Adventures with Complex Numbers
H. C. Rae
Department of Mathematics
King's College London
http://www.mth.kcl.ac.uk/
26 June 2003
Introduction
Most of you have probably not yet encountered complex numbers.
I have set myself a difficult task: In the space of one hour, to tell
you about complex numbers and to use them to solve an interesting
problem which I sometimes set as a revision exercise at the start
of a course taken by second year mathematics students at King's.
Before stating the problem, let's quickly review some familiar
ideas and introduce some basic material on complex numbers.
The first equations which one encounters in the algebra
class are equations which we can solve in the set of integers.
For example 2x=12, has the solution x=6. However, not all
equations of the form
|
qx=p, where p,q are integers, q ¹ 0, |
| (1) |
have integer solutions; 2x=7, for example, is in this category. In order to solve
such equations we have to extend the integers to the set of rational
numbers,which includes the integers as a subset.
In the set of rational numbers equation (1) has the
solution
However, the introduction of the rational numbers does not bring
an end to our problems. As soon as we study quadratic equations
we encounter equations which cannot be solved in the set of
rational numbers. Perhaps the simplest example is the quadratic
equation
It is easy to prove that this equation does not
have any solution in the set of rational numbers.
Perhaps you are
familiar with the argument. Suppose that x=[ p/q], is a
rational number in its lowest terms (so that p,q have no common
factors) and such that x2=2. Then
We conclude that p is even, and we may write p=2r, for
some integer r. It now follows that
This means
that q is also a multiple of 2. The fact that p,q are both
even contradicts our assumption that they have no common factors; it follows that there is no
rational number x such that x2=2.
As you know, this difficulty disappears when we further enlarge our
number system to the set of real numbers, which include the
rational numbers as a subset. The equation x2=2 has two
solutions in the system of real numbers, namely x=±Ö2.
Complex Numbers
What about the quadratic equation
The square of any
real number is non-negative, so this equation has no solutions in
the real number system.
Bearing in mind the progression from integers, to rational
numbers, to real numbers, it is entirely natural to further
extend our number system to the complex numbers. We
postulate the complex number i with the property that
and equation (6) then has solutions x=±i.
If we now take the set of all real numbers, adjoin the number i, and
then formally add and multiply the numbers in this extended system
according to the familiar rules of algebra, we obtain the set of
Complex Numbers, consisting of all numbers x+iy, where x and y are real.
Adding & Multiplying Complex Numbers
We add and multiply complex numbers according to the usual rules
of algebra:
|
(x1+iy1)+(x2+iy2)=(x1+x2)+i(y1+y2), |
|
(x1+iy1)(x2+iy2)
=x1x2+x1(iy2)+(iy1)(x2)+(iy1)(iy2)
=(x1x2-y1y2)+i(x1y2+y1x2),
where we've used the fact that i2=-1.
Equality of Complex Numbers
When are two complex numbers equal? Suppose that
Then
|
(x1-x2)=-i(y1-y2), (x1-x2)2=-(y1-y2)2 |
|
because i2=-1. We conclude that
When we add two non-negative real numbers and get the answer zero,
each of these numbers must be zero. It follows that
|
x1+iy1=x2+iy2Þ x1=x2 and y1=y2 |
| (7) |
The multiplicative inverse of a non-zero complex number
The multiplicative inverse of a complex number z=x+iy is the complex number u+iv
which is such that
Let's determine u+iv. Multiplying both sides of this equation by
the complex number x-iy we obtain
so that, provided z isn't zero (i.e. x and y aren't both
zero)
|
u+iv= |
x-iy
x2+y2
|
, (Notice: x2+y2 is real) |
|
Equivalently we may write
|
|
1
z
|
= |
1
x+iy
|
= |
(x-iy)
(x+iy)(x-iy)
|
= |
x-iy
x2+y2
|
|
|
A comment
Before continuing, a comment may be in order. Perhaps you are just
a little uneasy about the complex numbers we have introduced.
Maybe you are saying: What on earth is this number i he's
introduced with the magic property that i2=-1?
One might first observe that
you would have been justified in making a similar response
when you first encountered rational numbers and real numbers. What is the real number Ö2, for example?
What is it?
At some stage in your mathematical career it is indeed important
to discover deeper answers to such questions.
As regards the complex numbers that we have introduced I could
perhaps show you (in the unlikely event of there being time!)
that they have an interpretation as a set of 2×2
real matrices - maybe some of you have come across matrices at
GCSE level - and addition and multiplication of complex
numbers, as we have defined them correspond, in a very precise and natural
way,
to the addition and multiplication of the matrices representing
the complex numbers.
Geometric Representation of Complex Numbers
Imagine the Cartesian x-y plane with origin O. An application
of the equality principle (7)
shows that
every complex number corresponds to a unique point in the
Cartesian plane; in fact, the complex number z=x+iy is
represented by the point P with coordinates (x,y). The complex
number zero corresponds to the origin O.
In this picture the real numbers,
regarded as a subset of the complex numbers, correspond
to points on the x-axis.
The point with coordinates (x,y) may also be represented by
polar coordinates (r,q) (see diagram) and in fact
|
x=rcosq, y=rsinq, r= |
Ö
|
x2+y2
|
, |
|
so that any non-zero complex number
z=x+iy can be expressed in polar form
as
An important point
Notice that the angle q in (8) is not unique, and if q0
is an allowed value of q then so is q0+k(2p), where k is any integer. We're using radian measure
for our angles, so that 2p radians correspond to one full
rotation about O, and k(2p) corresponds to k full
rotations about O, bringing us back to where we started, whatever the value of the integer k.
Suppose now that two complex numbers are equal and that
when they are expressed in polar form we have
Bearing in mind our geometrical interpretation of complex numbers
we deduce that
|
r=R, q = f+k(2p), k=0,±1,±2,¼ |
| (9) |
Properties of cosq+isinq
We note the following property:
|
(cosq1 +isinq1)(cosq2 +isinq2) |
|
|
= (cosq1cosq2 -sinq1 sinq2) |
|
|
+i(sinq1cosq2+cosq1sinq2) |
|
using some well known formulae from trigonometry.
A useful definition
Formula (10) tells us that cosq+isinq
behaves like the exponential function. In fact if we make the
definition
equation (10) then reads as
If you like, you can regard the definition (11) purely as
a simplifying notation - although it does in fact have a deeper
significance.
In terms of our definition we can express any non-zero complex number z=x+iy
in polar form as
|
z=x+iy=r(cosq+isinq)=reiq |
|
(In this context we recall equation (8)).
Applications
We're now in a position to solve any quadratic equation with real
coefficients, x2+x+1=0, for example. We have
Therefore
and
|
(x+1/2)=±iÖ3/2, x=-1/2±iÖ3/2. |
|
In our approach we have introduced complex numbers in order to
ensure that we can solve any quadratic equation. One might
imagine that in order to solve higher order polynomial equations,
such as cubic or quartic equations, it might be necessary to
introduce hypercomplex numbers but thanks to the so-called
Fundamental Theorem of Algebra
we know that this is not
the case. In fact, any polynomial equation of degree n of the form
|
xn+a1xn-1+a2xn-2+¼+an-1x+an=0, |
|
where the ak (k=1,2,¼,n) are real or complex
numbers,
has n roots in the system of complex numbers, repeated
roots being counted separately. This theorem is usually proved in
university courses on Complex Analysis.
Now let's address the problem that I mentioned at the start of my
talk - the one that I give to the second year students for
revision purposes! It may seem a strange problem at first sight but I hope you'll
find the proof interesting. I'd like to convince you that
|
sin2 |
æ
è
|
|
p
2n+1
|
|
ö
ø
|
sin2 |
æ
è
|
|
2p
2n+1
|
|
ö
ø
|
¼sin2 |
æ
è
|
|
np
2n+1
|
|
ö
ø
|
|
|
where n is any positive integer.
First of all, we note that it's true for n=1 since in this
case it reduces to the well known formula sin2(p/3)=3/4.
What about n=2? In this case my formula gives
|
sin2 |
æ
è
|
|
p
5
|
|
ö
ø
|
sin2 |
æ
è
|
|
2p
5
|
|
ö
ø
|
= |
5
16
|
, |
|
which may be verified using
|
sin2 |
æ
è
|
|
p
5
|
|
ö
ø
|
= |
5-Ö5
8
|
, sin2 |
æ
è
|
|
2p
5
|
|
ö
ø
|
= |
5+Ö5
8
|
, |
|
two results which aren't very hard to prove.
Of course, the fact that my formula is true for n=1 and n=2
doesn't prove that it's true for all positive integers n. I'll
deal with case n=3 and it will then be apparent to you that the
same argument can in fact be used to establish (12) for general n.
We start by
solving the equation
The Fundamental Theorem of Algebra, to which we referred above,
tells us that this equation has seven roots in the system of complex
numbers; one of them is obviously z=1, but how do we find the
others?
Well, we showed above that any non-zero complex number can be
expressed in the form
z=reiq º r(cosq+isinq) . Substituting this into
(13) gives
Thinking back to the geometrical interpretation of the polar
representation of a complex number - remember equation (9) - we conclude that
|
r7=1 , 7q = 0+k(2p), k=0,±1,±2,¼ |
|
and hence r=1, q = 2kp/7, k=0,±1,±2,¼.
We obtain the seven roots of the equation z7=1 by allowing k
to take the values 0,±1,±2,±3; other values of the
integer k will also provide solutions, but they merely duplicate
the roots provided by the choice k=0,±1,±2,±3. So the
seven roots of the equation z7=1 are given by
|
z=zk=ei(2kp)/7=cos(2kp/7)+isin(2kp/7), |
|
where k=0,±1, ±2,±3 . Note that the choice k=0 gives
the root z=1.
We note the simple formulae
|
zk+z-k=2cos(2kp/7), zkz-k=1 |
| (14) |
which follow immediately from our definition of zk.
Now that we know the roots of the equation z7=1 we are in
a position to factorise the polynomial z7-1, of degree seven
- it's an idea with which you are familiar in the context of
quadratic equations.
We can write
z7-1=(z-z0)(z-z1)(z-z-1)
×(z-z2)(z-z-2)(z-z3)(z-z-3).
Using the fact that z0=1 and multiplying out the brackets in consecutive pairs we then obtain, for any complex z,
z7-1=(z-1)(z2-(z1+z-1)z+z1z-1)
×(z2-(z2+z-2)z+z2z-2)
×(z2-(z3+z-3)z+z3z-3).
Applying equation (14) we then obtain the representation,
valid for any complex number z,
(z7-1)=(z-1)(z2-2zcos(2p/7)+1)
|
×(z2-2zcos(4p/7)+1)(z2-2zcos(6p/7)+1) |
| (15) |
Dividing through by (z-1), z ¹ 1, and using the identity
we obtain
1+z+z2+¼+z6=(z2-2zcos(2p/7)+1)
|
×(z2-2zcos(4p/7)+1)(z2-2zcos(6p/7)+1) |
| (16) |
which is valid for all complex z.
Putting z=1 in this formula
gives
|
7=23(1-cos(2p/7))(1-cos(4p/7))(1-cos(6p/7)) |
|
An application of the well known identity
then yields
|
sin2(p/7)sin2(2p/7)sin2(3p/7)= |
7
26
|
= |
7
64
|
. |
|
which is none other than formula (12) for the case
n=3.
Notice that 7=2(3)+1. In fact, we can establish the
result (12) for general n by considering the equation
using the same arguments as we have just given in the particular case n=3.
We may note in passing that other interesting formulae may be
derived from (15). For example, equating the
coefficients of z on both sides of the equation we find
|
0=1+2(cos(2p/7)+cos(4p/7)+cos(6p/7)) |
|
which generates,
after an application of some trigonometric identities,
|
cos(p/7)cos(2p/7)cos(3p/7)= |
1
8
|
. |
|
I checked this formula using Excel and it is indeed correct!
Conclusion
I hope that I've persuaded you that complex numbers are exciting objects to
study. Even at an elementary level one can often use complex
numbers to solve problems which are otherwise not very tractable.
In mathematics we are constantly looking to generalise. Having
introduced complex numbers, one of the most natural things to do would be to study
functions of a complex variable; instead of f(x), where x is a real variable,
why not study f(z), where z=x+iy is a complex variable? In fact the
theory of functions of a complex variable turns out to be one of the most
beautiful branches of Pure Mathematics, with wide ranging
applications in many other areas of pure mathematics and, at a more mundane level,
in electrical theory, hydrodynamics, and particle physics.
Epilogue
One more thing ... ... especially for those who feel that complex numbers are
a bit of a con trick but are happy to work with 2×2 matrices!
With any complex number z=x+iy we can associate a 2×2
matrix Z given by
and conversely.
According to this rule we associate with the complex numbers z1=x1+iy1, z2=x2+iy2
the matrices Z1, Z2 given by
|
Z1= |
æ
ç
è
|
|
ö
÷
ø
|
, Z2= |
æ
ç
è
|
|
ö
÷
ø
|
|
|
First of all, we note that z1=z2 Û Z1=Z2. For,
so that two complex numbers are equal if and only if the
matrices corresponding to them are equal.
What matrix corresponds to the complex number z1+z2? Well,
|
z1+z2=x1+iy1+x2+iy2=(x1+x2)+i(y1+y2) |
|
so that the matrix associated with z1+z2 is, by definition, the matrix
|
= |
æ
ç
è
|
|
ö
÷
ø
|
+ |
æ
ç
è
|
|
ö
÷
ø
|
=Z1+Z2. |
|
We see that addition of complex numbers corresponds to adding the
corresponding matrices which represent the complex numbers.
Similarly,
|
=(x1x2-y1y2)+i(y1x2+x1y2) |
|
so that the matrix associated with z1z2 is, according to our
definition,
|
= |
æ
ç
è
|
|
ö
÷
ø
|
|
æ
ç
è
|
|
ö
÷
ø
|
=Z1Z2 |
|
as follows from the rules for multiplying 2×2 matrices, as
some of you will have learned at GSCE level. In other words, in
order to multiply two complex numbers z1,z2 we might as well multiply (in the same order)
the matrices
Z1,Z2 which represent them in the correspondence which we
have defined. In similar fashion, it is not hard to verify that if
z=x+iy ¹ 0 and Z is the matrix representing z, as defined by
equation (17), then the complex number 1/z=z-1
is represented by the matrix Z-1, the inverse of Z, which is
given by
Finally, I'd like to mention one of several important points that we've glossed over, mainly because of lack of time. Multiplication of
complex numbers is commutative so, continuing in the above
notation, z1z2=z2z1, for any complex numbers z1,z2. You probably know that when
multiplying matrices the commutative rule does not always hold.
However, you can easily check that Z1Z2=Z2Z1, where Z1,Z2 are the matrices representing z1,z2 respectively,
just by doing the matrix multiplication.
So, if you're happy to work with 2×2 matrices you should
not have any anxiety about working with complex numbers, according
to the rules we have explained in this talk!
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