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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 = z1 z 1 + z1 z 2 + z2 z 1 + z2 z 2
= |z1 |2 + z1 z 2 + z2 z 1 + |z2 |2
≤ |z1 |2 + 2 |z1 | |z2 | + |z2 |2
= (|z1 | + |z2 |)2
So, putting all this together gives,
|z1 + z2 |2 ≤ (|z1 | + |z2 |)2
Now, recalling that the modulus is always positive we can square root both sides and we’ll arrive
at the triangle inequality
...

© Paul Dawkins

– 14 –

Complex Number Primer

Polar & Exponential Form

Let’s first start by assuming that |z1 | ≥ |z2 |
...
So, let’s start with |z1 | and do some
work on it
...

|z1 + z2 | ≥ | |z1 | − |z2 | |

(27)

Also, if we replace z2 with −z2 in Equation 23 and Equation 27 we arrive at two more variations of
the triangle inequality
...


5 Polar & Exponential Form
Most people are familiar with complex numbers in the form z = a + bi, however there are some
alternate forms that are useful at times
...

Geometric Interpretation
Before we get into the alternate forms we should first take a very brief look at a natural geometric
interpretation of complex numbers since this will lead us into our first alternate form
...
We can think of this complex number as either the point
(a, b) in the standard Cartesian coordinate system or as the vector that starts at the origin and ends
at the point (a, b)
...


© Paul Dawkins

– 15 –

Complex Number Primer

Polar & Exponential Form

In this interpretation we call the x-axis the real axis and the y-axis the imaginary axis
...

Note as well that we can now get a geometric interpretation of the modulus
...
This interpretation also tells us that the inequality
|z1 | < |z2 | means that z1 is closer to the origin (in the complex plane) than z2 is
...
If we think of the non-zero
complex number z = a + bi as the point (a, b) in the xy-plane we also know that we can represent
this point by the polar coordinates (r, θ), where r is the distance of the point from the origin and θ
is the angle, in radians, from the positive x-axis to the ray connecting the origin to the point
...
Note that this means that there
are literally an infinite number of choices for θ
...
We will therefore only consider the
polar form of non-zero complex numbers
...

a = r cos(θ)

b = r sin(θ)

If we substitute these into z = a + bi and factor an r out we arrive at the polar form of the complex
number,
z = r (cos(θ) + i sin(θ))
(30)
Note as well that we also have the following formula from polar coordinates relating r to a and
b
...

Note as well that any two values of the argument will differ from each other by an integer multiple
of 2π
...

For a given complex number z pick any of the possible values of the argument, say θ
...
Since
it takes 2π radians to make one complete revolution you will be back at your initial starting point
when you reach θ + 2π and so have a new value of the argument
...


© Paul Dawkins

– 17 –

Complex Number Primer

Polar & Exponential Form

If you keep increasing the angle you will again be back at the starting point when you reach θ + 4π,
which is again a new value of the argument
...

Likewise, if you start at θ and decrease the angle you will be rotating the point about the origin in a
clockwise manner and will return to your original starting point when you reach θ−2π
...

So, we can see that if θ1 and θ2 are two values of arg z then for some integer k we will have,
θ1 − θ2 = 2πk

(34)



Note that we’ve also shown here that z = r cos(θ) + i sin(θ) is a parametric representation for
a circle of radius r centered at the origin and that it will trace out a complete circle in a counterclockwise direction as the angle increases from θ to θ + 2π
...
Note that the
inequalities at either end of the range tells that a negative real number will have a principal value
of the argument of Arg z = π
...

arg z = Arg z + 2πn
© Paul Dawkins

n = 0, ±1, ±2,
...


Example 1
Write down the polar form of each of the following complex numbers
...

r = |z| =

√

1+3=2

Now let’s find the argument of z
...
The principal value
of the argument will be the value of θ that is in the range −π < θ ≤ π, satisfies,
√
 √ 
3
tan(θ) =
⇒
θ = tan−1 − 3
−1
and is in the second quadrant since that is the location the complex number in the
complex plane
...
Recall that if your calculator returns a value of
θ1 then the second value that will also satisfy the equation will be θ2 = θ1 + π
...
You will need to compute both and the determine
which falls into the correct quadrant to match the complex number we have because
only one of them will be in the correct quadrant
...
Therefore, the principal value of the argument is,
Arg z =

© Paul Dawkins

2π
3
– 19 –

Complex Number Primer

Polar & Exponential Form

and all possible values of the argument are then
arg z =

2π
+ 2πn
3

n = 0, ±1, ±2,
...
Here is the polar form of
√
z = −1 + i 3
...
To get any of the other forms we
just need to compute a different value of the argument by picking n
...


 
 
8π
8π
z = 2 cos
+ i sin
n=1
3
3





16π
16π
z = 2 cos −
+ i sin −
n = −3
3
3
(b) z = −9
In this case we’ve already noted that the principal value of a negative real number is π
so we don’t need to compute that
...

arg z = π + 2πn = π (1 + 2n)
Now, r is,
r = |z| =

√

n = 0, ±1, ±2,
...
If we were to use Equation 33
to find the argument we would run into problems since the real part is zero and this
would give division by zero
...
In the complex plane purely
imaginary numbers are either on the positive y-axis or the negative y-axis depending
on the sign of the imaginary part
...
Therefore, the principal value and the general argument for this
complex number is,


π
π
1
Argz =
arg z = + 2πn = π
+ 2n
n = 0, ±1, ±2,
...
First, we’ll need Euler’s formula,
ei θ = cos(θ) + i sin(θ)

(36)

With Euler’s formula we can rewrite the polar form of a complex number into its exponential form
as follows
...
Also, because any two arguments for a
give complex number differ by an integer multiple of 2π we will sometimes write the exponential
form as,
z = rei (θ+2πn)
n = 0, ±1, ±2,
...

To get the value of r we can either use Equation 32 to write the exponential form or we can take a
more direct approach
...
Take the modulus of both sides and then do
a little simplification as follows,
q


 
p
 i θ
 i θ
2
|z| = re  = |r| e  = |r| |cos(θ) + i sin(θ)| = r + 0 cos2 (θ) + sin2 (θ) = r
and we see that r = |z|
...

Now that we’ve got the exponential form of a complex number out of the way we can use this
along with basic exponent properties to derive some nice facts about complex numbers and their
arguments
...
In the arithmetic section we gave
a fairly complex formula for the multiplicative inverse, however, with the exponential form of the
complex number we can get a much nicer formula for the multiplicative inverse
...
So, putting this together the
exponential form of the multiplicative inverse is,
1
z −1 = ei (−θ)
(37)
r
and the polar form of the multiplicative inverse is,


1
z −1 = cos (−θ) + i sin (−θ)
(38)
r
We can also get some nice formulas for the product or quotient of complex numbers
...
Since θ1 is any value of arg z1 and θ2 is any value of
arg z2 we can see that,
arg (z1 z2 ) = arg z1 + arg z2

arg

© Paul Dawkins

z1
z2

(43)


= arg z1 − arg z2

(44)

– 22 –

Complex Number Primer

Powers and Roots

Note that Equation 43 and Equation 44 may or may not work if you use the principal value of the
argument, Arg z
...
In this case we have z1 z2 = −i and
the principal value of the argument for each is,
Arg (i) =

π
2

Arg (−1) = π

Arg (−i) = −

π
2

However,

3π
π
6= −
2
2
and so Equation 43 doesn’t hold if we use the principal value of the argument
...


As an interesting side note, Equation 44 actually does work for this example if we use the principal
arguments
...
Suppose that we have two complex numbers given by their
exponential forms, z1 = r1 ei θ 1 and z2 = r2 ei θ 2
...
In this
case we have,
r1 ei θ 1 = r2 ei θ 2
Then we will have z1 = z2 if and only if,
r1 = r2

and

θ2 = θ1 + 2πk for some integer k (i
...
k = 0, ±1, ±2,
...

This may seem like a silly fact, but we are going to use this in the next section to help us find the
powers and roots of complex numbers
...

We’ll start with integer powers of z = reiθ since they are easy enough
...

© Paul Dawkins

– 23 –

Complex Number Primer

Powers and Roots

Example 1
Compute (3 + 3i)5
...
Instead we can convert to exponential form and then use
Equation 46 to quickly get the answer
...

r=

√

√
9+9=3 2

tan(θ) =

3
3

⇒

Arg z =

π
4

√ π
3 + 3i = 3 2ei 4
Note that we used the principal value of the argument for the exponential form, although we
didn’t have to
...

 √ 5 5π
(3 + 3i) = 3 2 ei 4

 
 
√
5π
5π
= 972 2 cos
+ i sin
4
4
√
√ !
√
2
2
= 972 2 −
−
i
2
2
5

= −972 − 972i

So, there really isn’t too much to integer powers of a complex number
...

(cos(θ) + i sin(θ))n = cos (n θ) + i sin (n θ)

n = 0, ±1, ±2,
...
We’ll start this off “simple” by
finding the nth roots of unity
...
are the distinct solutions to the
equation,
zn = 1
Clearly (hopefully) z = 1 is one of the solutions
...
To do this we will use the fact from the previous sections that states that z1 = z2 if and
© Paul Dawkins

– 24 –

Complex Number Primer

Powers and Roots

only if
r1 = r2

and

θ2 = θ1 + 2πk for some integer k (i
...
k = 0, ±1, ±2,
...
Doing this gives,

n
reiθ = 1 ei(0)
⇒
rn ei nθ = 1 ei(0)
So, according to the fact these will be equal provided,
rn = 1

k = 0, ±1, ±2,
...


k = 0, ±1, ±2,
...
So, our points will lie on the unit circle and they will be equally
spaced on the unit circle at every 2π
n radians
...
, n − 1 since we will get back to where we started once we reach
k=n
Therefore, there are n nth roots of unity and they are given by,






2πk
2πk
2πk
exp i
= cos
+ i sin
k = 0, 1, 2,
...
First define,


2π
ωn = exp i
n

(47)

(48)

then the nth roots of unity are,
ωnk =



k


2π
2πk
exp i
= exp i
n
n

k = 0, 1, 2,
...
, ωnn−1

(49)

where ωn is defined in Equation 48
...


Solution
We’ll start with n = 2
...
This should not be too surprising
as all we were doing was solving the equation
z2 = 1
and we all know that -1 and 1 are the two solutions
...
In this case we are really solving
z3 = 1
and in the world of real numbers we know that the solution to this is z = 1
...
The problem here is that
the remaining two are complex solutions and so are usually not thought about when solving
for real solution to this equation which is generally what we wanted up to this point
...



2π
ω3 = exp i
3
This gives,


1=1


2π
ω3 = exp i
3
 
 
2π
2π
= cos
+ i sin
3
3
√
1
3
=− +
i
2
2

ω32



4π
= exp i
3
 
 
4π
4π
= cos
+ i sin
3
3
√
1
3
=− −
i
2
2

I’ll leave it to you to check that if you cube the last two values you will in fact get 1
...
We’ll do this one much quicker than the previous cases
...
First let’s get some notation out of the way
...
So, if r0 = |z0 | and θ0 = arg z0 (note θ0 can be any value of
the argument, but we usually use the principal value) we have,

n
reiθ = r0 ei θ0
⇒
rn ei nθ = r0 ei θ0
So, this tells us that,
r=

√
n

r0

θ=

θ0 2πk
+
n
n

The distinct solutions to Equation 50 are then,
 

√
θ0 2πk
n
ak = r0 exp i
+
n
n

k = 0, ±1, ±2,
...
, n − 1

(51)

So, we can see that just as there were n nth roots of unity there are also n nth roots of z0
...
If a is any of the nth roots of z0 then all the
roots can be written as,
a, aωn , aωn2 ,
...


© Paul Dawkins

– 27 –

Complex Number Primer

Powers and Roots

Example 3
Compute all values of the following
...

 π
2i = 2 exp i
2
So, if we use θ0 =

π
2

we can use Equation 51 to write down the roots
...


(b)

√


1
3 −i 3

Here’s the exponential form of the number,
  π 
√
3 − i = 2 exp i −
6
Using Equation 51 the roots are,
 

√
π
2πk
3
ak = 2 exp i − +
18
3

© Paul Dawkins

k = 0, 1, 2

– 28 –

Complex Number Primer

Powers and Roots

Plugging in for k gives,
  π  √ 
 π
 π 
√
3
3
a0 = 2 exp i −
= 2 cos −
+ i sin −
= 1
...
21878 i
18
18
18







√
√
11π
11π
11π
3
3
a1 = 2 exp i
= 2 cos
+ i sin
= −0
...
18394 i
18
18
18







√
√
23π
23π
23π
3
3
a2 = 2 exp i
= 2 cos
+ i sin
= −0
...
96516 i
18
18
18
As with the previous part I’ll leave it to you to check that if you cube each of these you
√
will get 3 − i

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