The Field of Complex Numbers
S. F. Ellermeyer
The construction of the system of complex numbers begins by appending
to the system of real numbers a number which we call i with the property
that i2 = 1. (Note that there is no real number whose square is 1.)
The system of complex numbers consists of all numbers of the form a + bi
where a and b are real numbers. We de…ne addition and multiplication for
complex numbers in such a way that the rules of addition and multiplication
are consistent with the rules for real numbers. Thus, if z1 = a1 + b1 i and
z2 = a2 + b2 i, we de…ne
z1 + z2 = (a1 + a2 ) + (b1 + b2 ) i
and
z1 z2 = (a1 + b1 i) (a2 + b2 i)
= a1 a2 + a1 b 2 i + a2 b 1 i + b 1 b 2 i 2
= (a1 a2 b1 b2 ) + (a1 b2 + a2 b1 ) i.
To be formal, we de…ne the system of complex numbers to consist of the
set
C = f(a; b) j a and b are real numbersg
with addition and multiplication de…ned for z1 = (a1 ; b1 ) and z2 = (a2 ; b2 ) in
C by
z2 + z2 = (a1 + a2 ; b1 + b2 )
and
z1 z2 = (a1 a2
b1 b2 ; a1 b2 + a2 b1 ) .
Note that this de…nition of the complex numbers (in terms of ordered pairs
of real numbers) is similar to the formal de…nition that we gave for rings
of polynomials –the di¤erences being that elements of rings of polynomials
1
consist of in…nite sequences of elements (only a …nite number of which are
non–zero) and that the rule of multiplication is de…ned di¤erently. The
number i 2 C is formally de…ned as i = (0; 1). In order to show that C (with
the operations of addition and multiplication de…ned above) is a …eld, me
must show that C is a commutative ring with unity and that all non–zero
members of C are units.
0.1
C is closed under addition and multiplication
Let z1 and z2 2 C. Then z1 = (a1 ; b1 ) and z2 = (a2 ; b2 ) where a1 ; a2 ; b1 ; and
b2 are real numbers. Also (by the de…nitions of addition and multiplication),
we have
z2 + z2 = (a1 + a2 ; b1 + b2 )
and
z1 z2 = (a1 a2
b1 b2 ; a1 b2 + a2 b1 ) .
Since a1 + a2 , b1 + b2 , a1 a2 b1 b2 , and a1 b2 + a2 b1 are all real numbers, we
see that z1 + z2 and z1 z2 are both members of C. Thus C is closed under
addition and multiplication.
0.2
Associative Laws of Addition and Multiplication
Let z1 = (a1 ; b1 ), z2 = (a2 ; b2 ), and z3 = (a3 ; b3 ) be members of C. Then
z1 + (z2 + z3 ) = (a1 ; b1 ) + (a2 + a3 ; b2 + b3 )
= (a1 + (a2 + a3 ) ; b1 + (b2 + b3 ))
= ((a1 + a2 ) + a3 ; (b1 + b2 ) + b3 )
= (z1 + z2 ) + z3
which shows that the associative property of addition is satis…ed. (Note that
the associative property of addition for real numbers is used in showing this.)
Also
z1 (z2 z3 )
= (a1 ; b1 ) (a2 a3 b2 b3 ; a2 b3 + a3 b2 )
= (a1 (a2 a3 b2 b3 ) b1 (a2 b3 + a3 b2 ) ; a1 (a2 b3 + a3 b2 ) + b1 (a2 a3 b2 b3 ))
= (a1 a2 a3 a1 b2 b3 a2 b1 b3 a3 b1 b2 ; a1 a2 b3 + a1 a3 b2 + a2 a3 b1 b1 b2 b3 )
2
and
(z1 z2 ) z3
= (a1 a2 b1 b2 ; a1 b2 + a2 b1 ) (a3 ; b3 )
= ((a1 a2 b1 b2 ) a3 (a1 b2 + a2 b1 ) b3 ; (a1 a2 b1 b2 ) b3 + (a1 b2 + a2 b1 ) a3 )
= (a1 a2 a3 a1 b2 b3 a2 b1 b3 a3 b1 b2 ; a1 a2 b3 + a1 a3 b2 + a2 a3 b1 b1 b2 b3 )
which shows that the associative property of multiplication is satis…ed.
0.3
Commutative Properties of Addition and Multiplication
Let z1 = (a1 ; b1 ) and z2 = (a2 ; b2 ) be members of C. Then
z1 + z2 = (a1 + a2 ; b1 + b2 )
= (a2 + a1 ; b2 + b1 )
= z2 + z1
which shows that the commutative property of addition is satis…ed, and
z1 z2 = (a1 ; b1 ) (a2 ; b2 )
= (a1 a2 b1 b2 ; a1 b2 + a2 b1 )
= (a2 a1 b2 b1 ; b2 a1 + b1 a2 )
= (a2 ; b2 ) (a1 ; b1 )
= z2 z1
which shows that the commutative property of multiplication is satis…ed.
0.4
Existence of Additive and Multiplicative Identity
Elements
The complex number 0 = (0; 0) serves as an additive identity element for C
because if z = (a; b) is any complex number then
z + 0 = (a; b) + (0; 0) = (a + 0; b + 0) = (a; b) = z
and it is also obvious that 0 + z = z.
3
The complex number 1 = (1; 0) serves as a multiplicative identity element
for C because if z = (a; b) is any complex number then
1z = (1; 0) (a; b) = (1a
0b; 1b + 0a) = (a; b) = z
(and, since the commutative property of multiplication is satis…ed, it is also
true that z1 = z).
0.5
Existence of Additive Inverses
Let z = (a; b) be a complex number. Then
number and
z = ( a; b) is also a complex
z + ( z) = (a; b) + ( a; b) = (a + ( a) ; b + ( b)) = (0; 0) = 0
which shows that
0.6
z (as de…ned) is the additive inverse of z.
Distributive Property
Let z1 = (a1 ; b1 ), z2 = (a2 ; b2 ), and z3 = (a3 ; b3 ) be members of C. Then
z1 (z2 + z3 )
= (a1 ; b1 ) (a2 + a3 ; b2 + b3 )
= (a1 (a2 + a3 ) b1 (b2 + b3 ) ; a1 (b2 + b3 ) + b1 (a2 + a3 ))
= (a1 a2 + a1 a3 b1 b2 b1 b3 ; a1 b2 + a1 b3 + a2 b1 + a3 b1 )
= ((a1 a2 b1 b2 ) + (a1 a3 b1 b3 ) ; (a1 b2 + a2 b1 ) + (a1 b3 + a3 b1 ))
= z1 z2 + z1 z3
which shows that the distributive property is satis…ed.
0.7
Existence of Multiplicative Inverses
Let z = (a; b) be a complex number with z 6= 0. (This means that not both
a and b are zero and thus that a2 + b2 6= 0.) We claim that the complex
number
b
a
;
z 1=
2
2
2
a +b
a + b2
4
is the multiplicative inverse of z. To check that this correct, observe that
zz
1
1
b
a
;
2
2
+b
a + b2
b2
ab
ab
a2
+
;
+
=
a2 + b 2 a2 + b 2 a2 + b 2 a2 + b 2
= (1; 0)
= 1.
= (a; b)
a2
Some Terminology, De…nitions, Etc.
A complex number of the form (a; 0) can be identi…ed with the real number a
and we make the convention to write (a; 0) = a. (For example, (0; 0) = 0 and
(1; 0) = 1.) A complex number of the form (0; b) is called a purely imaginary
number. Thus, for example, the number (0; 1) = i is a purely imaginary
number. A complex number, (a; b), for which b 6= 0 is called an imaginary
number. Thus the system of complex numbers is made up of the real numbers
and the imaginary numbers (some of which are purely imaginary).
If z = (a; b) is any complex number, then
(a; b) = (a; 0) + (0; b) = (a; 0) + (b; 0) (0; 1) = a + bi
and we usually …nd it more convenient to write z = a + bi (instead of z =
(a; b)) in doing calculations. The usual rules of algebra of the real numbers
carry over exactly to the complex numbers. We just need to remember that
i2 = (0; 1) (0; 1) = ( 1; 0) =
1.
However, the view of a complex number as an ordered pair of real numbers
is useful for gaining a visual picture of the complex numbers. Each complex
number, (a; b), can be identi…ed with the point (a; b) in the Cartesian Plane.
When dealing with complex numbers, we call this the complex plane.
Just as we de…ne the absolute value of a real number to be the distance
from that number to the origin on the number line, we de…ne the modulus
(or absolute value) of a complex number, z = (a; b), to be the distance from
z to 0 = (0; 0) in the complex plane. The notation for the modulus of z
is the same notation used for the absolute value of a real number. Thus if
z = (a; b), then
p
jzj = a2 + b2 .
5
The location of a complex number in the complex plane is completely
determined by the modulus of the number and by an argument of the
number. Assuming that z = a + bi 6= 0, an argument of z is any number
(angle) such that
cos ( ) =
a
b
and sin ( ) =
.
jzj
jzj
The basic geometric picture relating z, jzj, and
is given below.
z=a+bi
b
|z|
θ
a
Any complex number, z = a + bi 6= 0, has in…nitely many arguments but
they all di¤er by multiples of 2 . Thus, for example, =4 is an argument
of the number z = 1 + i, but so is =4 + 2 = 9 =4 and so is =4 8 =
31 =4. The principal argument of a complex number z 6= 0 is the unique
argument, , such that
<
. The principal argument of z is denoted
by Arg (z) and the set of all arguments of z is denoted by arg (z). Thus
arg (z) = fArg (z) + 2k j k 2 Zg .
Using the notion of congruence modulo 2 , we can write this as
arg (z) = [Arg (z)]2 .
To illustrate these ideas, let us consider the complex number z = 1 + i.
6
For this number, we have
p
p
jzj = 12 + 12 = 2
Arg (z) =
arg (z) =
4
9 17
;
;::: [
4 4 4
;
7
;
4
15
;
4
23
;::: .
4
This example is illustrated in the picture below.
z=1+i
2
π /4
We remark that the complex number 0 is the only complex number for
which the argument is not de…ned.
For any non–zero complex number, z = a + bi, with argument (where
can be any member of arg (z)), we observe that
z = a + bi = jzj cos ( ) + jzj sin ( ) i = jzj (cos ( ) + sin ( ) i) .
In order to express this more brie‡y, we de…ne
cis ( ) = cos ( ) + sin ( ) i.
This allows us to write
z = a + bi = jzj cis ( ) .
For example, we can write
1+i=
p
2cis
7
4
.
Both ways of expressing a complex number (a + bi and jzj cis ( )) are useful.
In particular, we will see that the expression a + bi is useful in helping us to
gain a geometric understanding of addition of complex numbers; whereas the
expression jzj cis ( ) is useful in helping us to gain a geometric understanding
of multiplication of complex numbers.
Exercise 1 Let z =
3 + 4i and w = 4
i.
1. Compute z + w.
2. Compute zw.
3. Find w 1 .
4. Compute z=w (which means zw 1 ).
5. Find jzj, Arg (z), and arg (z) and write z in the form z = jzj cis ( )
(where is any member of arg (z)).
Exercise 2 Suppose that jzj = 3 and Arg (z) =
z = a + bi.
=6. Write z in the form
Exercise 3 If A is any set of real numbers and b is any real number, then
we de…ne the set A + b as
A + b = fa + b j a 2 Ag .
Thus, for example, if A = f 4; 1; 3; 35g, then A + 5 = f1; 6; 8; 40g.
Suppose that z is a non–zero complex number and that k is a non–zero real
number. Explain why arg (kz) = arg (z) if k > 0 and arg (kz) = arg (z) +
if k < 0.
2
2.1
Geometric Interpretations of Addition and
Multiplication of Complex Numbers
Addition
If z = a + bi and w = c + di are complex numbers, then
z + w = (a + c) + (b + d) i.
8
If we picture z and w as vectors in the complex plane, then z +w is the vector
sum of z and w. This is illustrated in the …gure below using the example
z = 3 + 4i
w=4 i
z + w = 1 + 3i.
8
6
4
2
-10
-5
5
10
-2
-4
-6
-8
2.2
Multiplication
If z = jzj cis ( ) and w = jwj cis ( ) are two non–zero complex numbers, then
zw
= jzj jwj (cos ( ) + sin ( ) i) (cos ( ) + sin ( ) i)
= jzj jwj (cos ( ) cos ( ) sin ( ) sin ( ) + (cos ( ) sin ( ) + sin ( ) cos ( )) i)
= jzj jwj (cos ( + ) + sin ( + ) i)
= jzj jwj cis ( + ) .
This shows that the modulus of zw is the product of the moduli of z and w
and that an argument of zw is the sum of an argument of z and an argument
of w.
p
To illustrate this idea, let us take z = 1+i and w = i. Note that jzj = 2,
Arg (z) = =4, jwj = 1, and Arg (w) = =2. Also
zw = (1 + i) i =
9
1+i
and we observe that
jzwj =
and that
Arg (zw) =
p
2 = jzj jwj
3
= Arg (z) + Arg (w) .
4
Exercise 4 Suppose that z is a non–zero complex number and suppose that
is an argument of z. Explain why
must be an argument of z 1 . (Hint:
What is an argument of the number 1?)
Exercise 5 Prove that if z = jzj cis ( ) and w = jwj cis ( ) are non–zero
complex numbers, then
z
jzj
=
cis (
).
w
jwj
Exercise 6 Explain why it is not always true (for any two non–zero complex
numbers, z and w) that
Arg (zw) = Arg (z) + Arg (w) :
Give an example of two complex numbers, z and w, for which the above
equation is true and then give an example where the above equation is not
true.
Exercise 7 If A and B are two sets, then we de…ne the set A + B as
A + B = fa + b j a 2 A and b 2 Bg .
For example, if A = f 4; 5; 6g and B = f1; 2; 5; 7g, then
A + B = f 3; 2; 1; 3; 6; 7; 10; 12; 8; 11; 13g .
Explain why it is true for any two non–zero complex numbers, z and w, that
arg (zw) = arg (z) + arg (w) .
10
3
De Moivre’s Formula and Roots of Complex Numbers
We will now prove (by induction) that if z = jzj cis ( ) is a non–zero complex
number and n is any positive integer, then
z n = jzjn cis (n ) .
This formula is clearly true when n = 1 and if we assume that it’s truth has
been proved for some positive integer n, then we obtain
z n+1
= zz n
= jzj cis ( ) jzjn cis (n )
= jzjn+1 (cos ( ) + sin ( ) i) (cos (n ) + sin (n ) i)
= jzjn+1 ((cos ( ) cos (n )
sin ( ) sin (n )) + (cos ( ) sin (n ) + sin ( ) cos (n )) i)
n+1
(cos ((n + 1) ) + sin ((n + 1) ) i)
n+1
cis ((n + 1) )
= jzj
= jzj
thus showing that the truth of the formula for n implies the truth of the
formula for n + 1. This formula is called De Moivre’s formula.
To illustrate
DeMoivre’s Formula, let us use it to compute (1 + i)8 . Since
p
j1 + ij = 2 and Arg (1 + i) = =4, we can write
p
1 + i = 2cis
.
4
Then by De Moivre’s formula,
p 8
(1 + i)8 =
2 cis 8
4
= 16 (cos (2 ) + sin (2 ) i)
= 16 (1 + 0i)
= 16.
We will now address the question of …nding the nth roots of a given non–zero
complex number, w = jwj cis ( ), where n is any given positive integer. Our
…rst observation is that one such root is the number
p
z0 = n jwjcis
n
11
because (by De Moivre’s formula)
z0n
=
p
n
p
n
n
jwjcis
n
n
jwj
=
cis n
n
= jwj cis ( ) = w.
We will show that w has exactly n distinct nth roots. To see this, let z be
any nth root of w (z n = w) and note that it must be true that
p
jzj = n jwj.
Also note that since
arg (w) = f + 2k j k 2 Zg ,
then it follows from De Moivre’s formula that
+ 2k
n
arg (z) =
For k = 0; 1; 2; : : : n
jk2Z .
1, the numbers
cis
+ 2k
n
are all distinct from each other but
+ 2j
n
cis
= cis
whenever j
k mod n because if j
integer t and we have
cis
+ 2j
n
+ 2k
n
k mod n, then j = k + tn for some
+ 2 (k + tn)
n
+ 2k
+ 2t
n
+ 2k
.
n
= cis
= cis
= cis
We conclude that if w = jwj cis ( ), then the n distinct nth roots of w are
z=
p
n
jwjcis
+ 2k
n
, k = 0; 1; 2; : : : n
12
1.
Note that, in the above discussion, can be any member of arg (w). In
particular, we can certainly
p take = Arg (w). The principal nth root of
w, which we denote by n w, is de…ned to be the root
p
n
w=
p
n
jwjcis
Arg (w)
n
.
Example 8 Let us …nd the three cube roots of 1.
Since j1j = 1 and Arg (1) = 0, the principal cube root of 1 is
p
3
1=
p
3
1cis
0
3
= cos (0) + sin (0) i = 1
and the other two cube roots of 1 are
p
3
1cis
0 + 2(1)
3
= cos
2
3
= cos
4
3
+ sin
2
3
+ sin
4
3
i=
p
1
3
+
i
2
2
i=
1
2
and
p
3
1cis
0 + 2 (2)
3
p
3
i.
2
p
Let us check by computation that 12 + 23 i is a cube root of 1. We will use
the formula (x + y)3 = x3 + 3x2 y + 3xy 2 + y 3 .
p !3
1
3
+
i
2
2
p !
p !2
p !3
3
2
1
1
3
1
3
3
+3
i +3
i +
i
=
2
2
2
2
2
2
p
p
1 3 3
9 3 3
=
+
i+
i
8
8
8
8
= 1.
Example 9 Let us …nd the two square roots of i.
Since jij = 1 and Arg (i) = =2, the principal square root of i is
p
p
p
p
=2
2
2
i = 1cis
= cos
+ sin
i=
+
i
2
4
4
2
2
13
and the other square root of i is
p
1cis
=2 + 2 (1)
2
= cos
5
4
+ sin
5
4
i=
p
2
2
p
2
i.
2
Remark 10 For any non–zero complex number,
w, the n nth roots of w are
p
spaced evenly around the circle of radius n jwj centered at the origin of the
complex plane.
Exercise 11 Find the four fourth roots of 1 and the four fourth roots of 16.
Exercise 12 Find the three cube roots of
1.
Exercise 13 Find the three cube roots of
8i.
4
Reducibility of Polynomials in C [x]
We have been studying the problem of reducibility of polynomials f 2 F [x]
where F is a given …eld (or perhaps even just a commutative ring with
unity). We have seen examples of polynomials that are not reducible in
Z [x] ; Q [x], and R [x] (where R is the …eld of real numbers). One of the
very important properties of the …eld of complex numbers, C, is that all
polynomials are reducible in C [x]! Furthermore (and this is actually an
equivalent statement), all polynomials can be factored into linear factors
in C [x]. This fact, which was discovered through the combined e¤orts of
many famous mathematicians (Gauss, LaGrange, Weierstrass, to name a few)
over a more than 100 year period is what is referred to as the Fundamental
Theorem of Algebra.
Theorem 14 (The Fundamental Theorem of Algebra) Let
f (x) = an xn + an 1 xn
1
+ : : : + a1 x + a0
be a polynomial in C [x]. Then f can be factored into n linear factors:
f (x) = an (x
r1 ) (x
r2 )
(x
rn )
where r1 ; r2 : : : ; rn are the (not necessarily distinct) roots of f .
14
Although we will not attempt to prove this theorem, we will illustrate it
with a simple example: If we consider the polynomial
f (x) = x5
3x4 + 2x3
6x2
8x + 24 = x2 + 4
x2
2 (x
3) ,
we see that this polynomial is reducible in Z [x] (which is shown by the factorization given above). However, since both x2 + 4 and x2 2 are irreducible
in Z [x], we see that f cannot be factored into linear factors in Z [x]. If we
consider f as a polynomial in R [x] (where R is the …eld of real numbers),
then
p
p
f (x) = x2 + 4 x
2 x + 2 (x 3)
which shows that f can be further factored in R [x] (in comparison with
Z [x]). Finally, if we consider f as a polynomial in C [x], then
p
p
f (x) = (x 2i) (x + 2i) x
2 x + 2 (x 3)
which is a complete factorization of f into linear factors.
Exercise 15 Factor the following polynomials as completely as possible in
Z [x], R [x], and C [x].
1. f (x) = 2x3
2. f (x) = x4
3x2 + 2x
7x2
3
18
3. f (x) = 4x4 + 5x2 + 1
Exercise 16 If z = a + bi is a complex number, then the complex conjugate
of z, denoted by z, is de…ned as z = a bi. For example, if z = 9 + 8i, then
z = 9 8i. Prove that the following facts are true for any complex numbers
z and w:
1. z + w = z + w
2. zw = z w.
3. If z is a real number then z = z.
4. If f is any polynomial with real coe¢ cients and z is a complex number
that is a root of f , then z is also a root of f .
15