrical
engineering, the Fourier transform is used to analyse varying voltages and currents.
Complex numbers are used in signal analysis and other fields for a convenient description
for periodically varying signals. This use is also extended into digital signal processing and
digital image processing, which utilise digital versions of Fourier analysis (and wavelet
analysis) to transmit, compress, restore, and otherwise process digital audio signals, still
images, and video signals. Knowledge of complex numbers is clearly essential for further
studies in so many engineering disciplines”. Bird, J., 2017. Higher engineering mathematics.
Routledge.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

Explain why the number system was expanded to include complex numbers.

Define a complex number.

Use an Argand diagram to represent a complex number pictorially.

Convert between different forms of complex numbers.
2
4.1.1 The Extended Number System
You have already learned that some quadratic equations have no real solutions. For
instance, the equation
x2 1  0
has no real solution. If we try to solve this equation, we get
x 2  1
so
x   1
A problem now arises in that we need to find the square root of a negative number. We know
from experience that squaring both positive and negative numbers yield a positive result
hence, there is no number x that can be squared to produce  1 . To overcome this
deficiency, mathematicians created an expanded system of numbers using an imaginary
unit j , defined as
j  1
a pure imaginary unit
where j  1 . By adding real numbers to real multiples of this imaginary unit, the set of
2
complex numbers is obtained. Each complex number can be written in the standard form
a  bj . The real number a is called the real part of the complex number a  bj , and the
number bj (where b is a real number) is called the imaginary part of the complex number.
The set of real numbers is a subset of the set of complex numbers, as shown in the figure
below. This is true because every real number a can be written as a complex number using
b  0 . That is, for every real number a , you can write a  a  0 j .
Real Numbers
Complex Numbers
Imaginary Numbers
3
This number system makes it possible to solve all quadratic equations hence, in this form,
the solutions to the equation x  1  0 can be represented as
2
x  0 j
or
x  0 j
Again, if the quadratic equation x  2 x  5  0 is solved using the quadratic formula, then,
2
x
 2  (2) 2  (4)(1)(5)
2(1)
 2   16  2  (16)(1)

2
2
 2  16  1  2  4  1


2
2
 1  2 j

The Complex Conjugate
If z  a  bj , we define its complex conjugate to be the number z  a  bj ; that is, we
change the sign of the imaginary part. For example, the complex conjugate of  7  j is
7 j

only change the sign of the imaginary part
ACTIVITY 1
1.
2.
Solve the following quadratic equations:
0,4  0,8 j 
 2  5 j 
1.1
5x 2  4 x  4
1.2
x 2  29  4 x  0
1.3
x2
2 x
2
1 1,73 j 
1.4
25x 2  10x  2  0
0,2  0,2 j 
Write down the complex conjugates of the following complex numbers:
2.1
3 4 j
2.2
 8 j  11
2.3
cos t  j sin t
3  4 j 
11 8 j 
cos t  j sin t 
4
Note: In a more general case, we usually use the letter z to denote a complex number with
real part a and imaginary part bj , so z  a  bj .
4.1.2 Graphical Representation of Complex Numbers
A complex number z  a  bj can be represented pictorially on rectangular or Cartesian
axes. The horizontal (or x) axis is used to represent the real axis Re  and the vertical (or y)
axis is used to represent the imaginary axis Im  . Such a diagram is called an Argand
diagram. In this context, the Cartesian plane will be referred to as the complex plane.
Example 4.1
Graphical Representation of Complex Numbers
The complex number z  2  3 j will be represented as:
Im
3
2
Re
Note that this complex number is plotted very much like plotting x and y coordinates. The
real part corresponds to the x - value and the imaginary part corresponds to the y - value.
ACTIVITY 2
Plot the following on an Argand diagram:
1.
z1  3  2 j
2.
z 2  2  4 j
3.
z 3  3  2 j
4.
z 4  2 2 j  1
5
4.1.3 The Rectangular Form of a Complex Number
In the diagram representing z  2  3 j (example 4.1), the shape formed by the dotted lines,
the imaginary axis and the real axis is a rectangle. The form z  a  bj of representing a
complex number is thus called the rectangular form, also referred to as standard form.
Example 4.2
1.
Complex Numbers and Radicals
Rewrite each of the following into rectangular form:
1.1
 5   7  5  (7)(1)  5  7   1  5  j 7
1.2
 12  3  (12)(1)  3  12   1  3  3  j 2 3
Note: Some of the properties of radicals that are true for real numbers turn out not to be
true for complex numbers. In particular, for positive real numbers a and b ,
b  ab
a
a
but
 b  (a) (b)
To avoid having to worry about this, always convert expressions of the form
equivalent form in terms of j before performing any operations.
For example:
 5  5  5(1) 5(1)
 5 j 5 j
 25 j 2
since j  1
2
 5 (1)  5
whereas,
(5)(5)  25  5
1.3
 3  12  j 3  j 12  j 2 36  6 (1)  6
6
 a to the
4.1.4 The Polar Form of a Complex Number
It is often useful to exchange the complex number z  a  bj , represented by rectangular
coordinates a ; b for polar coordinates r and
 as depicted on the figure below.
Im
z  a  bj
b
r

a
Re
From the given diagram, we note that:
cos 
a
r
b
r
and
sin  
and
b  r sin

  tan 1  
and so,
a  r cos
Furthermore,
tan  
b
a
Using Pythagoras’s theorem, we obtain r 
b
a
a 2  b 2 . By finding r and  we can express
the complex number z  a  bj in polar form as:
z  r cos   j sin 
 r cos   j sin  
z  r cos  j sin   is usually abbreviated to z  r  which is known as the polar form of
a complex number.
 Clearly, r is the ‘distance’ of the point
a ; b
from the origin and is called the
modulus of the complex number z . The modulus is always a non-negative number
and is denoted z .
 The angle is conventionally measured from the positive x axis. Angles measured in
an anticlockwise sense are regarded as positive while those measured in a clockwise
7
sense are regarded as negative. The angle
 is called the argument of z and is
expressed in degrees or radians.
Rectangular form:
z  a  bj
Polar form:
z  r cos  j sin    r 
z  r  a2  b2
a  r cos
b  r sin
tan  
b
a
Note that:
r    r cos( )  j sin(  )
 r cos   j sin  
z
If z  a  bj then z  a  bj and z  r (  )
Example 4.3
1.
Complex Numbers in Polar Form
Determine the modulus and argument (in radians) of the complex number z  1  j ,
and express z in polar form. Show all steps.
Solution:
The real part of z is  1 and the imaginary part is  1 . We then plot this on the Argand
diagram as:
Im
1
Re

r
1
Modulus:
z  r  a2  b2
 (1) 2  (1) 2  2
8
Argument:
 1
1
  tan (1)
 1
  tan 1 

, which is in the 3rd quadrant hence,
4
   
5
4
Polar form:
z 2
2.
5
4
z 2 
or
3
4
Convert the following complex numbers to rectangular form:
2.1
4 300
0
As shown in the diagram below, 4 30 lies in the first quadrant.
Im
b
4
300
a
a  r cos
Re
b  r sin 
 4 cos 300
 4 sin 30 0
 3,464
2
Hence,
4 300  3,464  2 j
9
2.2
7  1450
Im
a

Re
145
0
7
b
7 1450 is in the third quadrant as shown above.
Angle
  1800  1450  350
Hence:
a  r cos
 7 cos 350  5,734
a  5,734
and
b  r sin 
 7 sin 350  4,015
b  4,015
Therefore:
7  1450  5,734  4,015 j
10
Conversion Using a Scientific Calculator
Using the ‘Pol’ and ‘Rec’ functions on a calculator enables changing from rectangular to
polar and vice-versa to be achieved more quickly. This is how the conversion is done:
CASIO FX – 82ES
  : press Shift then +. Your calculator will show
From Rectangular a  bj  to Polar r
Pol( . Enter value a then Shift and then ) which will show Pol(a , and then enter the value b
and ) and then press  . The values of r and  will then be displayed.
   to Rectangular a  bj  : press Shift then –. Your calculator will show
From Polar r
Rec( . Enter value r then Shift and then ) which will show Rec(r , and then enter the value
 and also ) and then press  . The values of a and b will then be displayed as x  a and
y b.
SHARP CALCULATORS
  : Enter the value a then 2
From Rectangular a  bj  to Polar r
nd
function and then
STO. The calculator will show a , and then enter the value b and the calculator will show a ,
b after which you press 2nd Function and then 8. This gives you the value of r and then
press 2nd Function and then Exp which gives you the value of  .
   to Rectangular a  bj  : Enter the value a then 2
nd
From Polar r
function and then
STO. The calculator will show a , and then enter the value b and the calculator will show a ,
b after which you press 2nd Function and then 9. This gives you the value of x and then
press 2nd Function and then Exp which gives you the value of y.
Note: Make sure that your calculator is in the correct mode i.e. radian or degree. We will
from now onwards make use of this build-in functions on the scientific calculator to
do all the conversions.
Example 4.4
Evaluating Complex Numbers Using a Calculator
Use a calculator to confirm the answers we found in Example 4.3 above.
11
4.1.5 The Exponential Form of a Complex Number
There is still another way of expressing a complex number which we must deal with, for it too
has its uses. We shall arrive at it this way:
Many functions can be expressed as the sum of a sequence of terms involving integer
powers of x . For example,
x2 x3
e  1 x 
  ...
2! 3!
x
and this representation is valid for any real value of x . The expression on the RHS is an
infinite sum but its terms get smaller and smaller, and as more are included, the sum we
x
obtain approaches e . Other examples of power series include,
sin x  x 
x3 x5 x7
   ...
3! 5! 7!
(1)
cos x  1 
x2 x4 x6

  ...
2! 4! 6!
(2)
and
which are also valid for any real value of x . It is useful to extend the range of applicability of
these power series by allowing x to be a complex number. That is, we define the function
e z to be,
ez  1 z 
z2 z3
  ...
2! 3!
We have already seen that we can express a complex number in polar form:
z  r cos  j sin  
Using equations (1) and (2), we can write,
 2 4

z  r  1 

 ...  
2! 4!




3 5
j  

 ...  
3! 5!



2
3 4
5 
 r 1  j 
j 
 j ... 
2!
3! 4!
5! 

Furthermore, if we take the series for e and write j in place of x , we get:
x
12
e
j
2
3
4

j   j   j 
 1  j 


 ...
2!
 1  j 

2
2!
j
3!

3
3!

4!

4
4!
 ...
so that:
z  r cos   j sin  
 re j
This is called the exponential form of the complex number. It is obtained from the polar
 is the same in both. It is
important to note however that in the exponential form, the angle  must be in radians.
form quite easily since the r value is the same and the angle
Example 4.5
Converting Complex numbers
Complete the following table (questions are highlighted):
RECTANGULAR FORM
POLAR FORM
EXPONENTIAL FORM
2 j
2,236  153,4 
2,236 e 2, 678 j
0,73  2,91 j
3 1,325
3e1,325 j
2,135  1,315 j
2,507  0,552
2,507 e 0,552 j
 1 3 j
3,162 1,893
3,162 e1,893 j
1,732  j
2 cos 6  j sin 6


2e6
j
0,866  1,5 j
3 
3e3
4 4
4,472  0,464
4,472 e 0, 464 j
 0,574  0,819 j
cos125   j sin 125 
e 2,182 j
 3,394  7,415 j
3e 2
3 e1 2 j
3
13

j
ACTIVITY 3
1.
Write each of the following in polar form:
1.1
 3,5  0,25 j
1.2
33 3j
1.3
4j
2.

0
1 120 
1
3

j
2 2
0
Write each of the following in rectangular form:
2.1

 2 
 2  
3  cos 
  j sin 

 3 
 3 

 1,5  2,598 j 
2.2
1,7e1, 2 2,5 j
2.3
2 0,25
 4,52  3,38 j 
1,414  1,414 j 
2.4
4,86 cos 312,7 0  j sin 312,7 0 
3,296  3,572 j 
2.5
3.
0
2 j 2
1.4
1.5
3,51 3,07
6120 
41,57
2 45 
e
1

4
1,922  1,922 j 
j
Write each of the following in exponential form:
3.1
53j
3.2
12 cos1350  j sin 1350 
3.3
3 j
3.4
3 4 j
3.5
4,2 j  2,5
5,83e
12e
2e
5e
4,89e
j 0 , 54
2 , 36 j
0 , 52 j
2 , 21 j
14




2 ,11 j

4.2
ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS
Why it is important to understand: Algebraic Operations with Complex Numbers
To work with complex numbers, we will need to know how to add, subtract, multiply, and
divide them. Importantly, we can treat complex numbers of the form a  bj exactly as we
treat algebraic expressions of the form a  bx . We just need to remember that in this case,
j stands for the imaginary unit; it is not a variable that represents a real number. Just like
addition and subtraction, multiplication of complex numbers can be carried out by treating
a  bj in the same way we treat the algebraic expression a  bx . The key difference is that
we replace j 2 with  1 each time it occurs. However, to divide complex numbers, we will
multiply the numerator and denominator by the conjugate of the denominator.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

Perform algebraic operations addition and subtraction with complex numbers in
rectangular form.

Perform the operation multiplication with complex numbers in rectangular form.

Perform the operation division with complex numbers in rectangular form.

Do vector addition and subtraction of complex numbers.

Perform the operations multiplication and division with complex numbers in polar form.

Do basic arithmetic of complex numbers in exponential form.
15
4.2.1 Powers of j
POSITIVE INTEGER POWERS
Using the fact that j  1 , we can develop other quantities.
2
Example 4.1
Simplifying Expressions Containing j
Simplify the following expressions:
1.
j3
j3  j2. j
  1 . j   j
2.
j 12
j 12   j 2 
6
  1  1
6
3.
j6
j6   j2 
3
  1  1
3
4.
j 11
j 11   j 2  . j
5
  1 . j   j
5
NEGATIVE INTEGER POWERS
Negative powers follow from the reciprocal of j . Because
j 2  1
we know that by dividing both sides by j , j  
1
  j 1 so that:
j
j 1   j
j 2   j 2    1  1
j 3   j 2 . j 1   1 j   j
j  4   j  2    1  1
j  23   j  2  . j 1   1  j   j
j 32   j  2    1  1
1
11
2
16
11
16
1
2
16
4.2.2 Equality of Complex Numbers
Let us see what we can find out about two complex numbers which we are told are equal.
Let the numbers be
a  bj and c  dj
Then we have
a  bj  c  dj
Rearranging terms, we get
a  c  j d  b
In this last statement, the quantity of the LHS is entirely real, while that on the RHS is
entirely imaginary, i.e. a real quantity equals an imaginary quantity! This seems contradictory
and in general, it just cannot be true. But there is one special case for which the statement
can be true. That is when each side is zero.
a  c  j d  b
can be true only if
and if
ac  0

ac
d b  0

d b
So, we get this important result: If two complex numbers are equal:
(a)
The two real parts are equal
(b)
The two imaginary parts are equal
Example 4.2
1.
Equations Involving Complex Numbers
Find x and y so that x  6 j and 3  yj represent the same complex number.
If both quantities represent the same complex number, we have:
x  6 j  3  yj
Therefore
x3
2.
y  6
and
Solve for x and y if:
2.1
2 x  yj  4  3 j
2x  4
and
y3
x2
and
y3
17
x  y   x  y  j  4  5 j
2.2
x y 4
……
x y 5
……. (2)
(1)
Solving for x and y we have
x  y5
……. (3)
Substitute (3) into (1)
y 5 y  4
2 y  1
y  0,5
Sub y into (3)
x  4,5
4.2.3 Algebraic Operations with Complex Numbers in Rectangular Form
The basic arithmetic rules and properties of addition, subtraction, multiplication, and division
of real numbers apply to complex numbers.
4.2.3.1
Addition and subtraction
To add (or subtract) two complex numbers, we simply add (or subtract) the real and
imaginary parts of the numbers separately. That is: if a  bj and c  dj are two complex
numbers written in standard form, their sum and difference are defined as follows,
a  bj   c  dj   a  c  b  d  j
a  bj   c  dj   a  c  b  d  j
Sum:
Difference:
Example 4.3
Addition and Subtraction in Rectangular Form
If z1  3  4 j and z 2  4  2 j , find z1  z 2 and z1  z 2
Solution:
z1  z 2  3  4 j   4  2 j 
 3  4    4  2  j
72j
z1  z 2  3  4 j   4  2 j 
 3  4    4  2  j
 1  6 j
18
4.2.3.2
Vector addition and subtraction
Since complex numbers can be considered as vectors on an Argand diagram, we can add
and subtract complex numbers as vector addition and subtraction and use the familiar tail to
head method in Physics.
Example 4.4
1.
Complex Numbers on the Argand Diagram
Given z1  2  2 j and z 2  4  j , show z1  z 2 on the Argand diagram
Solution:
z1  z 2   2  4    2  1 j
 23j
Remark:

Fill in the units on the diagram

Use tail to head method

The resultant is the vector joining the first tail with the last head
19
2.
Show z1  z 2 on the Argand diagram, given z1  4  2 j and z 2  2  3 j
Solution:
z1  z 2  4  2   2  3 j
 2 j
3.
If z1  4  2 j and z 2  4 j  2 , show z1  z 2 on the Argand diagram
Solution:
z1  z 2   4  2 j   2  4 j 
  4  2   2  4  j  2  6 j
20
4.2.3.3
Multiplication
Multiplication of complex numbers in rectangular form is achieved by carrying out the
multiplication as if the numbers were ordinary polynomials with j as a variable and replace
j 2 by  1 . Then simplify by adding similar terms.
Example 4.5
1.
Multiplying Complex Numbers in Rectangular Form
Simplify each of the following:
1.1
34  6 j 
Solution:
34  6 j   12  18 j
1.2
2  2 j 3  4 j 
Solution:
2  2 j 3  4 j   6  8 j  6 j  8 j 2
 6  2 j  8(1)
 14  2 j
1.3
3  2 j 2
Solution:
3  2 j 2  3  2 j 3  2 j 
 9  6 j  6 j  4 j2
 9  12 j  4 1
 5  12 j
As mentioned earlier, the complex conjugate of a complex number is obtained by changing
the sign of the imaginary part. The product of a complex number and its complex conjugate
is always a real number. Thus, if z  a  bj , then
z z  a  bj a  bj 
 a 2  baj  baj  b 2 j 2
 a2  b2 j 2
 a 2  b 2 (1)
 a2  b2
21
For example: if z  3  2 j , find z z .
z z  3  2 j 3  2 j 
 9  6 j  6 j  4 j2
 9  4 1
 13
4.2.3.4
Division
To divide two complex numbers, it is necessary to make use of the complex conjugate. We
multiply both the numerator and denominator by the conjugate of the denominator and then
simplify the result.
Example 4.6
1.
Dividing Complex Numbers in Rectangular Form
Division of a complex number by a real number
1.1
15  12 j
3
Solution:
15  12 j 1
 15  12 j 
3
3
15 12 j
 
 54 j
3
3
2.
Division by a pure imaginary number
2.1
15  12 j
3j
Solution:
15  12 j  3 j  45 j  36 j 2


3j
3j
 9 j2
 45 j  36  1

 9  1
 36  45 j

9
 4  5 j
22
3
j5
2.2
Solution:
3
3
3
 2 2 
5
j
 j  . j  12 . j
3  j 3j
 

j  j  j2
3j

 3 j
  1
3.
Division by a complex number in rectangular form
25j
3 4 j
Solution:
2  5 j 2  5 j 3  4 j 


3  4 j 3  4 j 3  4 j 
6  8 j  15 j  20 j 2

32  4 2
 14  23 j

 0,56  0,92 j
25
The multiplication of two conjugates in the denominator allows a useful simplification. We
see that the effect of multiplying by the conjugate of the denominator is to make the
denominator of the solution purely real. The elimination of the complex number in the
denominator is called realisation (rationalisation) of the denominator.
4.2.4 Arithmetic with Complex Numbers in Polar Form
Complex numbers can be written in rectangular or polar form, each with an advantage.
Complex numbers in rectangular form are added, subtracted, also multiplied and divided.
Complex numbers in polar form are only multiplied and divided.
23
4.2.4.1
Multiplication
Suppose we want to multiply the complex numbers:
z1  r1 cos 1  j sin 1 
and
z 2  r2 cos  2  j sin  2 
We find:
z1 z 2  r1 cos 1  j sin 1   r2 cos  2  j sin  2 
 r1 r2 cos 1 cos  2  j cos 1 sin  2  j sin 1 cos  2  j 2 sin 1 sin  2 
 r1 r2 cos 1 cos  2  sin 1 sin  2   j sin 1 cos  2  sin  2 cos 1 
Using the compound angle addition and subtraction formulae:
cos 1   2   cos 1 cos  2  sin 1 sin  2
sin 1   2   sin 1 cos  2  cos 1 sin  2
This complex number can be written as:
z1 z 2  r1 r2 cos 1   2   j sin 1   2 
This is a new complex number which, if we compare with the general form
z  r cos  j sin   , we see has a modulus of r1 r2 and an argument of  1   2 . To
summarise: to multiply together two complex numbers in polar form,
(a)
Multiply their moduli
(b)
Add their arguments
That is:
z1 z 2  r1 r2 1   2
4.2.4.2
Division
A development similar to the one above shows that to divide two complex numbers in polar
form, we divide their moduli and subtract their arguments.
That is:
z1 r1
 1   2
z 2 r2
24
Example 4.7
1.
Combined Operations
Perform the indicated operation and express the results in rectangular form.
1.1
2,7 cos1450  j sin 1450 
0,3 cos 450  j sin 450 
Solution:
2,7 1450
0,3 45
0

2,7
1450  450
0,3
 9 1000  1,56  8,86 j
1.2
4,2 4,5
0,7 2,8
Solution:
4,2 4,5
0,7 2,8

4,2
4,5  2,8
0,7
 6 1,7  0,77  5,95 j
1.3
 2  j 2 3 2
 1  j 3 
32j
3 j


Solution:
 2  j 2 3 2
 1  j 3 

0
0
3  2 j 4  120  4  30

3 j
2 1200  2 1500


16  1500
4 2700
 4  4200
 2  3,46 j
25
2.
If z1  2,5 45
0
and z 2  4 70 , find z1  z 2 and z1  z 2 . Express the results in
0
polar form.
Solution:
z1  z 2  2,5 450  4 700
Must first convert to rectangular form
 1,77  1,77 j  1,37  3,76 j
 3,14  5,53 j
Convert to polar form
 6,36 60,410
z1  z 2  2,5 450  4 700
Must first convert to rectangular form
 1,77  1,77 j  1,37  3,76 j 
 0,4  1,99 j
Convert to polar form
 2,03  78,630
4.2.5
Arithmetic with Complex Numbers in Exponential Form
Complex numbers in exponential form are also used only in multiplication and division.
a  bj  r   re j , where  must be in radians when expressed in exponential form.
Consider the following complex numbers in exponential form:
z1  r1 1  r1 e j
z 2  r2  2  r2 e j
and
1
2
Multiplication:
z1  z 2  r1r2 e j  
1
2

Division:
z1 r1 j  
 e
z 2 r2
1
2

Since these two operations are commonly performed in polar form, we will rather look at the
application of complex numbers in exponential form. For instance:
If z  re
j

then ln z  ln re
j

 ln r  ln e j
ln z  ln r  j
26
Laws of logarithms
Example 4.8
1.
Evaluating Complex Numbers in Exponential Form
If z  4e
1, 3 j
1.1
, determine ln z in:
Rectangular Form
Solution:
If z  4e
1, 3 j

, then ln z  ln 4e
1, 3 j

 ln 4  1,3 j or
 1,386  1,3 j
1.2
Polar form
Solution:
1,9 43,17 0
2.
Given
1,9 0,753
or
z  3e1 j , find ln z in polar form. Express  in radians.
Solution:
If
z  3e1 j , then
ln z  ln 3e1 j 
 ln 3  ln e1 j
 ln 3  1  j
 ln 3  1  j
 2,0986  j
 2,325  0,445
3.
Determine, in polar form,
Solution:

ln3  4 j 

ln3  4 j   ln 5 0,927
 ln5e 0,927 j 
 ln 5  ln e 0,927 j
 ln 5  0,927 j
 1,857 29,950
or
1,857 0,523
27
ACTIVITY 4
1.
Solve the following complex equations:
1.1
2  j 3  2 j   a  bj
a  8 , b  1
1.2
2 j
 j x  yj 
1 j
3
1

 x  2 , y   2 
1.3
a  bj  3  j 3 4  6 j 3 
2  3 j   (a  bj)
x  2 yj    y  xj   2  j
a  18 , y  14
a  5 , b  12
x  3 , y  1
3y  2 j 2x  3 j

0
1 j
3j
9
4

x

,
y


2
3 
1.4
1.5
1.6
2.
Evaluate the following and show the results on an Argand diagram:
2.1
2.2
3.
4.
8  j 
 5  8 j 
3  2 j   5  j 
 2  6 j   3  2 j 
Evaluate the following:
3.1
2
1  j 4
3.2
1 3 j 

j 
1

2
j


 1
 2 
2
2
Perform the required operations and write the answer in rectangular form:
4.1
4.2
2  j 3  2 j 
0,2  1,6 j 
43j
0,88  0,42 j 
3
1

3 2 j 5 j
28
4.3
5.
6.
7.
9.

0,5  0,87 j 
Given z1  1  2 j , z 2  4  3 j , z 3  2  3 j and z 4  5  j , evaluate:
5.1
z1  z 3
z2  z4
 0,22  0,51 j 
5.2
z1 z 3
z1  z 3
0,12  1,58 j 
5.3
z2 
If
1,73  0,35 j 
z1
 z3
z4
z  7e 2,1 j , determine ln z in:
6.1
Rectangular form
6.2
Polar form,
If

ln 7  2,1 j 
2,86 0,82
in radians
z  4e1,52 j , determine ln z in:
7.1
8.

1 9   3
9  3
Polar form,

3,51 34,72 
0
in degrees
Determine in exponential form:
8.1
ln2  5 j 
8.2
ln 4  3 j 
2,06e 
4,11e 
0 , 615 j
1,17 j
Evaluate the following and leave the answers in standard form:
9.1
3e1 2 j  ( 2,1  5 j )
9.2
3e 2 j cos

3
 3 je2 j sin
 1,29  2,42 j 

1,74  2,45 j 
3
29
4.3
DE MOIVRE’S THEOREM
Why it is important to understand: De Moivre’s Theorem
“There are many, many examples of the use of complex numbers in engineering and
science. De Moivre’s theorem has several uses, including finding powers and roots of
complex numbers, solving polynomial equations, calculating trigonometric identities, and for
evaluating the sums of trigonometric series. The theorem is also used to calculate
exponential and logarithmic functions of complex numbers. De Moivre’s theorem has
applications in electrical engineering and physics”. Bird, J., 2017. Higher engineering
mathematics. Routledge.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

State De Moivre’s theorem.

Calculate powers of complex numbers.

Calculate roots of complex numbers.
30
4.3.1 Introduction
A very important result in complex number theory is De Moivre’s theorem. From the
multiplication of complex numbers in polar form, we concluded that:
r1 1  r2  2  r1r2 1   2
We can then say:
r  2  r   r   r 2 2
r  3  r   r   r   r 3 3
r  4  r   r   r   r   r 4 4
/
/
/
r  n  r 
 r   r   r  ... to n terms  r n n
This is what is called De Moivre’s theorem. In general, the theorem states:
r cos  j sin  
n
 r n cos n  j sin n 
or
r  
n
 r n n
The theorem is true for all positive, negative and fractional values of n (though we only
consider n a positive integer for this course). The theorem is used to determine powers and
roots of complex numbers.
31
4.3.2 Powers of Complex Numbers
Applying De Moivre’s theorem
Example 4.1
1.
Use De Moivre’s theorem to evaluate:
1.1

52j
3
Solution:

1.2

 
3
5  2 j  3 41,810

3
Complex number changed to polar form
 33 3  41,810
De Moivre’s theorem applied
 27125,430
Answer in polar form
 27e 2,189 j
Answer in exponential form
 15,652  22 j
Answer in rectangular form
3  1  2 j 5
4
2  3 j 2  2  j 
32j
Solution:

  1  2 j   2,646  0,857 2,236 2,034
3,606 0,983 1,732  0,615
2  3 j   2  j 
32j
3
2
3
5
4
2


5
4
18,525  2,57155,89310,172
13,0031,9668,999  2,462
1035,418 7,601
117,014  0,496
Converting to polar form
De Moivre’s theorem
Using r1 1  r2  2  r1 r2 1   2
r1 1
r
 1 1   2
r2  2 r2
 8,849 8,097
Using
 8,849 e 8, 097
Exponential form
 2,129  8.589 j
Rectangular form
32
4.3.3 Roots of Complex Numbers
De Moivre’s theorem can also be used to find the nth roots of complex numbers. An nth root
of the complex number z is a complex number w such that
wn  z
Writing these two numbers in polar form as
w  scos  j sin  
z  r cos   j sin  
and
and using De Moivre’s theorem, we get
s n cos n  j sin n   r cos  j sin  
The equality of these two complex numbers shows that

sn  r
cos n  cos 
and
sr
1
n
sin n  sin 
and
From the fact that sine and cosine have period 2 , it follows that
n    2k


  2k
n
1
    2k 
   2k 
w  r n cos
  j sin 

 n 
  n 
Thus
Since this expression gives a different value of w for k  0 , 1 , 2 , ..... , n  1 , we have the
following:
r cos  j sin  
1
n
1
1


 r  cos   2k   j sin   2k 
n
n


1
n
or
r  
r
or
r  
 rn
1
n
1
n
1
n
1
  2k 
n
if
 is in radians
1
1
  k 3600 
n
if
 is in degrees
where k  0 , 1 , 2 , ..... , n  1
33
Example 4.2
1.
Finding the Roots of Complex Numbers
Find all the cube roots of 3  4 j and express the answer in rectangular form.
Solution:
3  4 j  3  4 j 3
1
3


1
 5  0,927 3
5
1
3
1
 0,927  2k 
3
k 0 ,1, 2
 1,71  0,309  2,094k
The roots are:
2.
k  0:
Root 1
R1  1,71  0,309  1,629  0,52 j
k  1:
Root 2
R2  1,711,785  0,363  1,671 j
k  2:
Root 3
R3  1,713,879  1,266  1,15 j
Find all the rectangular square roots of  1  j 2
Solution:


 1,732 2,186
1 j 2  1 j 2
1
2
1
2
 1,732 2
1
1
2,186  2k 
2
 1,3161,093  k
k 0 ,1
The roots are:
k  0:
Root 1
R1  1,3161,093  0,605  1,169 j
k  1:
Root 2
R2  1,316 4.235  0,605  1,169 j
34
ACTIVITY 5
1.
2.
Use De Moivre’s theorem to simplify the following:
1.1
1  j 4
1.2


3,156 2150


 3 Cos   Sin  j 


6
6
1.3
 2  2 j 5
1  3 j 3 4  j 2
1.4
 3,1 cos125  j sin125 
 2,4 cos 35  j sin 35 


(answer in rectangular form)



2

(answer in polar form)
 4  0 j 
3,32 370 
0
(answer in exponential form)
0,648e


 1,08  1,87 j 
3
(in rectangular form)
16, 262 j

Given z1  4   4 and z 2  1  2 j , use De Moivre’s Theorem to evaluate:
z1 3 z 2 2
z 
4
17,89 5,25
(answer in polar form)
2
3.
4.
Perform the indicated operation and express the answer in rectangular form.
3.1
Find the cube roots of  1
3.2
Find the cube roots of  2  2 j
3.3
Find all the roots of:
0,5  0,87 j ; 1; 0,5  0,87 j 
0,99  j ; 0,38  0,36 j ;  1,37  0,35 j 
3.3.1
z 3  64  0
2  3,46 j ;  4  0 j ; 2  3,46 j 
3.3.2
z3  2  j 3  0
1,34  0,326 j ;  0,954  j ;  0,39  1,327 j 
If  1  6 j is one root of the quadratic equation ax 2  bx  c  0 , then the other root is
 1 6 j 
equal to?
35
4.4
APPLICATIONS OF COMPLEX NUMBERS
Why it is important to understand: Applications of Complex Numbers
The truth is that the numbers we studied in this chapter were not received very well when
they were invented, as you can guess from the names they were given: complex and
imaginary. These names are not exactly ringing endorsements. Still, complex numbers
eventually came into widespread use in areas like electrical engineering, physics, chemistry,
statistics, and aeronautical engineering.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

Apply the theory of determinants to solve complex numbers.

Apply complex numbers in alternating current theory.
36
4.4.1
Theory of Determinants Applied to Solve Complex Numbers
Example 4.1
1.
Applying Determinants to Solve Complex Equations
Solve for x and y if:
1.1
j5
3
j
 3
xj
Solution:
xj 2  3 j 2  3
x  1  3  1  3
x 3  3
x0
1.2
x  yj
23j
23j
1 j
0
Solution:
x  yj 1  j   2  3 j 2  3 j   0  0 j
x  xj  yj  y   4  12 j  9  0  0 j
x  y  5  x  y  12 j  0  0 j
x y5 0
Real parts equal
x  y  12  0
Imaginary parts equal
Solving the two equations simultaneously gives:
x  8,5
y  3,5
and
37
2.
Evaluate the following:
2.1
1 j
2j
 3 j 1 4 j
Solution:
1 j
2j
 3 j 1 4 j
 1  j 1  4 j   2 j  3 j 
 1 4 j  j  4 j2  6 j2
 1 4 j  j  4  6
 1  3 j
2j
2.2
1 j 3
1 j
1
j
0
4j
5
Solution:
Using the first column, the value of the determinant is:
2 j 
1
j
4j 5
 1  j 
1 j 3
4j
5
 0
1 j 3
1
j
 2 j 5  4 j 2   1  j 5  5 j  12 j   0
 2 j 9   1  j 5  7j
 18 j  5  7 j  5 j  7 j 2 
 18 j   2  12 j 
 2  30 j
38
4.4.2
Application of Complex Numbers in Alternating Current Theory
Ohm’s law is written as V  IR
or
R
V
I
Where: R is the resistance (measured in Ohms) of some object or resistor. V is the
potential difference across the resistor and I is the current that flows through it.
Resistors can be connected in series or in parallel with the following accompanied formulas:
Resistors in series
V  V1  V2  total voltage
I  I 1  I 2  total current
R  R1  R2  total resistance
Resistors in parallel
V  V1  V2  total voltage
I  I 1  I 2  total current
1
1
1
1



R R1 R2 total resistance
Example 4.2
1.
Application in Electrical Alternating Current Theory
Calculate the voltage produced if 0,32  0,8 j amps flow through a circuit containing
two resistors in series with resistance’s
3,6  12,5 j and 5,4  9,6 j ohms
respectively.
Solution:
Total resistance
(R)
 3,6  12,5 j  5,4  9,6 j
 9  22,1 j
Voltage
(V )
IR
 0,32  0,8 j 9  22,1 j 
 4,65  6,35 j
39
2.
Calculate the voltage in a circuit if two resistors are connected in parallel with
individual currents 0,7  0,2 j A and 0,9  0,3 j A and resistance’s 20  25 j  and
14  20 j  respectively.
Solution:
V I R
I  I1  I 2
 0,7  0,2 j  0,9  0,3 j
 1,6  0,5 j
 1,6763 0,3029
R


R1 R2
R1  R2
20  25 j 14  20 j 
20  25 j  14  20 j
781,6  1,85617
56,4  0,9237
 13,858  0,93247
V I R
V  1,6763 0,30288  13,858  0,93247
 23,23  0,62959
 18,78  13,678 j Volt
40