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© 2022

Complex Number Primer

Introduction

1 Introduction
Before I get started on this let me first make it clear that this document is not intended to teach you
everything there is to know about complex numbers
...
The purpose of this document is to give you a brief overview of complex
numbers, notation associated with complex numbers, and some of the basic operations involving
complex numbers
...
If you don’t remember how to do arithmetic I will show an example or two to remind you how to do arithmetic, but I’m going to assume that you don’t need more
than that as a reminder
...
Problems tend to arise however because most instructors seem to
assume that either students will see beyond this exposure in some later class or have already seen
beyond this in some earlier class
...

That is the purpose of this document
...
We’ll also be seeing a slightly different way of looking at some of the basics
that you probably didn’t see when you were first introduced to complex numbers and proving some
of the basic facts
...
It is presented solely for those who might be
interested
...

Also included in this section is a more precise definition of subtraction and division than is normally
given when a person is first introduced to complex numbers
...

The remaining sections are the real point of this document and involve the topics that are typically
not taught when students are first exposed to complex numbers
...


© Paul Dawkins

–2–

Complex Number Primer

The Definition

2 The Definition
As I’ve already stated, I am assuming that you have seen complex numbers to this point and that
√
you’re aware that i = −1 and so i2 = −1
...
We’ll also take a look
at how we define arithmetic for complex numbers
...
Also note
that this section is not really required to understand the remaining portions of this document
...

So, let’s give the definition of a complex number
...
The number a is called
the real part of z and the number b is called the imaginary part of z and are often denoted as,
Rez = a

Imz = b

There are a couple of special cases that we need to look at before proceeding
...

Next, let’s take a look at a complex number that has a zero imaginary part,
z = a + 0i = a
In this case we can see that the complex number is in fact a real number
...

We next need to define how we do addition and multiplication with complex numbers
...
However,
the multiplication formula that you were given at that point in time required the use of i2 = −1 to
completely derive and for this section we don’t yet know that is true
...

Above we noted that we can think of the real numbers as a subset of the complex numbers
...
For instance, given the two complex numbers,
z1 = a + 0i

z2 = c + 0i

the formulas yield the correct formulas for real numbers as seen below
...
However, before we do that we need to acknowledge that powers of complex
numbers work just as they do for real numbers
...
Doing this gives,
i2 = i · i
= (0 + 1i) (0 + 1i)
= ((0) (0) − (1) (1)) + ((0) (1) + (0) (1)) i
= −1
So, by defining multiplication as we’ve done above we get that i2 = −1 as a consequence of the
√
definition instead of just stating that this is a true fact
...
e
...


3 Arithmetic
Before proceeding in this section let me first say that I’m assuming that you’ve seen arithmetic with
complex numbers at some point before and most of what is in this section is going to be a review for
you
...


© Paul Dawkins

–4–

Complex Number Primer

Arithmetic

In the previous section we defined addition and multiplication of complex numbers and showed that
i2 = −1 is a consequence of how we defined multiplication
...
In practice we tend to just multiply two complex
numbers much like they were polynomials and then make use of the fact that we now know that
i2 = −1
...


Example 1
Compute each of the following
...

(a) (58 − i) + (2 − 17i) = 58 − i + 2 − 17i = 60 − 18i
(b) (6 + 3i) (10 + 8i) = 60 + 48i + 30i + 24i2 = 60 + 78i + 24 (−1) = 36 + 78i
(c) (4 + 2i) (4 − 2i) = 16 − 8i + 8i − 4i2 = 16 + 4 = 20

It is important to recall that sometimes when adding or multiplying two complex numbers the result
might be a real number as shown in the third part of the previous example!
The third part of the previous example also gives a nice property about complex numbers
...

Let’s now take a look at the subtraction and division of two complex numbers
...

Let’s take a quick look at an example of both to remind us how they work
...

(a) (58 − i) − (2 − 17i)
(b)

6+3i
10+8i

(c)

5i
1−7i

Solution
(a) (58 − i) − (2 − 17i)
There really isn’t too much to do here so here is the work,
(58 − i) − (2 − 17i) = 58 − i − 2 + 17i = 56 + 16i

(b)

6 + 3i
10 + 8i
Recall that with division we just need to eliminate the i from the denominator and using
Equation 1 we know how to do that
...

6 + 3i
(6 + 3i) (10 − 8i)
=
10 + 8i
(10 + 8i) (10 − 8i)
60 − 48i + 30i − 24i2
=
100 + 64
84 − 18i
=
164
84
18
=
−
i
164 164
21
9
=
− i
41 82

© Paul Dawkins

–6–

Complex Number Primer

(c)

Arithmetic

5i
1 − 7i
We’ll do this one a little quicker
...
However, let’s take a look at a more precise and mathematical definition of both of these
...

The remainder of this document involves topics that are typically first taught in a Abstract/Modern
Algebra class
...
Also note that we’re going to be skipping some of the
ideas and glossing over some of the details that don’t really come into play in complex numbers
...
The definitions I’ll be giving below are
correct for complex numbers, but in a more general setting are not quite correct
...
I just wanted to make
it clear that I’m skipping some of the more general definitions for easier to work with definitions that
are valid in complex numbers
...

Technically, the only arithmetic operations that are defined on complex numbers are addition and
multiplication
...
We’ll start with subtraction since it is (hopefully) a little easier to
see
...
An additive inverse is some element
typically denoted by −z so that
z + (−z) = 0
(4)
Now, in the general field of abstract algebra, −z is just the notation for the additive inverse and in
many cases is NOT given by −z = (−1) z ! Luckily for us however, with complex variables that
is exactly how the additive inverse is defined and so for a given complex number z = a + bi the
additive inverse, −z, is given by,
−z = (−1) z = −a − bi
It is easy to see that this does meet the definition of the additive inverse and so that won’t be
shown
...
Given two
complex numbers z1 = a + bi and z2 = c + di we define the subtraction of them as,
z1 − z2 = z1 + (−z2 )
© Paul Dawkins

(5)
–7–

Complex Number Primer

Arithmetic

Or, in other words, when subtracting z2 from z1 we are really just adding the additive inverse of z2
(which is denoted by −z2 ) to z1
...

z1 − z2 = z1 + (−z2 ) = (a + bi) + (−c − di) = (a − c) + (b − d) i
So, that wasn’t too bad I hope
...
Or, to
put it another way, you’ve always been taught that −z is just a shorthand notation for (−1) z, but
in the general topic of abstract algebra this does not necessarily have to be the case
...
Okay, now that we have subtraction out of the way, let’s
move on to division
...
This time we’ll need
a multiplicative inverse
...
It isn’t! It is just a notation that is used to denote the multiplicative inverse
...
However, with complex numbers this will not be the case! In fact, let’s see just what
the multiplicative inverse for a complex number is
...
Now, we know that we must have
z z −1 = 1
so, let’s actual do the multiplication
...

So, now that we have the definition of the multiplicative inverse we can finally define division of
two complex numbers
...
Note as well that this actually does match with the process that we
used above
...
So, let’s start out with the following division
...

As a final topic let’s note that if we don’t want to remember the formula for the multiplicative inverse
we can get it by using the process we used in the original multiplication
...


© Paul Dawkins

–9–

Complex Number Primer

Conjugate and Modulus

4 Conjugate and Modulus
In the previous section we looked at algebraic operations on complex numbers
...
We’ll also
take a look at quite a few nice facts about these operations
...
Given the complex
number z = a + bi the complex conjugate is denoted by z and is defined to be,
z = a − bi

(8)

In other words, we just switch the sign on the imaginary part of the number
...

z=z

(9)

z1 ± z2 = z 1 ± z 2

(10)

z 1 z2 = z 1 z 2

z1
z1
=
z2
z2

(11)



(12)

The first one just says that if we conjugate twice we get back to what we started with originally and
hopefully this makes some sense
...

So, just so we can say that we worked a number example or two let’s do a couple of examples
illustrating the above facts
...

(a) z for z = 3 − 15i
(b) z1 − z2 for z1 = 5 + i and z2 = −8 + 3i
(c) z1 − z2 for z1 = 5 + i and z2 = −8 + 3i

Solution
There really isn’t much to do with these other than to so the work so,
(a) z = 3 + 15i

⇒

z = 3 + 15i = 3 − 15i = z

Sure enough we can see that after conjugating twice we get back to our original number
...


There is another nice fact that uses conjugates that we should probably take a look at
...
We’ll start with a complex number z = a + bi and
then perform each of the following operations
...
Given a complex number z = a + bi the modulus is denoted by |z| and is defined by
p
|z| = a2 + b2
(14)
Notice that the modulus of a complex number is always a real number and in fact it will never be
negative since square roots always return a positive number or zero depending on what is under
the radical
...
e
...
So, from
this we can see that for real numbers the modulus and absolute value are essentially the same
thing
...
To see this let’s square both sides of Equation 14 and use the fact that
Re z = a and Im z = b
...
Finally, for any real number a we also know that a ≤ |a| (absolute value
...
Let’s
start with a complex number z = a + bi and take a look at the following product
...

Notice as well that in computing the modulus the sign on the real and imaginary part of the complex
number won’t affect the value of the modulus and so we can also see that,
|z| = |z|

(18)

|−z| = |z|

(19)

and

We can also now formalize the process for division from the previous section now that we have
the modulus and conjugate notations
...
Then
using Equation 17 we can simplify the notation a little
...

10 + 8i

Solution
In this case we have z1 = 6 + 3i and z2 = 10 + 8i
...

If |z| = 0 then z = 0
|z1 z2 | = |z1 | |z2 |
 
 z1  |z1 |
 =
 z2  |z2 |

(20)
(21)
(22)

Property Equation 20 should make some sense to you
...

To verify Equation 21 consider the following,
|z1 z2 |2 = (z1 z2 ) (z1 z2 )

using property (17)

= (z1 z2 ) (z 1 z 2 )

using property (11)

= z1 z 1 z2 z 2

rearranging terms

2

= |z1 | |z2 |

2

using property (17) again (twice)

So, from this we can see that
|z1 z2 |2 = |z1 |2 |z2 |2
Finally, recall that we know that the modulus is always positive so take the square root of both
sides to arrive at
|z1 z2 | = |z1 | |z2 |
Property Equation 22 can be verified using a similar argument
...
We now need to take a
look at a similar relationship for sums of complex numbers
...

The triangle inequality is actually fairly simple to prove so let’s do that
...

|z1 + z2 |2 = (z1 + z2 ) (z1 + z2 ) = (z1 + z2 ) (z1 + z2 )
Now multiply out the right side to get,
|z1 + z2 |2 = z1 z 1 + z1 z 2 + z2 z 1 + z2 z 2

(24)

Next notice that,
z2 z 1 = z 2 z1 = z 2 z1
and so using Equation 13, Equation 15 and Equation 18 we can write middle two terms of the right
side of Equation 24 as
z1 z 2 + z2 z 1 = z1 z 2 + z1 z 2 = 2Re (z1 z 2 ) ≤ 2 |z1 z 2 | = 2 |z1 | |z 2 | = 2 |z1 | |z2 |
Also use Equation 17 on the first and fourth term in Equation 24 to write them as,
z1 z 1 = |z1 |2

z2 z 2 = |z2 |2

With the rewrite on the middle two terms we can now write Equation 24 as
|z1 + z2 |2

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