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Complex Numbers Primer
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. That is a subject that can (and does) take a whole course to cover. 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.
This document has been written with the assumption that you’ve seen complex numbers at some point in the past, know (or at least knew at some point in time) that complex
numbers can be solutions to quadratic equations, know (or recall) i = − 1 , and that you’ve seen how to do basic arithmetic with 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.
For most students the assumptions I’ve made above about their exposure to complex numbers is the extent of their exposure. 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. Students are then all of a sudden expected to know more than basic arithmetic of complex numbers but often haven’t actually seen it anywhere and have to quickly pick it up on their own in order to survive in the class.
That is the purpose of this document. We will go beyond the basics that most students have seen at some point and show you some of the notation and operations involving complex numbers that many students don’t ever see once they learn how to deal with complex numbers as solutions to quadratic equations. 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.
The first section is a more mathematical definition of complex numbers and is not really required for understanding the remainder of the document. It is presented solely for those who might be interested.
The second section (arithmetic) is assumed to be mostly a review for those reading this document and can be read if you need a quick refresher on how to do basic arithmetic with complex numbers. 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. Again, understanding these definitions is not required for the remainder of the document they are only presented so you can say you’ve seen it.
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.
So, with that out of the way, let’s get started…
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 i^2 = − 1. This is an idea that most people first
see in an algebra class (or wherever they first saw complex numbers) and i = − 1 is defined so that we can deal with square roots of negative numbers as follows,
− 100 = (^) ( (^100) )( − (^1) ) = 100 − 1 = 100 i = 10 i
What I’d like to do is give a more mathematical definition of a complex numbers and
show that i^2 = − 1 (and hence i = − 1 ) can be thought of as a consequence of this definition. We’ll also take a look at how we define arithmetic for complex numbers.
What we’re going to do here is going to seem a little backwards from what you’ve probably already seen but is in fact a more accurate and mathematical definition of complex numbers. Also note that this section is not really required to understand the remaining portions of this document. It is here solely to show you a different way to define complex numbers.
So, let’s give the definition of a complex number.
Given two real numbers a and b we will define the complex number z as,
z = a + bi (1)
Note that at this point we’ve not actually defined just what i is at this point. 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,
Re z = a Im z = b (2)
There are a couple of special cases that we need to look at before proceeding. First, let’s take a look at a complex number that has a zero real part, z = 0 + bi = bi In these cases, we call the complex number a pure imaginary number.
Next, let’s take a look at a complex number that has a zero imaginary part, z = a + 0 i = a In this case we can see that the complex number is in fact a real number. Because of this we can think of the real numbers as being a subset of the complex numbers.
We next need to define how we do addition and multiplication with complex numbers. Given two complex numbers z 1 = a + bi and z (^) 2 = c + di we define addition and
multiplication as follows,
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. I am also going to be introducing subtraction and division in a way that you probably haven’t seen prior to this point, but the results will be the same and aren’t important for the remaining sections of this document.
In the previous section we defined addition and multiplication of complex numbers and
showed that i^2^ = − 1 is a consequence of how we defined multiplication. However, in practice, we generally don’t multiply complex numbers using the definition. 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 i^2^ = − 1.
Just so we can say that we’ve worked an example let’s do a quick addition and multiplication of complex numbers.
Example 1 Compute each of the following.
(a) (^) ( 58 − i (^) ) + (^) ( 2 − 17 i ) (b) (^) ( 6 + 3 i (^) ) ( 10 + 8 i ) (c) (^) ( 4 + 2 i (^) ) ( 4 − 2 i )
Solution As noted above, I’m assuming that this is a review for you and so won’t be going into great detail here.
(a) (^) ( 58 − i (^) ) + (^) ( 2 − 17 i (^) )= 58 − i + 2 − 17 i = 60 − 18 i
(b) (^) ( 6 + 3 i (^) ) ( 10 + 8 i (^) ) = 60 + 48 i + 30 i + 24 i^2 = 60 + 78 i + (^24) ( − (^1) ) = 36 + 78 i
(c) (^) ( 4 + 2 i (^) ) ( 4 − 2 i (^) ) = 16 − 8 i + 8 i − 4 i^2 = 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.
(^ a^ +^ bi^ )( a^ −^ bi^ ) =^ a^^2 +^ b^2 (5)
We’ll be using this fact with division and looking at it in slightly more detail in the next section.
Let’s now take a look at the subtraction and division of two complex numbers. Hopefully, you recall that if we have two complex numbers, (^) z 1 = a + bi and (^) z 2 = c + di
then you subtract them as,
z 1 (^) − z 2 = (^) ( a + bi (^) ) − (^) ( c + di (^) ) = (^) ( a − c (^) ) + (^) ( b − d i ) (6)
And that division of two complex numbers,
1 2
z a bi z c di
can be thought of as simply a process for eliminating the i from the denominator and writing the result as a new complex number u + vi.
Let’s take a quick look at an example of both to remind us how they work.
Example 2 Compute each of the following.
(a) (^) ( 58 − i (^) ) − (^) ( 2 − 17 i )
(b)
i i
(c)
i − i Solution (a) There really isn’t too much to do here so here is the work,
( 58 −^ i^ ) −^ ( 2 −^17 i^ )=^58 −^ i^ −^2 +^17 i^ =^56 +^16 i
(b) Recall that with division we just need to eliminate the i from the denominator and using (1) we know how to do that. All we need to do is multiply the numerator and denominator by (^10) − 8 i and we will eliminate the i from the denominator.
( ) ( )
( ) ( ) 2
i^ i^ i i i i i i i
i
i i
+ +^ −
(c) We’ll do this one a little quicker.
( )
( ) ( )
i i i i i i i i
Now, for the most part this is all that you need to know about subtraction and multiplication of complex numbers for this rest of this document. However, let’s take a look at a more precise and mathematical definition of both of these. If you aren’t interested in this then you can skip this and still be able to understand the remainder of this document.
The remainder of this document involves topics that are typically first taught in a Abstract/Modern Algebra class. Since we are going to be applying them to the field of
shorthand notation for (^) ( − (^1) ) z , but in the general topic of abstract algebra this does not
necessarily have to be the case. It’s just that in all of the examples where you are liable to run into the notation − z in “real life”, whatever that means, we really do mean
− z = ( − 1 ) z.
Okay, now that we have subtraction out of the way, let’s move on to division. As with subtraction we first need to define an inverse. This time we’ll need a multiplicative inverse. A multiplicative inverse for a non-zero complex number z is an element denoted by z −^1 such that
z z −^1 = 1
Now, again, be careful not to make the assumption that the “exponent” of -1 on the notation is in fact an exponent. It isn’t! It is just a notation that is used to denote the multiplicative inverse. With real (non-zero) numbers this turns out to be a real exponent and we do have that
4 1 1 4
for instance. 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. Let’s start out with the
complex number z = a + bi and let’s call its multiplicative inverse z −^1 = u + vi. Now, we know that we must have
z z −^1 = 1
so, let’s actual do the multiplication.
( )( ) ( ) ( )
1
zz a bi u vi au bv av bu i
This tells us that we have to have the following, au − bv = 1 av + bu = 0 Solving this system of two equations for the two unknowns u and v (remember a and b are known quantities from the original complex number) gives,
2 2 2 2
a b u v a b a b
Therefore, the multiplicative inverse of the complex number z is,
1 2 2 2 2
a b z i a b a b
As you can see, in this case, the “exponent” of -1 is not in fact an exponent! Again, you really need to forget some notation that you’ve become familiar with in other math courses.
So, now that we have the definition of the multiplicative inverse we can finally define division of two complex numbers. Suppose that we have two complex numbers z 1 and
z 2 then the division of these two is defined to be, 1 1 1 2 2
z z z z
In other words, division is defined to be the multiplication of the numerator and the multiplicative inverse of the denominator. Note as well that this actually does match with the process that we used above. Let’s take another look at one of the examples that we looked at earlier only this time let’s do it using multiplicative inverses. So, let’s start out with the following division.
( )( )
i i i i
We now need the multiplicative inverse of the denominator and using (6) this is,
( )
1 2 2 2 2
i i i
Now, we can do the multiplication,
( )( ) ( )
i i i i i i i i i i
Notice that the second to last step is identical to one of the steps we had in the original working of this problem and, of course, the answer is the same.
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. In other words, to get the multiplicative inverse we can do the following
( ) ( ) ( )
1 2 2
i i i i i
As you can see this is essentially the process we used in doing the division initially.
There is another nice fact that uses conjugates that we should probably take a look at. However, instead of just giving the fact away let’s derive it. We’ll start with a complex number z = a + bi and then perform each of the following operations.
( ) ( ) 2 2
z z a bi a bi z z a bi a bi a bi
Now, recalling that Re z = a and Im z = b we see that we have,
Re Im 2 2
z z z z z z i
Modulus The other operation we want to take a look at in this section is the modulus of a complex
number. Given a complex number z = a + bi the modulus is denoted by z and is
defined by
z = a^2^ + b^2 (7)
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.
Notice that if z is a real number ( i.e. z = a + 0 i ) then,
z = a^2 = a
where the ⋅ on the z is the modulus of the complex number and the ⋅ on the a is the
absolute value of a real number (recall that in general for any real number a we have
a^2 = a ). So, from this we can see that for real numbers the modulus and absolute
value are essentially the same thing.
We can get a nice fact about the relationship between the modulus of a complex numbers and its real and imaginary parts. To see this let’s square both sides of (7) and use the fact that Re z = a and Im z = b. Doing this we arrive at
( ) ( )
(^2 2 2 2 ) z = a + b = Re z + Im z
Since all three of these terms are positive we can drop the Im z part on the left which gives the following inequality,
( ) ( ) ( )
2 2 2 2 z = Re z + Im z ≥ Re z
If we then square root both sides of this we get,
z ≥ Re z
where the ⋅ on the z is the modulus of the complex number and the ⋅ on the Re z are
absolute value bars. Finally, for any real number a we also know that a ≤ a (absolute
value…) and so we get,
z ≥ Re z ≥ Re z (8)
We can use a similar argument to arrive at,
z ≥ Im z ≥ Im z (9)
There is a very nice relationship between the modulus of a complex number and it’s conjugate. Let’s start with a complex number z = a + bi and take a look at the following product.
z z = (^) ( a + bi (^) )( a − bi (^) ) = a^2^ + b^2
From this product we can see that 2 z z = z (10)
This is a nice and convenient fact on occasion.
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 (11)
and
− z = z (12)
We can also now formalize the process for division from the previous section now that we have the modulus and conjugate notations. In order to get the i out of the denominator of the quotient we really multiplied the numerator and denominator by the conjugate of the denominator. Then using (10) we can simplify the notation a little. Doing all this gives the following formula for division,
1 1 2 1 2 2 (^2 2 2 )
z z z z z z z z (^) z
Here’s a quick example illustrating this,
Example 2 Evaluate
i i
Solution In this case we have (^) z 1 (^) = 6 + 3 i and (^) z 2 (^) = 10 + 8 i. Then computing the various parts of the
formula gives, (^2 2 ) z 2 (^) = 10 − 8 i z 2 = 10 + 8 = 164
The quotient is then,
6 3 (^6 3 )(^10 8 )^60 48 30 24 221 10 8 164 164 41 82
i i^ i i i i i i
+ +^ − − + −
Here are some more nice facts about the modulus of a complex number.
If z = 0 then z = 0 (13)
z z 2 1 (^) = z z 2 1 (^) = z z 2 1
and so using (6), (8) and (11) we can write middle two terms of the right side of (17) as
z z 1 2 (^) + z z 2 1 (^) = z z 1 2 (^) + z z 1 2 (^) = 2 Re (^) ( z z 1 2 (^) )≤ 2 z z 1 (^) 2 = 2 z 1 (^) z (^) 2 = 2 z 1 (^) z 2
Also use (10) on the first and fourth term in (17) to write them as, 2 2 z z 1 1 (^) = z 1 (^) z z 2 (^) 2 = z 2
With the rewrite on the middle two terms we can now write (17) as
( )
2 1 2 1 1 1 2 2 1 2 2 2 2 1 1 2 2 1 2 2 2 1 1 2 2 2 1 2
z z z z z z z z z z
z z z z z z
z z z z
z z
So, putting all this together gives,
( )
2 2 z 1 (^) + z 2 (^) ≤ z 1 (^) + z 2
Now, recalling that the modulus is always positive we can square root both sides and we’ll arrive at the triangle inequality.
z 1 (^) + z 2 (^) ≤ z 1 (^) + z 2
There are several variations of the triangle inequality that can all be easily derived.
Let’s first start by assuming that z 1 (^) ≥ z 2. This is not required for the derivation, but
will help to get a more general version of what we’re going to derive here. So, let’s start
with z 1 and do some work on it.
1 1 2 2 1 2 2 1 2 2
Using triangle inequality
z z z z z z z z z z
Now, rewrite things a little and we get,
z 1 (^) + z 2 (^) ≥ z 1 (^) − z 2 ≥ 0 (18)
If we now assume that z 1 (^) ≤ z 2 we can go through a similar process as above except this
time switch z 1 and z 2 and we get,
z 1 (^) + z 2 (^) ≥ z 2 (^) − z 1 (^) = − (^) ( z 1 (^) − z 2 )≥ 0 (19)
Now, recalling the definition of absolute value we can combine (18) and (19) into the following variation of the triangle inequality.
z 1 (^) + z 2 (^) ≥ z 1 (^) − z 2 (20)
Also, if we replace z 2 with − z 2 in (16) and (20) we arrive at two more variations of the
triangle inequality.
z 1 (^) − z 2 (^) ≤ z 1 (^) + z 2 (21)
z 1 (^) − z 2 (^) ≥ z 1 (^) − z 2 (22)
On occasion you’ll see (22) called the reverse triangle inequality.
Complex Numbers Primer
When working with complex numbers we assume that r is positive and that θ can be any of the possible (both positive and negative) angles that end at the ray. Note that this
means that there are literally an infinite number of choices for θ.
We excluded z = 0 since θ is not defined for the point (0,0). We will therefore only consider the polar form of non-zero complex numbers.
We have the following conversion formulas for converting the polar coordinates ( r , θ)
into the corresponding Cartesian coordinates of the point, ( a b , ).
a = r cos θ b = r sinθ
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 sinθ) (1)
Note as well that we also have the following formula from polar coordinates relating r to a and b.
r = a^2 + b^2 but, the right side is nothing more than the definition of the modulus and so we see that,
r = z (2)
So, sometimes the polar form will be written as,
z = z (^) ( cos θ + i sinθ) (3)
The angle θ is called the argument of z and is denoted by,
θ = arg z
The argument of z can be any of the infinite possible values of θ each of which can be found by solving
tan
b a
© 2006 Paul Dawkins 16 http://tutorial.math.lamar.edu/terms.aspx
Complex Numbers Primer
and making sure that θ is in the correct quadrant.
Note as well that any two values of the argument will differ from each other by an integer
multiple of 2 π. This makes sense when you consider the following.
For a given complex number z pick any of the possible values of the argument, say θ. If you now increase the value of θ , which is really just increasing the angle that the point makes with the positive x -axis, you are rotating the point about the origin in a counter-
clockwise manner. Since it takes 2 π radians to make one complete revolution you will
be back at your initial starting point when you reach θ + 2 π and so have a new value of
the argument. See the figure below.
If you keep increasing the angle you will again be back at the starting point when you
reach θ + 4 π, which is again a new value of the argument. Continuing in this fashion we
can see that every time we reach a new value of the argument we will simply be adding
multiples of 2 π onto the original 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 − 2 π. Continuing in this fashion and we can again see that each new value of the
argument will be found by subtracting a multiple of 2 π from the original value of the
argument.
So we can see that if θ 1 and θ 2 are two values of arg z then for some integer k we will
have,
θ 1 −θ 2 = 2 π k (5)
© 2006 Paul Dawkins 17 http://tutorial.math.lamar.edu/terms.aspx
Complex Numbers Primer
1 2
The first one is in quadrant four and the second one is in quadrant two and so is the one that we’re after. Therefore, the principal value of the argument is, 2 Arg 3
z
and all possible values of the argument are then 2 arg 2 0, 1, 2, 3
z n n
Now, let’s actually do what we were originally asked to do. Here is the polar form of
z = − + 1 i 3. 2 2 2 cos sin 3 3
z i
^ ^ ^
Now, for the sake of completeness we should acknowledge that there are many more equally valid polar forms for this complex number. To get any of the other forms we just need to compute a different value of the argument by picking n. Here are a couple of other possible polar forms.
8 8 2 cos sin 1 3 3 16 16 2 cos sin 3 3 3
z i n
z i n
^ ^ ^
^ ^ ^
(b) 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. For completeness sake here are all possible values
of the argument of any negative number.
arg z = π + 2 π n = π ( 1 + 2 n (^) ) n = 0, ±1, ±2,
Now, r is,
r = z = 81 + 0 = 9
The polar form (using the principal value) is,
z = 9 cos( ( π) + i sin( π))
Note that if we’d had a positive real number the principal value would be (^) Arg z = 0
(c) This another special case much like real numbers. If we were to use (4) to find the argument we would run into problems since the real part is zero and this would give division by zero. However, all we need to do to get the argument is think about where this complex number is in the complex plane. 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.
© 2006 Paul Dawkins 19 http://tutorial.math.lamar.edu/terms.aspx
Complex Numbers Primer
For our case the imaginary part is positive and so this complex number will be on the positive y -axis. Therefore, the principal value and the general argument for this complex number is, 1 Arg arg 2 2 0, 1, 2, 2 2 2
z z n n n
Also, in this case r = 12 and so the polar form (again using the principal value) is,
12 cos sin 2 2
z i
^ ^ ^
Exponential Form Now that we’ve discussed the polar form of a complex number we can introduce the second alternate form of a complex number. First, we’ll need Euler’s formula,
e i^^ θ^ = cos θ + i sinθ (7)
With Euler’s formula we can rewrite the polar form of a complex number into its exponential form as follows.
z = r e i^ θ
where θ = arg z and so we can see that, much like the polar form, there are an infinite
number of possible exponential forms for a given complex number. Also, because any
two arguments for a give complex number differ by an integer multiple of 2 π we will
sometimes write the exponential form as,
z = r e^ i^^ (^ θ^ +^2 π n ) n = 0, ±1, ±2,
where θ is any value of the argument although it is more often than not the principal value of the argument.
To get the value of r we can either use (3) to write the exponential form or we can take a more direct approach. Let’s take the direct approach. Take the modulus of both sides and then do a little simplification as follows,
z = r e i^^ θ = r e i θ = r cos θ + i sin θ = r^2 + 0 cos^2 θ + sin^2 θ= r
and so we see that r = z.
Note as well that because we can consider z = r (^) ( cos θ + i sinθ)as a parametric
representation of a circle of radius r and the exponential form of a complex number is
really another way of writing the polar form we can also consider z = r e i^ θa parametric representation of a circle of radius r.
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.
© 2006 Paul Dawkins 20 http://tutorial.math.lamar.edu/terms.aspx
