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Transcript
```Chapter 5
Complex numbers
Why be one-dimensional when you can be two-dimensional?
?
−3
−2
−1
0
1
2
3
?
We begin by returning to the familiar number line, where I have placed
the question marks there appear to be no numbers. I shall rectify this by
defining the complex numbers which give us a number plane rather than just
a number line. Complex numbers play a fundamental rôle in mathematics.
In this chapter, I shall use them to show how e and π are connected and
how certain primes can be factorized. They are also fundamental to physics
where they are used in quantum mechanics.
5.1
Complex number arithmetic
In the set of real numbers we can add, subtract, multiply and divide, but
we cannot always extract square roots. For example, the real number 1 has
125
126
CHAPTER 5. COMPLEX NUMBERS
the two real square roots 1 and −1, whereas the real number −1 has no real
square roots, the reason being that the square of any real non-zero number is
always positive. In this section, we shall repair this lack of square roots and,
as we shall learn, we shall in fact have achieved much more than this. Complex numbers were first studied in the 1500’s but were only fully accepted
and used in the 1800’s.
√
Warning! If r is a positive real number then r is usually interpreted to
mean the positive square root. If I want
√ to emphasize that both square roots
need to be considered I shall write ± r.
When the discriminant of a quadratic equation is strictly less than zero,
we know that it has no real roots. In this section, we shall show that in
this case the equation has two complex roots. This will mean that quadratic
equations will always have two roots. The key step is the following
We introduce a new number, denoted by i, whose defining property
is that i2 = −1. We shall assume that in all other respects it
satisfies the usual axioms of high-school algebra. This assumption
will be justified later.
We shall now explore the consequences of this definition which turns out to
be a profound one for mathematics.
It follows that i and −i are the two missing square roots of 1. In all other
respects the number i will behave like a real number. Thus if b is any real
number then bi is a number, and if a is any real number then a + bi is a
number.
A complex number is a number of the form a + bi where a, b ∈ R. We
denote the set of complex numbers by C. Complex numbers are sometimes
called imaginary numbers. This is not such a good term: they are not
figments of our imagination like unicorns or dragons. Like all numbers they
are, however, products of our imagination: no one has seen the complex
number number i but, then again, no one has seen the number 2. If z =
a + bi then we call a the real part of z, denoted Re(z), and b the complex or
imaginary part of z, denoted Im(z).
Two complex numbers a + bi and c + di are equal precisely when
a = c and b = d. In other words, when their real parts are equal
and when their complex parts are equal.
5.1. COMPLEX NUMBER ARITHMETIC
127
We can think of every real number as being a special kind of complex
number because if a is real then a = a + 0i. Thus R ⊆ C. Complex numbers
of the form bi are said to be purely imaginary.
Now we show that we can add, subtract, multiply and divide complex
numbers. Addition, subtraction and multiplication are all easy.
Let a + bi, c + di ∈ C. To add these numbers means to calculate (a + bi) +
(c+di). We assume that the order in which we add complex numbers doesn’t
matter and that we may bracket sums of complex numbers how we like and
still get the same answer and so we can rewrite this as a + c + bi + di. Next
we assume that multiplication of complex numbers distributes over addition
of complex numbers to get (a + c) + (b + d)i. Thus
(a + bi) + (c + di) = (a + c) + (b + d)i.
The definition of subtraction is similar and justified in the same way
(a + bi) − (c + di) = (a − c) + (b − d)i.
To multiply our numbers means to calculate (a + bi)(c + di). We first
assume complex multiplication distributes over complex addition to get (a +
bi)(c + di) = ac + adi + bic + bidi. Next we assume that the order in which
we multiply complex numbers doesn’t matter to get ac + adi + bic + bidi =
ac + adi + bci + bdi2 . Now we use the fact that i2 = −1 to get ac + adi + bci +
bdi2 = ac + adi + bci − bd. We now rearrange the terms to get the following
definition of multiplication
(a + bi)(c + di) = (ac − bd) + (ad + bc)i.
Examples 5.1.1. Carry out the following calculations.
1. (7 − i) + (−6 + 3i). We add together the real parts to get 1; adding
together −i and 3i we get 2i. Thus the solution is 1 + 2i.
2. (2 + i)(1 + 2i). First we multiply out the brackets as usual to get
2 + 4i + i + 2i2 . We now use the fact that i2 = −1 to get 2 + 4i + i − 2.
Finally we simplify to get 0 + 5i = 5i.
2
√
. Multiply out and simplify to get −i.
3. 1−i
2
128
CHAPTER 5. COMPLEX NUMBERS
The final operation is division. We have to show that when a + ib 6= 0
the reciprocal
1
a + ib
is also a complex number. We use an idea that can also be applied in other
situations called rationalizing the denominator. It is convenient first to define
a new operation on complex numbers. Let z = a + bi ∈ C. Define
z̄ = a − bi.
The number z̄ is called the complex conjugate of z. Why is this operation
useful? Let’s calculate z z̄. We have
z z̄ = (a + bi)(a − bi) = a2 − abi + abi − b2 i2 = a2 + b2 .
Notice that z z̄ = 0 if and only if z = 0. Thus for non-zero complex numbers
z, the number z z̄ is a positive real number. Let’s see how we can use the
complex conjugate to define division of complex numbers. Our goal is to
calculate
1
a + bi
where a+bi 6= 0. The first step is to multiply top and bottom by the complex
conjugate of a + bi. We therefore get
a − bi
1
a − bi
= 2
= 2
(a − bi) .
2
(a + bi)(a − bi)
a +b
a + b2
Examples 5.1.2. Carry out the following calculations.
1+i
.
i
1.
The complex conjugate of i is −i. Multiply top and bottom of the
fraction to get −i+1
= 1 − i.
1
2.
i
.
1−i
The complex conjugate of 1 − i is 1 + i. Multiply top and bottom
of the fraction to get i(1+i)
= i−1
.
2
2
3.
4+3i
.
7−i
The complex conjugate of 7 − i is 7 + i. Multiply top and bottom
of the fraction to get (4+3i)(7+i)
= 1+i
.
50
2
We shall need the following properties of the complex conjugate later on.
Lemma 5.1.3.
5.1. COMPLEX NUMBER ARITHMETIC
129
1. z1 + . . . + zn = z1 + . . . + zn .
2. z1 . . . zn = z1 . . . zn .
3. z is real if and only if z = z.
We now introduce a way of thinking about complex numbers that enables
us to visualize them. A complex number z = a + bi has two components: a
and b. It is irresistible to plot these as a point in the plane. The plane used
in this way is called the complex plane: the x-axis is the real axis and the
y-axis is interpreted as the complex axis.
z = a + ib
ib
a
Although a complex number can be thought of as labelling a point in the
complex plane, it can also be regarded as labelling the directed line segment
from
√ the origin to the point. By Pythagoras’ theorem, the length of this line
is a2 + b2 . We define
√
|z| = a2 + b2
where z = a + bi. This is called the modulus1 of the complex number z.
Observe that
√
|z| = z z̄.
We shall use the following important property of moduli.
Lemma 5.1.4. |wz| = |w| |z|.
Proof. Let
p− bd) + (ad + bc)i. Now
p w = a + bi and z = c + di. Then wz = (ac
|wz| = (ac − bd)2 + (ad + bc)2 whereas |w| |z| = (a2 + b2 )(c2 + d2 ). But
(ac − bd)2 + (ad + bc)2 = (ac)2 + (bd)2 + (ad)2 + (bc)2 = (a2 + b2 )(c2 + d2 ).
Thus the result follows.
1
Plural: moduli
130
CHAPTER 5. COMPLEX NUMBERS
The complex numbers were obtained from the reals by simply throwing in
one new number, i, a square root of −1. Remarkably, every complex number
has a square root.
Theorem 5.1.5. Every nonzero complex number has exactly two square
roots.
Proof. Let z = a + bi be a nonzero complex number. We want to find a
complex number w so that w2 = z. Let w = x + yi. Then we need to find
real numbers x and y such that (x+yi)2 = a+bi. Thus (x2 −y 2 )+2xyi = a+bi,
and so equating real and imaginary parts, we have to solve the following two
equations
x2 − y 2 = a and 2xy = b.
Now we actually have enough information to solve our problem, but we can
make life easier for ourselves by adding one extra equation. To get it, we use
the modulus function. From (x+yi)2 = a+bi
we get that |x + yi|2 = |a + bi|.
√
2
Now |x + yi| = x2 + y 2 and |a + bi| = a2 + b2 . We therefore have three
equations
√
x2 − y 2 = a and 2xy = b and x2 + y 2 = a2 + b2 .
If we add the first and third equation together we get
√
√
a2 + b 2
a + a2 + b 2
a
2
=
.
x = +
2
2
2
We can now solve for x and therefore for y.
Example 5.1.6. Every negative real number has√two square roots. We have
that the square roots of −r, where r > 0 are ±i r.
Example 5.1.7. Find both square roots of 3 + 4i and check your answers.
We assume that there is a complex number x + yi where both x and y are
real such that
(x + yi)2 = 3 + 4i.
Squaring and comparing real and imaginary parts we get that the following
two equations must be satisfied by x and y
x2 − y 2 = 3 and 2xy = 4.
5.1. COMPLEX NUMBER ARITHMETIC
131
We also have a third equation by taking moduli
x2 + y 2 = 5.
Adding the first and third equation together we get x = ±2. Thus y = 1 if
x = 2 and y = −1 if x = −2. The roots we want are therefore 2 + i and
−2 − i. Of course, one root will be minus the other. Now square either root
to check your answer: (2 + i)2 = 4 + 4i − 1 = 3 + 4i, as required.
Remark Notice that the two square roots of a non-zero complex number will
have the form w and −w; in other words, one root will be −1 times the other.
If we combine our method for solving quadratics with our method for
determining the square roots of complex numbers, we have a method for
finding the roots of quadratics with any coefficients, whether they be real or
complex.
Example 5.1.8. Solve the quadratic equation
4z 2 + 4iz + (−13 − 16i) = 0.
The complex numbers obey the same algebraic laws as the reals and so we
can solve this equation by completing the square or we can simply plug the
numbers into the formula for the roots of a quadratic. Here I shall complete
the square. First, we convert the equation into a monic one
z 2 + iz +
(−13 − 16i)
= 0.
4
Next, we observe that
2
1
i
= z 2 + iz − .
z+
2
4
Thus
2
i
1
z + iz = z +
+ .
2
4
2
Our equation therefore becomes
2
i
1
13
z+
+ + − − 4i = 0.
2
4
4
132
CHAPTER 5. COMPLEX NUMBERS
We therefore have
2
i
z+
= 3 + 4i.
2
Taking square roots of both sides using a previous calculation, we have that
z+
It follows that z = 2 +
work.
i
2
i
= 2 + i or − 2 − i.
2
or − 2 −
3i
.
2
Now check that these roots really do
Every quadratic equation ALWAYS has has exactly two roots.
Exercises 5.1
1. Solve the following problems in complex number arithmetic. In each
case, the answer should be in the form a + ib where a and b are real.
(a) (2 + 3i) + (4 + i).
(b) (2 + 3i)(4 + i).
(c) (8 + 6i)2 .
(d)
(e)
(f)
2+3i
.
4+i
3
1
+ 1+i
.
i
3+4i
3−4i
− 4+4i
.
3−4i
2. Find the square roots of each of the following complex numbers and
(a) −i.
√
(b) −1 + i 24.
(c) −13 − 84i.
(a) x2 + x + 1 = 0.
(b) 2x2 − 3x + 2 = 0.
(c) x2 − (2 + 3i)x − 1 + 3i = 0.
5.2. THE FUNDAMENTAL THEOREM OF ALGEBRA
5.2
133
The fundamental theorem of algebra
I want to describe now a result which is one of the most important consequences of the properties of complex numbers: the fundamental theorem of
algebra. It should be understood that this is a misnomer since algebra has
expanded beyond all bounds since this theorem was first proved. Nevertheless, it is an important result playing a key role in calculus where it is used
(in its real version which I also describe) to prove that any rational function
can be integrated using partial fractions.
In this section, we shall work with arbitrary polynomials and I shall now
recall some terminology for handling them. An expression
an xn + an−1 xn−1 + . . . + a1 x + a0
where ai are complex numbers, called the coefficients, is called a polynomial.
We assume an 6= 0. The degree of this polynomial is n. We abbreviate this
to deg. If an = 1 the polynomial is said to be monic. The term a0 is called
the constant term and the term an xn is called the leading term. Polynomials
can be added, subtracted and multiplied.
Two polynomials are equal if they have the same degree and the coefficients of terms of the same degree are equal.
• Polynomials of degree 1 are said to be linear;
• those of degree 2, quadratic;
• those of degree 3, cubic;
• those of degree 4, quartic;
• those of degree 5, quintic.
There are special terms for polynomials of degree higher than 5, if you want
them.
Why are polynomials interesting? There are two answers to this question.
First, they have widespread applications such as in helping to solve linear
differential equations and in studying matrices. Second, a polynomial defines
a function which is calculated in a very simple way using the operations of
addition, subtraction and multiplication. However many, more complicated,
functions can be usefully approximated by polynomial ones.
134
CHAPTER 5. COMPLEX NUMBERS
We denote by C[x] the set of polynomials with complex coefficients and
by R[x], the set of polynomials with real coefficients. I will write F [x] to
mean F = R or F = C.
5.2.1
The remainder theorem
The addition, subtraction and multiplication of polynomials is easy. We shall
therefore concentrate in this section on division.
Let f (x), g(x) ∈ F [x]. We say that g(x) divides f (x), denoted by
g(x) | f (x),
if there is a polynomial q(x) ∈ F [x] such that f (x) = g(x)q(x). We say that
g(x) is a factor.
Example 5.2.1. Let f (x) = x4 + 2x + 1 and g(x) = x + 1. Then
x + 1 | x4 + 2x + 1
since x4 + 2x + 1 = (x + 1)(x3 − x2 + x + 1).
In multiplying and dividing polynomials the following result is key.
Lemma 5.2.2. Let f (x), g(x) ∈ F [x] be non-zero polynomials. Then
deg f (x)g(x) = deg f (x) + deg g(x).
Proof. Let f (x) have leading term am xm and let g(x) have leading term bn xn .
Then the leading term of f (x)g(x) is am bn xm+n . Now am bn 6= 0 and so the
degree of f (x)g(x) is m + n, as required.
The following result is analogous to the remainder theorem for integers
Lemma 4.1.1
Lemma 5.2.3 (Remainder theorem). Let f (x) and g(x) be polynomials in
F [x] where deg f (x) ≥ deg g(x). Then either
g(x) | f (x)
or
f (x) = g(x)q(x) + r(x)
where deg r(x) < deg g(x).
5.2. THE FUNDAMENTAL THEOREM OF ALGEBRA
135
Example 5.2.4. Let f (x) = x3 + x + 3 and g(x) = x2 + x. Then x3 + x + 3 =
(x − 1)(x2 + x) + (2x + 3). Here x − 1 is the quotient and 2x + 3 is the
remainder.
The following example is a reminder of how to carry out long division of
polynomials. Remember that answers can always be checked by multiplying
out.
Example 5.2.5. Divide 6x4 + 5x3 + 4x2 + 3x + 2 by 2x2 + 4x + 5 and so find
the quotient and remainder. We set out the computation in the following
form.
2x2 + 4x + 5 6x4 + 5x3 + 4x2 + 3x + 2
To get the term involving 6x4 we would have to multiply the lefthand side
by 3x2 . As a result we write down the following
3x2
2x2 + 4x + 5 6x4 + 5x3 + 4x2 + 3x + 2
6x4 + 12x3 + 15x2
We now subtract the lower righthand side from the upper and we get
3x2
2x2 + 4x + 5 6x4 + 5x3 + 4x2 + 3x + 2
6x4 + 12x3 + 15x2
−7x3 − 11x2 + 3x + 2
The procedure is now repeated with the new polynomial.
3x2 − 72 x
2x2 + 4x + 5 6x4 + 5x3 + 4x2 + 3x + 2
6x4 + 12x3 + 15x2
−7x3 − 11x2 + 3x + 2
−7x3 − 14x2 − 35
x
2
3x2 + 41
x
+
2
2
136
CHAPTER 5. COMPLEX NUMBERS
The procedure is repeated one more time with the new polynomial
3x2 − 72 x + 32 quotient
2x2 + 4x + 5 6x4 + 5x3 + 4x2 + 3x + 2
6x4 + 12x3 + 15x2
−7x3 − 11x2 + 3x + 2
−7x3 − 14x2 − 35
x
2
41
2
3x + 2 x + 2
x + 15
3x2 + 12
2
2
29
11
x
−
remainder
2
2
This is the end of the line because the new polynomial we obtain has degree
strictly less than the polynomial we are dividing by. What we have shown is
that
3
29
11
7
4
3
2
2
2
+
x−
.
6x + 5x + 4x + 3x + 2 = 2x + 4x + 5 3x − x +
2
2
2
2
You can verify this is true by multiplying out the righthand side.
5.2.2
Roots of polynomials
Let f (x) ∈ F [x]. A number r ∈ F is said to be a root or zero of f (x) if
f (r) = 0. The roots of f (x) are the solutions of the equation f (x) = 0.
Example 5.2.6. The number 1 is a root of x100 −2x98 +1 because 1−2+1 = 0.
Checking whether a number is a root is easy, but finding a root in the
first place is trickier. The next result tells us that when we find roots of polynomials we are in fact determining linear factors. It is crucial to eveything
we shall do.
Proposition 5.2.7. Let r ∈ F . Then r is a root of f (x) ∈ F [x] if and only
if (x − r) | f (x).
Proof. Suppose that (x − r) | f (x). Then by definition f (x) = (x − r)q(x)
for some polynomial q(x). If we now calculate f (r) we see immediately that
it must be zero.
We now prove the converse. Suppose that r is a root of f (x). By the
remainder theorem, either (x − r) | f (x) or f (x) = q(x)(x − r) + r(x) where
deg(r(x)) < deg(x − r) = 1. If the former then we are done. If the latter
5.2. THE FUNDAMENTAL THEOREM OF ALGEBRA
137
then it follows that r(x) is in fact a constant (that is, just a number). Call
this number a. If we calculate f (r) we get a. It follows that in fact a = 0
and so (x − r) | f (x).
Example 5.2.8. We have seen that the number 1 is a root of x100 − 2x98 + 1.
Thus by the above result (x − 1) | x100 − 2x98 + 1.
A root r of a polynomial f (x) is said to have multiplicity m if
(x − r)m | f (x)
but (x − r)m+1 does not divide f (x). A root is always counted according to
its multiplicity.
Example 5.2.9. The polynomial x2 + 2x + 1 has −1 as a root and no other
roots. However (x + 1)2 = x2 + 2x + 1 and so the root −1 occurs with
multiplicity 2. Thus the polynomial has two roots counting multiplicities.
This is the sense in which we can say that a quadratic equation always has
two roots.
The following result is extremely useful. It provides an upper bound to
the number of roots a polynomial may have.
Theorem 5.2.10. A non-constant polynomial of degree n has at most n
roots.
Proof. Let f (x) be a non-zero polynomial of degree n > 0. Suppose that
f (x) has a root a. Then f (x) = (x − a)f1 (x) by Proposition 5.2.7 and the
degree of f1 (x) is n − 1. This argument can be repeated and we reach the
desired conclusion.
5.2.3
The fundamental theorem of algebra
The big question I have so far not dealt with is whether a polynomial need
have a root at all. This is answered by the following theorem whose name reflects its importance when first discovered, and not its significance in modern
algebra. We shall not give a proof because that would require more advanced
methods than are covered in this book. It was first proved by Gauss.
Theorem 5.2.11 (Fundamental theorem of algebra (FTA)). Every nonconstant polynomial of degree n with complex coefficients has a root.
138
CHAPTER 5. COMPLEX NUMBERS
This theorem has the following important consequence using Theorem ??.
Corollary 5.2.12. Every polynomial with complex coefficients of degree n
has exactly n complex roots (counting multiplicities). Thus every such polynomial can be written as a product of linear polynomials.
Proof. Let f (x) be a non-constant polynomial of degree n. By the FTA, this
polynomial has a root r1 . Thus f (x) = (x − r1 )f1 (x) where f1 (x) is a polynomial of degree n − 1. This argument can be repeated and we eventually end
up with f (x) = a(x − r1 ) . . . (x − rn ) where a is the last quotient, necessarily
a complex number.
Example 5.2.13. It can be checked that the quartic x4 − 5x2 − 10x − 6 has
roots −1, 3, i − 1 and −1 − i. We can therefore write
x4 − 5x2 − 10x − 6 = (x + 1)(x − 3)(x + 1 + i)(x + 1 − i).
In many practical examples, our polynomials will have real coefficients
and we will want any factors of the polynomial to be likewise real. The result
above doesn’t do that because it could produce complex factors. However,
we can rectify this situation at a very small price. We shall use the notion of
the complex conjugate of a complex number that we introduced earlier. We
may now prove the following key lemma.
Lemma 5.2.14. Let f (x) be a polynomial with real coefficients. If the complex number z is a root then so too is z.
Proof. Let
f (x) = an xn + an−1 xn−1 + . . . + a1 x + a0
where the ai are real numbers. Let z be a complex root. Then
0 = an z n + an−1 z n−1 + . . . + a1 z + a0 .
Take the complex conjugate of both side and use the properties of the complex
conjugate to get
0 = an z̄ n + an−1 z̄ n−1 + . . . + a1 z̄ + a0
and so z̄ is also a root.
5.2. THE FUNDAMENTAL THEOREM OF ALGEBRA
139
Example 5.2.15. We saw above that
x4 − 5x2 − 10x − 6 = (x + 1)(x − 3)(x + 1 + i)(x + 1 − i).
Observe that the complex roots −1 − i and −1 + i are complex conjugates
of each other.
Lemma 5.2.16. Let z be a complex number which is not real. Then
(x − z)(x − z̄)
is an irreducible quadratic with real coefficients.
On the other hand, if x2 + bx + c is an irreducible quadratic with real
coefficients then its roots are complex conjugates of each other.
Proof. To prove the first claim, we multiply out to get
(x − z)(x − z̄) = x2 − (z + z̄)x + z z̄.
Observe that z + z̄ and z z̄ are both real numbers. The discriminant of this
polynomial is (z − z̄)2 . You can check that if z is complex and non-real then
z − z̄ is purely complex. It follows that its square is negative. We have
therefore shown that our quadratic is irreducible.
The proof of the second claim follows from the formula for the roots of a
quadratic combined with the fact that the square root of a negative real will
have the form ±αi where α is real.
Example 5.2.17. We saw above that
x4 − 5x2 − 10x − 6 = (x + 1)(x − 3)(x + 1 + i)(x + 1 − i).
Multiply out (x + 1 + i)(x + 1 − i) and we get x2 + 2x + 2. Thus
x4 − 5x2 − 10x − 6 = (x + 1)(x − 3)(x2 + 2x + 2)
with all the polynomials involved being real.
The following theorem is the one that we can use to help us solve problems
involving real polynomials.
Theorem 5.2.18 (Fundamental theorem of algebra for real polynomials).
Every non-constant polynomial with real coefficients can be written as a product of polynomials with real coefficients which are either linear or irreducible
140
CHAPTER 5. COMPLEX NUMBERS
Proof. We can write the polynomial as a product of linear polynomials. Bring
the real linear factors to the front. The remaining linear polynomials will
have complex coefficients. They correspond to roots that come in complex
conjugate pairs. Multiplying together those complex linear factors corresponding to complex conjugate roots we get real quadratics and the result is
proved.
In fact, we can write any real polynomial as a real number times a product
of monic linear and quadratic factors. This result is the basis of the method
of partial fractions used in integrating rational functions in calculus.
Finding the exact roots of a polynomial is difficult, in general. However,
the following result tells us how to find the rational roots of polynomials with
integer coefficients. It is a nice, and perhaps unexpected, application of the
number theory we developed in Chapter 4.
Theorem 5.2.19 (Rational root theorem). Let
f (x) = an xn + an−1 xn−1 + . . . + a1 x + a0
be a polynomial with integer coefficients. If rs is a root with r and s coprime
then r | a0 and s | an . In particular, if the polynomial is monic then any
rational roots must be integers and divide the constant term.
Proof. Substituting
r
s
into f (x) we have, by assumption, that
r
r
r
0 = an ( )n + an−1 ( )n−1 + . . . + a1 ( ) + a0 .
s
s
s
Multiply through by sn to get
0 = an rn + an−1 srn−1 + . . . + sn−1 r + a0 sn .
We now make two observations. First, r | a0 sn . I claim that r and sn are
coprime. We may now deduce that r | a0 from a previous exercise. It only
remains to prove the claim. Let p be any prime that divides r and sn . Then
by Euclid’s lemma, p divides r and s which is a contradiction since r and s
are coprime. It follows that r | a0 . Second, s | an rn . By a similar argument
to the previous case s | an .
Example 5.2.20. Find all the roots of the following polynomial
x4 − 8x3 + 23x2 − 28x + 12.
5.2. THE FUNDAMENTAL THEOREM OF ALGEBRA
141
The polynomial is monic and so the only possible rational roots are integers
and must divide 12. Thus the only possible rational roots are
±1, ±2, ±3, ±4, ±6, ±12.
We find immediately that 1 is a root and so (x−1) must be a factor. Dividing
out by this factor we get the quotient
x3 − 7x2 + 16x − 12.
We check this polynomial for rational roots and find 2 works. Dividing out
by (x − 2) we get the quotient
x2 − 5x + 6.
Once we get down to a quadratic we can solve it directly. In this case it
factorizes as (x − 2)(x − 3). We therefore have that
x4 − 8x3 + 23x2 − 28x + 12 = (x − 1)(x − 2)2 (x − 3).
At this point, I usually multiply out the righthand side and check that I
really do have an equality. In this case, all roots are rational and are 1,2,2,3.
Exercises 5.2
1. Find the quotient and remainder when the first polynomial is divided
by the second.
(a) x3 − 7x − 1 and x − 2.
(b) x4 − 2x2 − 1 and x2 + 3x − 1.
(c) 2x3 − 3x2 + 1 and x.
2. Find all roots using the information given.
(a) 4 is a root of 3x3 − 20x2 + 36x − 16.
(b) −1, −2 are both roots of x4 + 2x3 + x + 2.
3. Find a cubic having roots 2, −3, 4.
142
CHAPTER 5. COMPLEX NUMBERS
4. Find a quartic having roots i, −i, 1 + i and 1 − i.
5. The cubic x3 + ax2 + bx + c has roots α, β and γ. Show that a, b, c can
each be written in terms of the roots.
√
6. 3 + i 2 is a root of x4 + x3 − 25x2 + 41x + 66. Find the remaining
roots.
√
7. 1 − i 5 is a root of x4 − 2x3 + 4x2 + 4x − 12. Find the remaining roots.
8. Find all the roots of the following polynomials.
(a) x3 + x2 + x + 1.
(b) x3 − x2 − 3x + 6.
(c) x4 − x3 + 5x2 + x − 6.
9. Write each of the following polynomials as a product of linear or quadratic
real factors.
(a) x3 − 1.
(b) x4 − 1.
(c) x4 + 1.
5.3
Complex number geometry
We have proved that every non-zero complex number has two square roots
and from the fundamental theorem of algebra (FTA), we know that every
non-zero complex number has three cube roots, and four fourth roots, and
more generally n nth roots. However, we didn’t prove the FTA. The main
goal of this section is to prove that every non-zero complex number has n
nth-roots. To do this, we shall think about complex numbers in a geometric,
rather than an algebraic, way. Throughout this section we shall not assume
FTA. We shall only need Theorem 5.2.10: every polynomial of degree n has
at most n roots.
5.3. COMPLEX NUMBER GEOMETRY
5.3.1
143
sin and cos
We first recall some well-known properties of the trigonometric functions sin
and cos. First the addition formulae
sin(α + β) = sin α cos β + cos α sin β
and
cos(α + β) = cos α cos β − sin α sin β.
These formulae were important historically because they enabled unknown
values of sin’s and cos’s to be calculated from known ones, and so they were
useful in constructing trig tables in the days before calculators
In university mathematics, angles are usually measured in radians rather
than degrees. This is because radians are a natural unit of angle measurement
whereas the system of angle measurement based on degrees is an historical
accident. Why 360 degrees in a circle? Ask the Ancient Babylonians. Positive angles are measures in an anticlockwise direction.
The sin and cos functions are periodic functions with period 2π. This
means that for all angles θ
sin(θ + 2πn) = sin θ and cos(θ + 2πn) = cos θ
for all n ∈ Z. This fact will be crucial in what follows.
The following table of values will be useful. I leave it as an exercise to
justify it.
θ
0◦
30◦
45◦
sin θ cos θ
0
1
√
60◦
90◦
5.3.2
1
2
√1
√2
3
2
1
3
2
1
√
2
1
2
0
The complex plane
In this section, we shall describe in more detail an alternative way of thinking
about complex numbers which turns out to be very fruitful. Recall that a
complex number z = a + bi has two components: a and b. We can plot these
144
CHAPTER 5. COMPLEX NUMBERS
as a point in the plane. The plane used in this way is called the complex
plane: the x-axis is the real axis and the y-axis is interpreted as the complex
axis. Although a complex number can be thought of as labelling a point in
the complex plane, it can more usefully be regarded as labelling the directed
line segment from the origin to the point. This is how we shall regard it.
Let z = a + bi be a non-zero complex number and let θ be the angle that it
makes with the positive reals. The length of z as a directed line segment in
the complex plane is |z|, and by basic trig a = |z| cos θ and b = |z| sin θ. It
follows that
z = |z| (cos θ + i sin θ) .
z
i |z| sin θ
θ
|z| cos θ
Observe that |z| is a non-negative real number. This way of writing complex
numbers is called the polar form.
At this point, I need to clarify the only feature of complex numbers that
causes confusion. I have already mentioned that the functions sin and cos
are periodic. For that reason, there is not just one number θ that yields
the complex number z but infinitely many of them: namely, all the numbers
θ + 2πk where k ∈ Z. For this reason, we define the argument of z, denoted
by arg z, not merely to be the single angle θ but the set of all angles θ + 2πk
where k ∈ Z. The angle θ is chosen so that 0 ≤ θ < 2π and is called, for
convenience, the principal argument. But note that books vary on what they
choose to call the principal argument. This feature of the argument plays a
crucial role when we come to calculate nth roots.
Observe that complex numbers of the form
cos θ + i sin θ
are precisely the complex numbers of unit length. Thus the set of all such
numbersdescribe the unit circle with centre the origin in the complex plane.
5.3. COMPLEX NUMBER GEOMETRY
145
Thus every non-zero complex number is a real number times a complex number lying on the unit circle.
Let w = r (cos θ + i sin θ) and z = s (cos φ + i sin φ) be two non-zero
complex numbers. We shall calculate wz. We have that
wz = rs (cos θ + i sin θ) (cos φ + i sin φ)
= rs[(cos θ cos φ − sin θ sin φ) + (sin θ cos φ + cos θ sin φ)i]
but using the properties of the sin and cos functions this reduces to
wz = rs (cos(θ + φ) + i sin(θ + φ)) .
We thus have the following important result:
when two non-zero complex numbers are multiplied together their
lengths are multiplied and their arguments are added.
This result helps us to understand the meaning of i. Multiplication by i
is the same as a rotation about the origin by a right angle. Multiplication by
i2 is therefore the same as a rotation about the origin by two right angles.
But this is exactly the same as multiplication by −1.
i
−1
1
−i
We may apply similar reasoning to explain geometrically why −1 × −1 = 1.
We of course proved this algebraically in Chapter 2. Multiplication by −1 is
interpreted as rotation about the origin by 180◦ . It follows that doing this
twice takes us back to where we started and so is equivalent to multiplication
by 1.
The proof of the next theorem follows by induction from the result we
proved above.
146
CHAPTER 5. COMPLEX NUMBERS
Theorem 5.3.1 (De Moivre). Let n be a positive integer. If z = r (cos θ + i sin θ)
then
z n = rn (cos nθ + i sin nθ) .
This result has nice applications in painlessly obtaining trigonometric
identities.
Example 5.3.2. Express cos 3θ in terms of cos θ and sin θ using De Moivre’s
Theorem. We have that
(cos θ + i sin θ)3 = cos 3θ + i sin 3θ.
However, we can expand the lefthand side to get
cos3 θ + 3i cos2 θ sin θ + 3 sin θ(i sin θ)2 + (i sin θ)3
which simplifies to
cos3 θ − 3 cos θ sin2 θ + i 3 cos2 θ sin θ − sin3 θ
where we use the fact that i2 = −1 and i3 = −i and i4 = 1. Equating real
and imaginary parts we get
cos 3θ = cos3 θ − 3 cos θ sin2 θ.
We also get the formula
sin 3θ = 3 cos2 θ sin θ − sin3 θ
5.3.3
Arbitrary roots of complex numbers
In this section, we shall prove that every non-zero complex number has n
nth roots: thus it has three cube roots, and four fourth roots and so on. We
begin with a special case that turns out to give us almost all the information
we need to solve the general case.
The nth roots of unity
5.3. COMPLEX NUMBER GEOMETRY
147
We shall show that the number 1 has n nth roots — these are called the
n roots of unity. We know that the equation z n − 1 = 0 has at most n roots,
so all we need do is find n roots and we are home and dry. We begin with a
motivating example.
Example 5.3.3. We find the three cube roots of 1. There are two ways of
we mean an algebraic expression obtained from the rationals by carrying
out the four basic operations of addition, multiplication, subtraction and
division together with the extraction of nth roots. Divide the unit circle in
the complex plane into an equilateral triangle with 1 as one of its vertices.
Then the other two roots are ω1 = cos 120◦ + i sin 120◦ obtained by dividing
. If we put ω = ω1
2π by 3 and ω2 = cos 240◦ + i sin 240◦ which is twice 2π
3
2
then in fact ω2 = ω . This is the trigonometric form of the roots.
ω
1
ω2
In this case, it is easy to write down their radical forms as well. We have
that
√ √ 1
1
ω=
−1 + i 3 and ω 2 = − 1 + i 3 .
2
2
The general case is solved in a similar way to our example above using
regular n-gons in the complex plane where one of the vertices is 1.
Theorem 5.3.4 (Roots of unity). The n roots of unity are given by the
following formula
2kπ
2kπ
cos
+ i sin
n
n
for k = 1, 2, . . . , n. These complex numbers are arranged uniformly on the
unit circle and form a regular polygon with n sides: the cube roots of unity
form an equilateral triangle, the fourth roots form a square, the fifth roots
form a pentagon, and so on.
148
CHAPTER 5. COMPLEX NUMBERS
There is only one point here that is a little confusing. It is always possible
and easy to write down the trigonometric form of the nth roots of unity. It
is also always possible to write down the radical form of the nth roots of
unity but this is far from easy in general. In fact, it forms part of the
advanced subject known as Galois theory.
Example 5.3.5. As part of showing that the 17-gon could be constructed
using only a ruler and compass, Gauss proved the following result which is
highly non-trivial. You can verify that it is true by using a calculator — at
least up to the limits of your calculator.
q
√
√
2π
= −1 + 17 + 34 − 2 17
16 cos
17
s
q
q
√
√
√
√
+
68 + 12 17 − 16 34 + 2 17 − 2(1 − 17) 34 − 2 17
Arbitrary nth roots
The nth roots of unity play an important role in finding arbitrary nth
roots. We begin with an example to illustrate the idea.
Example 5.3.6. We√
find the three cube roots of 2. If you use your calculator
you will simply find 3 2, a real number. There should be two others: where
are they? The explanation is that the other two cube roots are complex. Let
ω be the complex cube root of 1 that we described above. Then the three
cube roots of 2 are the following
√
√
√
3
3
3
2, ω 2, ω 2 2.
The above example generalizes.
Theorem 5.3.7 (nth roots). Let z = r (cos θ + i sin θ) be a non-zero complex
number. Put
√
θ
θ
n
,
u = r cos + i sin
n
n
the obvious nth root, and put
ω = cos
2π
2π
+ i sin ,
n
n
5.3. COMPLEX NUMBER GEOMETRY
149
the first interesting nth root of unity. Then the nth roots of z are as follows
u, uω, . . . , uω n−1 .
It follows that the nth roots of z = r (cos θ + i sin θ) can be written in the
form
√
θ
2kπ
θ
2kπ
n
r cos
+
+ i sin
+
n
n
n
n
for k = 0, 1, 2, . . . , n − 1.
This is the reason why every non-zero number has two square roots that
differ by a multiple of −1: the two square roots of 1 are 1 and -1.
5.3.4
Euler’s formula
We have seen that every real number can be written as a whole number plus
a possibly infinite decimal part. It turns out that many functions can also be
written as a sort of decimal. I shall illustrate this by means of an example.
it is equal to its derivative and e0 = 1. We would like to write
ex = a0 + a1 x + a2 x2 + a3 x3 + . . .
where the ai are real numbers that we have yet to determine. We can work
out the value of a0 easily by putting x = 0. This tells us that a0 = 1. To get
the value of a1 we first differentiate our expression to get
ex = a1 + 2a2 x + 3a3 x2 + . . .
Now put x = 0 again and this time we get that a1 = 1. To get the value of
a2 we differentiate our expression again to get
ex = 2a2 + 3 · 2 · a3 x + . . .
Now put x = 0 and we get that a2 = 21 . Continuing in this way we quickly
spot the pattern for the values of the coefficient an . We find that an = n!1
where n! = n(n − 1)(n − 2) . . . 2 · 1. What we have done for ex we can also
do for sin x and cos x and we obtain the following series expansions of each
of these functions.
150
CHAPTER 5. COMPLEX NUMBERS
• ex = 1 + x +
x2
2!
+
x3
3!
+
x4
4!
+ . . ..
• sin x = x −
x3
3!
+
x5
5!
−
x7
7!
+ . . ..
• cos x = 1 −
x2
2!
+
x4
4!
−
x6
6!
+ . . ..
There are interesting connections between these three series. We shall
now show that complex numbers help to explain them. Without worrying
about the validity of doing so, we calculate the infinite series expansion of
eiθ . We have that
eiθ = 1 + (iθ) +
1
1
(iθ)2 + (iθ)3 + . . .
2!
3!
that is
1 2 1 3
1
θ − θ i + θ4 + . . .
2!
3!
4!
By separating out real and complex parts, and using the infinite series we
obtained above, we get Euler’s remarkable formula
eiθ = 1 + iθ −
eiθ = cos θ + i sin θ.
Thus the complex numbers enable us to find the hidden connections between
the three most important functions of calculus: the exponential function and
the sine and cosine functions. It follows that every non-zero complex number
can be written in the form reiθ . If we put θ = π in Euler’s formula, we get
the following result, which is widely regarded as one of the most amazing in
mathematics.
Theorem 5.3.8 (Euler’s identity).
eπi = −1.
This result shows us that the real numbers π, e and −1 are connected,
but that to establish that connection we have to use the complex number i.
This is one of the important roles of the complex numbers in mathematics in
that they enable us to make connections between topics that look different:
they form a mathematical hyperspace.
5.4. MAKING SENSE OF COMPLEX NUMBERS
151
Exercises 5.3
1. Express cos 5x and sin 5x in terms of cos x and sin x.
2. Prove the following where x is real.2
(a) sin x =
(b) cos x =
1
(eix − e−ix ).
2i
1 ix
(e + e−ix ).
2
Hence show that cos4 x = 18 [cos 4x + 4 cos 2x + 3].
3. Find the 4th roots of unity.
4. Find the 6th roots of unity.
5. Find the 8th roots of unity.
6. Solve x3 = −8i.
7. Determine all the values of ii . What do you notice?
5.4
Making sense of complex numbers
In this chapter, I have assumed that complex numbers exist and that they
obey the usual high-school rules of algebra. In this section, I shall sketch out
a proof of this.
We start with the set R × R whose elements are ordered pairs (a, b) where
a and b are real numbers. It will be helpful to denote these ordered pairs by
bold letters so a = (a1 , a2 ). We define 0 = (0, 0), 1 = (1, 0) and i = (0, 1).
We now define operations as follows
• If a = (a1 , a2 ) and b = (b1 , b2 ), define a + b = (a1 + b1 , a2 + b2 ).
• If a = (a1 , a2 ) define −a = (−a1 , −a2 ).
• If a = (a1 , a2 ) and b = (b1 , b2 ), define
ab = (a1 b1 − a2 b2 , a1 b2 + a2 b1 ).
2
Compare (a) and (b) below with sinh x = 12 (ex − e−x ) and cosh x = 21 (ex + e−x ).
152
CHAPTER 5. COMPLEX NUMBERS
• If a = (a1 , a2 ) 6= 0 define
−a2
a1
,p 2
).
a−1 = ( p 2
2
a1 + a2
a1 + a22
It is now a long exercise to check that all the usual axioms of high-school
algebra hold. Observe now that the element (a1 , a2 ) can be written
(a1 , 0)1 + (a2 , 0)i
and that
ii = (0, 1)(0, 1) = (−1, 0) = −1.
The elements of the form (a, 0) can be identified with the real numbers. This
proves that the complex numbers as I described them earlier in this chapter
really do exist.
5.5
Gaussian integers and factorizing primes
Complex numbers may be used to factorize some primes. For example,
5 = (1 − 2i)(1 + 2i).
To develop this example further, we shall need some definitions. The integers
Z are a subset of the reals R. We define the Gaussian integers, denoted
by Z[i], to be all complex numbers of the form m + in where m and n are
integers. What our example shows is that some primes can be factorized using
Gaussian integers. The question is: which ones? Observe that 5 = 12 + 22 .
In other words, it can be written as a sum of two squares. Another example
of a prime that can be written as a sum of two squares is 13. We have that
13 = 9 + 4 = 32 + 22 .
This prime can also be factorizes using Gaussian integers
13 = (3 + 2i)(3 − 2i).
In fact, any prime p that can be written as a sum of two squares p = a2 + b2 ,
cane also be factorized using Gaussian integers
p = (a + ib)(a − ib).
This raises the question of exactly which primes can be written as a sum of
two squares.
153
Lemma 5.5.1. Let p be an odd prime that can be written as a sum of two
squares. Then p ≡ 1 (mod 4).
Proof. Let p = a2 + b2 . Since p is assumed odd, we must have that one of a2
and b2 is even and the other odd. Without loss of generality, we may assume
that a2 is odd and b2 is even. But from Chapter 2, this implies that a is
odd and b is even. We may therefore write a = 2u and b = 2v + 1 for some
natural numbers u and v. But then p = 4u2 + 4v 2 + 2v + 1. It follows that
p ≡ 1 (mod 4).
Lemma 5.5.2. Each odd prime p satisfies either p ≡ 1 (mod 4) or p ≡ 3
(mod 4).
Proof. The possible remainder when p is divided by 4 are 0, 1, 2, 3. Since p
is a prime both 0 and 2 are impossible and the result follows.
The lemma above tells us that each odd prime belongs to exactly one of
two camps. The obvious question is whether both of these camps are infinite.
Proposition 5.5.3.
1. There are infinitely many primes p such that p ≡ 3 (mod 4).
2. There are infinitely many primes p such that p ≡ 1 (mod 4).
We have proved that if an odd prime p can be written as a sum of two
squares then p ≡ 1 (mod 4). The hard question is whether the converse is
true.
Theorem 5.5.4 (Euler, 1754). An odd prime p can be written as a sum of
two squares if, and only if, p ≡ 1 (mod 4).
We may deduce from this theorem that every odd prime p ≡ 1 (mod 4)
can be factorized by means of Gaussian integers.
5.6
By the fundamental theorem of algebra, we know that any non-constant
polynomial has roots. Such roots can be explicitly calculated using rootfinding algorithms, such as the Jenkins-Traub algorithm. So if your interest
in the roots is practical — for example, this problem is an essential part of
154
CHAPTER 5. COMPLEX NUMBERS
solving linear differential equations — then you might think that there is
no more to be said. But it turns out, as so often happens in maths, that
asking a more precise question can lead to new ideas. In this section, I shall
survey the nature of the solutions of a polynomial equation not merely their
existence. We begin by describing the way in which cubics and quartics may
be solved purely algebraically.
5.6.1
Cubic equations
Let
f (x) = a3 x3 + a2 x2 + a1 x + a0
where a3 6= 0. I shall assume all coefficients are real though the theory
works in general. We shall find all the roots of f (x). This problem can be
simplified in two ways. First, we may divide through by a3 and so, without
loss of generality, we may assume that f (x) is monic. That is a3 = 1. Second,
by means of a substitution we may obtain a cubic in which the coefficient of
the term in x2 is zero. Put x = y − a33 . You should do this and check that
you get a polynomial of the form
g(y) = y 3 + py + q.
We say that such a cubic is reduced. It follows that without loss of generality,
we need only solve the cubic
g(x) = x3 + px + q.
To do this needs what looks like a minor miracle. Let u and v be two complex
variables. Let ω = cos 2π
+ i sin 2π
, one of the complex cube roots of unity.
3
3
You should now check that the following cubic
t(x) = x3 − 3uv − (u3 + v 3 )
has the roots
u + v,
uω + vω 2 ,
uω 2 + vω.
Now we can solve
x3 + px + q = 0
if we can find u and v such that
p = −3uv,
q = −u3 − v 3 .
155
Now if we cube the first equation, we get the following two equations
−p
= u3 v 3 ,
27
−q = u3 + v 3 .
If we regard u3 and v 3 as the unknowns we know their sum and we know their
product. This means that u3 and v 3 are the roots of the quadratic equation
x2 + qx −
p3
= 0.
27
We therefore have that
1
u3 =
2
−q +
and
1
v3 =
2
−q −
r
r
27q 2 + 4p3
27
27q 2 + 4p3
27
!
!
.
To find u we have to take a cube root of the number u3 and there are three
possible such roots. Choose one such value for u. We then choose the value
of v so that p = −3uv.
Example 5.6.1. Find the roots of x3 − 9x − 2 = 0. Here p = 9 and q = −2.
The quadratic equation we have to solve is therefore
x2 − 2x − 27 = 0.
√
√
This has roots 1 ± 2 7. Put u3 = +2 7. We may choose a real cube root
in this case to get
q
√
3
u = 1 + 28.
We must then choose v to be
q
√
3
u = 1 − 28.
We may now write down the three roots of our original cubic.
The following cubic equation was studied by Bombelli in 1572 and had
an important influence on the development of complex numbers.
156
CHAPTER 5. COMPLEX NUMBERS
Example 5.6.2. Consider the cubic
x3 − 15x − 4 = 0.
The associated quadratic in this case is
x2 + 4x + 125 = 0.
This gives the two solutions that Bombelli would have written in a way
equivalent to the following
√
x = 2 ± −121.
We would write this as
x = 2 ± 11i.
Thus
u3 = 2 + 11i and v 3 = 2 − 11i.
There are
√ three cube roots of 2 + 11i all complex. Let’s press on√regardless.
Write 3 2 + 11i to represent one of those cube roots. Write 3 2 − 11i to
be the corresponding cube root such that their product is 5. Thus at least
symbolically we may write
√
√
u + v = 3 2 + 11i + 3 2 − 11i.
What is surprising is that for some choice of these cube roots this value must
be real. The reason is that the graph of our cubic has one real root which
can easily be checked to be 4. To see why, observe that
(2 + i)3 = 2 + 11i and (2 − i)3 = 2 − 11i.
If we choose 2 + i as one of the cube roots of 2 + 11i then we have to choose
2 − i as the corresponding cube root of 2 − 11i. In this way, we get
4 = (2 + 11i) + (2 − 11i)
as a root. It was the fact that real roots arose in this way that provided
the first inkling that there was a number system, the complex numbers,
that extended the so-called real numbers, but had every much as tangible
existence.
5.6.2
157
Quartic equations
Let
f (x) = a4 x4 + a3 x3 + a2 x2 + a1 x + a0 .
As usual, we may assume that a4 = 1. By means of a suitable substitution,
which is left as an exercise, we may eliminate the cubed term. We therefore
end up with a reduced quartic which it is convenient to write in the following
way
x4 = ax2 + bx + c.
Suppose that we could write the righthand side as a perfect square (dx + e)2 .
Then our quartic could be written as the product of two quadratics
x2 − (dx + e) x2 + dx + e .
The roots of each these two quadratics will be the four roots of our original
quartic. It is not true that we can always do this, but by means of another
miracle we can transform the equation into one with the same roots where
we can. Let t be a new variable whose value will be determined later. We
may write
(x2 + t)2 = (a + 2t)x2 + bx + (c + t2 ).
We now want to choose a value of t so that the righthand side is a perfect
square. This happens when the discriminant of the quadratic (a + 2t)x2 +
bx + (c + t2 ) is zero. That is when
b2 − 4(a + 2t)(c + t2 ) = 0.
Now this is a cubic in t. We now use the method of the previous section to
find a specific value of t say t1 . We then get
2
b
2
2
.
(x + t1 ) = (a + 2t1 ) x +
2(a + 2t1 )
It follows that the roots of the original quartic are the roots of the following
√
b
2
(x + t1 ) − a + 2t1 x +
=0
2a + 4t1
and
2
(x + t1 ) +
√
a + 2t1 x +
b
2a + 4t1
= 0.
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CHAPTER 5. COMPLEX NUMBERS
Example 5.6.3. Solve the quartic
x4 = 1 − 4x.
We shall find a value of t below
(x2 + t)2 = t4 + 2x2 t + t2 = 2x2 t − 4x + (1 + t2 )
which makes the righthand side a perfect square. This requires us to find a
root of the cubic
t3 + t − 2 = 0.
Here t = 1 works. Our quartic with t therefore becomes
(x2 + 1)2 = 2(x − 1)2 .
Therefore the roots of our original quartic are the roots of the following two
√
√
(x2 + 1) − 2(x − 1) = 0 and x2 + 1 + 2(x − 1) = 0.
The roots of our original quartic are therefore
p√
p√
1±i
8+1
−1 ±
8−1
√
√
and
.
2
2
5.6.3
Symmetries and particles
Although quadratic equations had been solved in antiquity, it was not until
the 16th century that cubics and quartics were first solved. This great leap
forward in the development of algebra was centred on a group of Italian
mathematicians — Scipione del Ferro (1465–1525), Niccolo Tartaglia (1500–
1557), Girolamo Cardano (1501–1576), Ludovico Ferrari (1522–1562), Rafael
Bombelli (1526–1572) — whose antics are worthy of an opera or Shakespeare
comedy but the importance of their work cannot be overemphasized. But
two points arise. First, the solution of quadratics, cubics and quartics seem
to rely on mathematical miracles. Second, we appear to see a pattern: to
solve cubics we need to solve an associated quadratic and to solve quartics
we need to solve an associated cubic. These two points were investigated by
a number of mathematicians in great depth: in particular, Lagrange (1736–
1813), Ruffini (1765–1822) and Abel (1802–1829). The expectation was high
159
that quintics should be solvable by using quartics in a way that continued
the pattern. Then came the great surprize. Ruffini and Abel proved that
the pattern does not continue and that one cannot always describe the roots
of a quintic in radical form — there are, of course, five roots — the point
is that these roots cannot in general be written down using an algebraic
formula. The question is why and the answer to this question also explains
the algebraic miracles we used above. It was discovered by Evariste Galois
(1811–1832). I shall not go into the details of his biography — he was killed,
for instance, in a duel — since you will find much more written about him
elsewhere, some of it accurate, instead I shall focus on his mathematics.
By building on the work of Lagrange, he explained the miracles above and
much more. His approach was new: to determine whether the roots of a
polynomial could be expressed in algebraic terms as radical expressions, he
studied the symmetries of the polynomial. Just what this means is explained
in a subject known as Galois theory after its founder. Crucially, this is not
a mere extrapolation of existing algebraic manipulation, instead it involves
working at a higher level of abstraction. As so often happens in mathematics,
a development in one area led to developments in other areas. Sophus Lie
(1811–1832) realized that symmetries could also be used to help understand
the tricks that were used to solve differential equations. It was in this way
that symmetry came to play a fundamental role in physics. If you hear a
particle physicist talking about symmetries, they are paying an unconscious
tribute to Galois’ bold work in studying the nature of the roots of polynomial
equations.
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CHAPTER 5. COMPLEX NUMBERS
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