Chapter 5 Complex numbers - h. Wmarkl/teaching/ALGEBRA/Chapter5.pdf · 128 CHAPTER 5. COMPLEX...

Chapter 5

Complex numbers

Why be one-dimensional when you can be two-dimensional?

0 1 2 3−1−2−3

?

?We begin by returning to the familiar number line, where I have placedthe question marks there appear to be no numbers. I shall rectify this bydefining the complex numbers which give us a number plane rather than justa number line. Complex numbers play a fundamental role in mathematics.In this chapter, I shall use them to show how e and π are connected andhow certain primes can be factorized. They are also fundamental to physicswhere they are used in quantum mechanics.

5.1 Complex number arithmetic

In the set of real numbers we can add, subtract, multiply and divide, butwe 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 realsquare roots, the reason being that the square of any real non-zero number isalways 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. Com-plex numbers were first studied in the 1500’s but were only fully acceptedand 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 rootsneed 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 inthis case the equation has two complex roots. This will mean that quadraticequations will always have two roots. The key step is the following

We introduce a new number, denoted by i, whose defining propertyis that i2 = −1. We shall assume that in all other respects itsatisfies the usual axioms of high-school algebra. This assumptionwill be justified later.

We shall now explore the consequences of this definition which turns out tobe a profound one for mathematics.

It follows that i and −i are the two missing square roots of 1. In all otherrespects the number i will behave like a real number. Thus if b is any realnumber then bi is a number, and if a is any real number then a + bi is anumber.

A complex number is a number of the form a + bi where a, b ∈ R. Wedenote the set of complex numbers by C. Complex numbers are sometimescalled imaginary numbers. This is not such a good term: they are notfigments of our imagination like unicorns or dragons. Like all numbers theyare, however, products of our imagination: no one has seen the complexnumber 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 orimaginary part of z, denoted Im(z).

Two complex numbers a+ bi and c+ di are equal precisely whena = c and b = d. In other words, when their real parts are equaland 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 complexnumber because if a is real then a = a+ 0i. Thus R ⊆ C. Complex numbersof the form bi are said to be purely imaginary.

Now we show that we can add, subtract, multiply and divide complexnumbers. 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’tmatter and that we may bracket sums of complex numbers how we like andstill get the same answer and so we can rewrite this as a+ c+ bi+ di. Nextwe assume that multiplication of complex numbers distributes over additionof 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 firstassume complex multiplication distributes over complex addition to get (a+bi)(c + di) = ac + adi + bic + bidi. Next we assume that the order in whichwe 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 followingdefinition 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; addingtogether −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 get2 + 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.

3.(

1−i√2

)2

. Multiply out and simplify to get −i.


The final operation is division. We have to show that when a + ib 6= 0the reciprocal

1

a+ ib

is also a complex number. We use an idea that can also be applied in othersituations called rationalizing the denominator. It is convenient first to definea new operation on complex numbers. Let z = a+ bi ∈ C. Define

z = a− bi.

The number z is called the complex conjugate of z. Why is this operationuseful? Let’s calculate zz. We have

zz = (a+ bi)(a− bi) = a2 − abi+ abi− b2i2 = a2 + b2.

Notice that zz = 0 if and only if z = 0. Thus for non-zero complex numbersz, the number zz is a positive real number. Let’s see how we can use thecomplex conjugate to define division of complex numbers. Our goal is tocalculate

1

a+ bi

where a+bi 6= 0. The first step is to multiply top and bottom by the complexconjugate of a+ bi. We therefore get

a− bi(a+ bi)(a− bi) =

a− bia2 + b2

=1

a2 + b2(a− bi) .

Examples 5.1.2. Carry out the following calculations.

1. 1+ii

. The complex conjugate of i is −i. Multiply top and bottom of thefraction to get −i+1

1= 1− i.

2. i1−i . The complex conjugate of 1− i is 1 + i. Multiply top and bottom

of the fraction to get i(1+i)2

= i−12

.

3. 4+3i7−i . The complex conjugate of 7− i is 7 + i. Multiply top and bottom

of the fraction to get (4+3i)(7+i)50

= 1+i2

.

We shall need the following properties of the complex conjugate later on.

Lemma 5.1.3.


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 enablesus to visualize them. A complex number z = a + bi has two components: aand b. It is irresistible to plot these as a point in the plane. The plane usedin this way is called the complex plane: the x-axis is the real axis and they-axis is interpreted as the complex axis.

a

ibz = a+ ib

Although a complex number can be thought of as labelling a point in thecomplex plane, it can also be regarded as labelling the directed line segmentfrom the origin to the point. By Pythagoras’ theorem, the length of this lineis√a2 + b2. We define

|z| =√a2 + b2

where z = a + bi. This is called the modulus1 of the complex number z.Observe that

|z| =√zz.

We shall use the following important property of moduli.

Lemma 5.1.4. |wz| = |w| |z|.Proof. Let w = a+ bi and z = c+di. Then wz = (ac− bd) + (ad+ bc)i. Now|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.1Plural: moduli


The complex numbers were obtained from the reals by simply throwing inone new number, i, a square root of −1. Remarkably, every complex numberhas a square root.

Theorem 5.1.5. Every nonzero complex number has exactly two squareroots.

Proof. Let z = a + bi be a nonzero complex number. We want to find acomplex number w so that w2 = z. Let w = x + yi. Then we need to findreal numbers x and y such that (x+yi)2 = a+bi. Thus (x2−y2)+2xyi = a+bi,and so equating real and imaginary parts, we have to solve the following twoequations

x2 − y2 = a and 2xy = b.

Now we actually have enough information to solve our problem, but we canmake life easier for ourselves by adding one extra equation. To get it, we usethe modulus function. From (x+yi)2 = a+bi we get that |x+ yi|2 = |a+ bi|.Now |x+ yi|2 = x2 + y2 and |a+ bi| =

√a2 + b2. We therefore have three

equations

x2 − y2 = a and 2xy = b and x2 + y2 =√a2 + b2.

If we add the first and third equation together we get

x2 =a

2+

√a2 + b2

2=a+√a2 + b2

2.

We can now solve for x and therefore for y.

Example 5.1.6. Every negative real number has two square roots. We havethat 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 arereal such that

(x+ yi)2 = 3 + 4i.

Squaring and comparing real and imaginary parts we get that the followingtwo equations must be satisfied by x and y

x2 − y2 = 3 and 2xy = 4.


We also have a third equation by taking moduli

x2 + y2 = 5.

Adding the first and third equation together we get x = ±2. Thus y = 1 ifx = 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 rootto 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 willhave 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 fordetermining the square roots of complex numbers, we have a method forfinding the roots of quadratics with any coefficients, whether they be real orcomplex.

Example 5.1.8. Solve the quadratic equation

4z2 + 4iz + (−13− 16i) = 0.

The complex numbers obey the same algebraic laws as the reals and so wecan solve this equation by completing the square or we can simply plug thenumbers into the formula for the roots of a quadratic. Here I shall completethe square. First, we convert the equation into a monic one

z2 + iz +(−13− 16i)

4= 0.

Next, we observe that

(z +

i

2

)2

= z2 + iz − 1

4.

Thus

z2 + iz =

(z +

i

2

)2

+1

4.

Our equation therefore becomes

(z +

i

2

)2

+1

4+

(−13

4− 4i

)= 0.


We therefore have (z +

i

2

)2

= 3 + 4i.

Taking square roots of both sides using a previous calculation, we have that

z +i

2= 2 + i or − 2− i.

It follows that z = 2 + i2

or − 2 − 3i2

. Now check that these roots really dowork.

Every quadratic equation ALWAYS has has exactly two roots.

Exercises 5.1

1. Solve the following problems in complex number arithmetic. In eachcase, 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) 2+3i4+i

.

(e) 1i

+ 31+i

.

(f) 3+4i3−4i− 3−4i

4+4i.

2. Find the square roots of each of the following complex numbers andcheck your answers.

(a) −i.(b) −1 + i

√24.

(c) −13− 84i.

3. Solve the following quadratic equations and check your answers.

(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 133

5.2 The fundamental theorem of algebra

I want to describe now a result which is one of the most important conse-quences of the properties of complex numbers: the fundamental theorem ofalgebra. It should be understood that this is a misnomer since algebra hasexpanded beyond all bounds since this theorem was first proved. Neverthe-less, 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 functioncan be integrated using partial fractions.

In this section, we shall work with arbitrary polynomials and I shall nowrecall some terminology for handling them. An expression

anxn + an−1x

n−1 + . . .+ a1x+ 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 thisto deg. If an = 1 the polynomial is said to be monic. The term a0 is calledthe constant term and the term anx

n is called the leading term. Polynomialscan be added, subtracted and multiplied.

Two polynomials are equal if they have the same degree and the coeffi-cients 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 wantthem.

Why are polynomials interesting? There are two answers to this question.First, they have widespread applications such as in helping to solve lineardifferential equations and in studying matrices. Second, a polynomial definesa function which is calculated in a very simple way using the operations ofaddition, subtraction and multiplication. However many, more complicated,functions can be usefully approximated by polynomial ones.


We denote by C[x] the set of polynomials with complex coefficients andby R[x], the set of polynomials with real coefficients. I will write F [x] tomean F = R or F = C.

5.2.1 The remainder theorem

The addition, subtraction and multiplication of polynomials is easy. We shalltherefore 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 thatg(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 amxm and let g(x) have leading term bnx

n.Then the leading term of f(x)g(x) is ambnx

m+n. Now ambn 6= 0 and so thedegree of f(x)g(x) is m+ n, as required.

The following result is analogous to the remainder theorem for integersLemma 4.1.1

Lemma 5.2.3 (Remainder theorem). Let f(x) and g(x) be polynomials inF [x] where deg f(x) ≥ deg g(x). Then either

g(x) | f(x)

orf(x) = g(x)q(x) + r(x)

where deg r(x) < deg g(x).


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 theremainder.

The following example is a reminder of how to carry out long division ofpolynomials. Remember that answers can always be checked by multiplyingout.

Example 5.2.5. Divide 6x4 + 5x3 + 4x2 + 3x+ 2 by 2x2 + 4x+ 5 and so findthe quotient and remainder. We set out the computation in the followingform.

2x2 + 4x+ 5 6x4 + 5x3 + 4x2 + 3x+ 2

To get the term involving 6x4 we would have to multiply the lefthand sideby 3x2. As a result we write down the following

3x2

2x2 + 4x+ 5 6x4 + 5x3 + 4x2 + 3x+ 26x4 + 12x3 + 15x2

We now subtract the lower righthand side from the upper and we get

3x2

2x2 + 4x+ 5 6x4 + 5x3 + 4x2 + 3x+ 26x4 + 12x3 + 15x2

−7x3 − 11x2 + 3x+ 2

The procedure is now repeated with the new polynomial.

3x2 − 72x

2x2 + 4x+ 5 6x4 + 5x3 + 4x2 + 3x+ 26x4 + 12x3 + 15x2

−7x3 − 11x2 + 3x+ 2−7x3 − 14x2 − 35

2x

3x2 + 412x+ 2


The procedure is repeated one more time with the new polynomial

3x2 − 72x+ 3

2quotient

2x2 + 4x+ 5 6x4 + 5x3 + 4x2 + 3x+ 26x4 + 12x3 + 15x2

−7x3 − 11x2 + 3x+ 2−7x3 − 14x2 − 35

2x

3x2 + 412x+ 2

3x2 + 122x+ 15

2292x− 11

2remainder

This is the end of the line because the new polynomial we obtain has degreestrictly less than the polynomial we are dividing by. What we have shown isthat

6x4 + 5x3 + 4x2 + 3x+ 2 =(2x2 + 4x+ 5

)(3x2 − 7

2x+

3

2

)+

(29

2x− 11

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) iff(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 thefirst place is trickier. The next result tells us that when we find roots of poly-nomials we are in fact determining linear factors. It is crucial to eveythingwe shall do.

Proposition 5.2.7. Let r ∈ F . Then r is a root of f(x) ∈ F [x] if and onlyif (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 thatit must be zero.

We now prove the converse. Suppose that r is a root of f(x). By theremainder theorem, either (x− r) | f(x) or f(x) = q(x)(x− r) + r(x) wheredeg(r(x)) < deg(x − r) = 1. If the former then we are done. If the latter


then it follows that r(x) is in fact a constant (that is, just a number). Callthis number a. If we calculate f(r) we get a. It follows that in fact a = 0and 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 toits multiplicity.

Example 5.2.9. The polynomial x2 + 2x+ 1 has −1 as a root and no otherroots. However (x + 1)2 = x2 + 2x + 1 and so the root −1 occurs withmultiplicity 2. Thus the polynomial has two roots counting multiplicities.This is the sense in which we can say that a quadratic equation always hastwo roots.

The following result is extremely useful. It provides an upper bound tothe number of roots a polynomial may have.

Theorem 5.2.10. A non-constant polynomial of degree n has at most nroots.

Proof. Let f(x) be a non-zero polynomial of degree n > 0. Suppose thatf(x) has a root a. Then f(x) = (x − a)f1(x) by Proposition 5.2.7 and thedegree of f1(x) is n − 1. This argument can be repeated and we reach thedesired conclusion.

5.2.3 The fundamental theorem of algebra

The big question I have so far not dealt with is whether a polynomial needhave a root at all. This is answered by the following theorem whose name re-flects its importance when first discovered, and not its significance in modernalgebra. We shall not give a proof because that would require more advancedmethods than are covered in this book. It was first proved by Gauss.

Theorem 5.2.11 (Fundamental theorem of algebra (FTA)). Every non-constant polynomial of degree n with complex coefficients has a root.


This theorem has the following important consequence using Theorem ??.

Corollary 5.2.12. Every polynomial with complex coefficients of degree nhas exactly n complex roots (counting multiplicities). Thus every such poly-nomial can be written as a product of linear polynomials.

Proof. Let f(x) be a non-constant polynomial of degree n. By the FTA, thispolynomial has a root r1. Thus f(x) = (x−r1)f1(x) where f1(x) is a polyno-mial of degree n− 1. This argument can be repeated and we eventually endup with f(x) = a(x− r1) . . . (x− rn) where a is the last quotient, necessarilya complex number.

Example 5.2.13. It can be checked that the quartic x4− 5x2− 10x− 6 hasroots −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 coefficientsand we will want any factors of the polynomial to be likewise real. The resultabove 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 ofthe complex conjugate of a complex number that we introduced earlier. Wemay now prove the following key lemma.

Lemma 5.2.14. Let f(x) be a polynomial with real coefficients. If the com-plex number z is a root then so too is z.

Proof. Let

f(x) = anxn + an−1x

n−1 + . . .+ a1x+ a0

where the ai are real numbers. Let z be a complex root. Then

0 = anzn + an−1z

n−1 + . . .+ a1z + a0.

Take the complex conjugate of both side and use the properties of the complexconjugate to get

0 = anzn + an−1z

n−1 + . . .+ a1z + a0

and so z is also a root.


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 conjugatesof each other.

Lemma 5.2.16. Let z be a complex number which is not real. Then

(x− z)(x− z)

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− z) = x2 − (z + z)x+ zz.

Observe that z + z and zz are both real numbers. The discriminant of thispolynomial is (z− z)2. You can check that if z is complex and non-real thenz − z is purely complex. It follows that its square is negative. We havetherefore shown that our quadratic is irreducible.

The proof of the second claim follows from the formula for the roots of aquadratic combined with the fact that the square root of a negative real willhave 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 problemsinvolving 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 prod-uct of polynomials with real coefficients which are either linear or irreduciblequadratic.


Proof. We can write the polynomial as a product of linear polynomials. Bringthe real linear factors to the front. The remaining linear polynomials willhave complex coefficients. They correspond to roots that come in complexconjugate pairs. Multiplying together those complex linear factors corre-sponding to complex conjugate roots we get real quadratics and the result isproved.

In fact, we can write any real polynomial as a real number times a productof monic linear and quadratic factors. This result is the basis of the methodof 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 withinteger coefficients. It is a nice, and perhaps unexpected, application of thenumber theory we developed in Chapter 4.

Theorem 5.2.19 (Rational root theorem). Let

f(x) = anxn + an−1x

n−1 + . . .+ a1x+ a0

be a polynomial with integer coefficients. If rs

is a root with r and s coprimethen r | a0 and s | an. In particular, if the polynomial is monic then anyrational roots must be integers and divide the constant term.

Proof. Substituting rs

into f(x) we have, by assumption, that

0 = an(r

s)n + an−1(

r

s)n−1 + . . .+ a1(

r

s) + a0.

Multiply through by sn to get

0 = anrn + an−1sr

n−1 + . . .+ sn−1r + a0sn.

We now make two observations. First, r | a0sn. I claim that r and sn are

coprime. We may now deduce that r | a0 from a previous exercise. It onlyremains to prove the claim. Let p be any prime that divides r and sn. Thenby Euclid’s lemma, p divides r and s which is a contradiction since r and sare coprime. It follows that r | a0. Second, s | anrn. By a similar argumentto the previous case s | an.

Example 5.2.20. Find all the roots of the following polynomial

x4 − 8x3 + 23x2 − 28x+ 12.


The polynomial is monic and so the only possible rational roots are integersand 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. Dividingout by this factor we get the quotient

x3 − 7x2 + 16x− 12.

We check this polynomial for rational roots and find 2 works. Dividing outby (x− 2) we get the quotient

x2 − 5x+ 6.

Once we get down to a quadratic we can solve it directly. In this case itfactorizes 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 Ireally 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 dividedby 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.


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 caneach be written in terms of the roots.

6. 3 + i√

2 is a root of x4 + x3 − 25x2 + 41x + 66. Find the remainingroots.

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 quadraticreal 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 rootsand from the fundamental theorem of algebra (FTA), we know that everynon-zero complex number has three cube roots, and four fourth roots, andmore generally n nth roots. However, we didn’t prove the FTA. The maingoal of this section is to prove that every non-zero complex number has nnth-roots. To do this, we shall think about complex numbers in a geometric,rather than an algebraic, way. Throughout this section we shall not assumeFTA. We shall only need Theorem 5.2.10: every polynomial of degree n hasat most n roots.

5.3. COMPLEX NUMBER GEOMETRY 143

5.3.1 sin and cos

We first recall some well-known properties of the trigonometric functions sinand cos. First the addition formulae

sin(α + β) = sinα cos β + cosα sin β

and

cos(α + β) = cosα cos β − sinα sin β.

These formulae were important historically because they enabled unknownvalues of sin’s and cos’s to be calculated from known ones, and so they wereuseful in constructing trig tables in the days before calculators

In university mathematics, angles are usually measured in radians ratherthan degrees. This is because radians are a natural unit of angle measurementwhereas the system of angle measurement based on degrees is an historicalaccident. Why 360 degrees in a circle? Ask the Ancient Babylonians. Posi-tive angles are measures in an anticlockwise direction.

The sin and cos functions are periodic functions with period 2π. Thismeans 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 sin θ cos θ

0◦ 0 1

30◦ 12

√3

2

45◦ 1√2

1√2

60◦√

32

12

90◦ 1 0

5.3.2 The complex plane

In this section, we shall describe in more detail an alternative way of thinkingabout complex numbers which turns out to be very fruitful. Recall that acomplex number z = a+ bi has two components: a and b. We can plot these


as a point in the plane. The plane used in this way is called the complexplane: the x-axis is the real axis and the y-axis is interpreted as the complexaxis. Although a complex number can be thought of as labelling a point inthe complex plane, it can more usefully be regarded as labelling the directedline 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 itmakes with the positive reals. The length of z as a directed line segment inthe complex plane is |z|, and by basic trig a = |z| cos θ and b = |z| sin θ. Itfollows that

z = |z| (cos θ + i sin θ) .

i |z| sin θ

|z| cos θ

z

θ

Observe that |z| is a non-negative real number. This way of writing complexnumbers is called the polar form.

At this point, I need to clarify the only feature of complex numbers thatcauses confusion. I have already mentioned that the functions sin and cosare periodic. For that reason, there is not just one number θ that yieldsthe complex number z but infinitely many of them: namely, all the numbersθ + 2πk where k ∈ Z. For this reason, we define the argument of z, denotedby arg z, not merely to be the single angle θ but the set of all angles θ+ 2πkwhere k ∈ Z. The angle θ is chosen so that 0 ≤ θ < 2π and is called, forconvenience, the principal argument. But note that books vary on what theychoose to call the principal argument. This feature of the argument plays acrucial 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 suchnumbersdescribe the unit circle with centre the origin in the complex plane.


Thus every non-zero complex number is a real number times a complex num-ber lying on the unit circle.

Let w = r (cos θ + i sin θ) and z = s (cosφ+ i sinφ) be two non-zerocomplex numbers. We shall calculate wz. We have that

wz = rs (cos θ + i sin θ) (cosφ+ i sinφ)

= rs[(cos θ cosφ− sin θ sinφ) + (sin θ cosφ+ cos θ sinφ)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 theirlengths are multiplied and their arguments are added.

This result helps us to understand the meaning of i. Multiplication by iis the same as a rotation about the origin by a right angle. Multiplication byi2 is therefore the same as a rotation about the origin by two right angles.But this is exactly the same as multiplication by −1.

1

i

−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 isinterpreted as rotation about the origin by 180◦. It follows that doing thistwice takes us back to where we started and so is equivalent to multiplicationby 1.

The proof of the next theorem follows by induction from the result weproved above.


Theorem 5.3.1 (De Moivre). Let n be a positive integer. If z = r (cos θ + i sin θ)then

zn = rn (cosnθ + i sinnθ) .

This result has nice applications in painlessly obtaining trigonometricidentities.

Example 5.3.2. Express cos 3θ in terms of cos θ and sin θ using De Moivre’sTheorem. 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 realand 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

for free.

5.3.3 Arbitrary roots of complex numbers

In this section, we shall prove that every non-zero complex number has nnth roots: thus it has three cube roots, and four fourth roots and so on. Webegin with a special case that turns out to give us almost all the informationwe need to solve the general case.

The nth roots of unity


We shall show that the number 1 has n nth roots — these are called then roots of unity. We know that the equation zn− 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 amotivating example.

Example 5.3.3. We find the three cube roots of 1. There are two ways ofwriting these roots: trigonometric form and radical form. By radical formwe mean an algebraic expression obtained from the rationals by carryingout the four basic operations of addition, multiplication, subtraction anddivision together with the extraction of nth roots. Divide the unit circle inthe 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 dividing2π by 3 and ω2 = cos 240◦ + i sin 240◦ which is twice 2π

3. If we put ω = ω1

then in fact ω2 = ω2. This is the trigonometric form of the roots.

1

ω

ω2

In this case, it is easy to write down their radical forms as well. We havethat

ω =1

2

(−1 + i

√3)

and ω2 = −1

2

(1 + i

√3).

The general case is solved in a similar way to our example above usingregular 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 thefollowing formula

cos2kπ

n+ i sin

2kπ

n

for k = 1, 2, . . . , n. These complex numbers are arranged uniformly on theunit circle and form a regular polygon with n sides: the cube roots of unityform an equilateral triangle, the fourth roots form a square, the fifth rootsform a pentagon, and so on.


There is only one point here that is a little confusing. It is always possibleand easy to write down the trigonometric form of the nth roots of unity. Itis also always possible to write down the radical form of the nth roots ofunity but this is far from easy in general. In fact, it forms part of theadvanced subject known as Galois theory.

Example 5.3.5. As part of showing that the 17-gon could be constructedusing only a ruler and compass, Gauss proved the following result which ishighly non-trivial. You can verify that it is true by using a calculator — atleast up to the limits of your calculator.

16 cos2π

17= −1 +

√17 +

√34− 2

√17

+

√(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 nthroots. 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 calculatoryou 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. Letω be the complex cube root of 1 that we described above. Then the threecube roots of 2 are the following

3√

2, ω3√

2, ω2 3√

2.

The above example generalizes.

Theorem 5.3.7 (nth roots). Let z = r (cos θ + i sin θ) be a non-zero complexnumber. Put

u = n√r

(cos

θ

n+ i sin

θ

n

),

the obvious nth root, and put

ω = cos2π

n+ i sin

2π

n,


the first interesting nth root of unity. Then the nth roots of z are as follows

u, uω, . . . , uωn−1.

It follows that the nth roots of z = r (cos θ + i sin θ) can be written in theform

n√r

(cos

(θ

n+

2kπ

n

)+ i sin

(θ

n+

2kπ

n

))

for k = 0, 1, 2, . . . , n− 1.

This is the reason why every non-zero number has two square roots thatdiffer 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 plusa possibly infinite decimal part. It turns out that many functions can also bewritten as a sort of decimal. I shall illustrate this by means of an example.Consider the function ex. All you need to know about this function is thatit is equal to its derivative and e0 = 1. We would like to write

ex = a0 + a1x+ a2x2 + a3x

3 + . . .

where the ai are real numbers that we have yet to determine. We can workout the value of a0 easily by putting x = 0. This tells us that a0 = 1. To getthe value of a1 we first differentiate our expression to get

ex = a1 + 2a2x+ 3a3x2 + . . .

Now put x = 0 again and this time we get that a1 = 1. To get the value ofa2 we differentiate our expression again to get

ex = 2a2 + 3 · 2 · a3x+ . . .

Now put x = 0 and we get that a2 = 12. Continuing in this way we quickly

spot the pattern for the values of the coefficient an. We find that an = 1n!

where n! = n(n − 1)(n − 2) . . . 2 · 1. What we have done for ex we can alsodo for sin x and cosx and we obtain the following series expansions of eachof these functions.


• ex = 1 + x+ x2

2!+ x3

3!+ x4

4!+ . . ..

• sinx = x− x3

3!+ x5

5!− x7

7!+ . . ..

• cosx = 1− x2

2!+ x4

4!− x6

6!+ . . ..

There are interesting connections between these three series. We shallnow show that complex numbers help to explain them. Without worryingabout the validity of doing so, we calculate the infinite series expansion ofeiθ. We have that

eiθ = 1 + (iθ) +1

2!(iθ)2 +

1

3!(iθ)3 + . . .

that is

eiθ = 1 + iθ − 1

2!θ2 − 1

3!θ3i +

1

4!θ4 + . . .

By separating out real and complex parts, and using the infinite series weobtained above, we get Euler’s remarkable formula

eiθ = cos θ + i sin θ.

Thus the complex numbers enable us to find the hidden connections betweenthe three most important functions of calculus: the exponential function andthe sine and cosine functions. It follows that every non-zero complex numbercan be written in the form reiθ. If we put θ = π in Euler’s formula, we getthe following result, which is widely regarded as one of the most amazing inmathematics.

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 inthat 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 cosx and sin x.

2. Prove the following where x is real.2

(a) sinx = 12i

(eix − e−ix).(b) cos x = 1

2(eix + e−ix).

Hence show that cos4 x = 18[cos 4x+ 4 cos 2x+ 3].

3. Find the 4th 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 theyobey the usual high-school rules of algebra. In this section, I shall sketch outa proof of this.

We start with the set R×R whose elements are ordered pairs (a, b) wherea and b are real numbers. It will be helpful to denote these ordered pairs bybold 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 = (a1b1 − a2b2, a1b2 + a2b1).

2Compare (a) and (b) below with sinhx = 12 (ex − e−x) and coshx = 1

2 (ex + e−x).


• If a = (a1, a2) 6= 0 define

a−1 = (a1√a2

1 + a22

,−a2√a2

1 + a22

).

It is now a long exercise to check that all the usual axioms of high-schoolalgebra hold. Observe now that the element (a1, a2) can be written

(a1, 0)1 + (a2, 0)i

and thatii = (0, 1)(0, 1) = (−1, 0) = −1.

The elements of the form (a, 0) can be identified with the real numbers. Thisproves that the complex numbers as I described them earlier in this chapterreally 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 integersZ are a subset of the reals R. We define the Gaussian integers, denotedby Z[i], to be all complex numbers of the form m + in where m and n areintegers. What our example shows is that some primes can be factorized usingGaussian 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 exampleof 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 oftwo squares.

5.6. RADICAL SOLUTIONS 153

Lemma 5.5.1. Let p be an odd prime that can be written as a sum of twosquares. 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 assumethat a2 is odd and b2 is even. But from Chapter 2, this implies that a isodd and b is even. We may therefore write a = 2u and b = 2v + 1 for somenatural numbers u and v. But then p = 4u2 + 4v2 + 2v + 1. It follows thatp ≡ 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 pis 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 oftwo 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 twosquares then p ≡ 1 (mod 4). The hard question is whether the converse istrue.

Theorem 5.5.4 (Euler, 1754). An odd prime p can be written as a sum oftwo 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 Radical solutions

By the fundamental theorem of algebra, we know that any non-constantpolynomial has roots. Such roots can be explicitly calculated using root-finding algorithms, such as the Jenkins-Traub algorithm. So if your interestin the roots is practical — for example, this problem is an essential part of


solving linear differential equations — then you might think that there isno more to be said. But it turns out, as so often happens in maths, thatasking a more precise question can lead to new ideas. In this section, I shallsurvey the nature of the solutions of a polynomial equation not merely theirexistence. We begin by describing the way in which cubics and quartics maybe solved purely algebraically.

5.6.1 Cubic equations

Letf(x) = a3x

3 + a2x2 + a1x+ a0

where a3 6= 0. I shall assume all coefficients are real though the theoryworks in general. We shall find all the roots of f(x). This problem can besimplified in two ways. First, we may divide through by a3 and so, withoutloss 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 ofthe term in x2 is zero. Put x = y − a3

3. You should do this and check that

you get a polynomial of the form

g(y) = y3 + 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 complexvariables. Let ω = cos 2π

3+ i sin 2π

3, one of the complex cube roots of unity.

You should now check that the following cubic

t(x) = x3 − 3uv − (u3 + v3)

has the rootsu+ v, uω + vω2, uω2 + vω.

Now we can solvex3 + px+ q = 0

if we can find u and v such that

p = −3uv, q = −u3 − v3.


Now if we cube the first equation, we get the following two equations

−p27

= u3v3, −q = u3 + v3.

If we regard u3 and v3 as the unknowns we know their sum and we know theirproduct. This means that u3 and v3 are the roots of the quadratic equation

x2 + qx− p3

27= 0.

We therefore have that

u3 =1

2

(−q +

√27q2 + 4p3

27

)

and

v3 =1

2

(−q −

√27q2 + 4p3

27

).

To find u we have to take a cube root of the number u3 and there are threepossible such roots. Choose one such value for u. We then choose the valueof 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 rootin this case to get

u =3

√1 +√

28.

We must then choose v to be

u =3

√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 hadan important influence on the development of 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 wayequivalent to the following

x = 2±√−121.

We would write this asx = 2± 11i.

Thusu3 = 2 + 11i and v3 = 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 leastsymbolically 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 mustbe real. The reason is that the graph of our cubic has one real root whichcan 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 choose2− 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 providedthe first inkling that there was a number system, the complex numbers,that extended the so-called real numbers, but had every much as tangibleexistence.


5.6.2 Quartic equations

Letf(x) = a4x

4 + a3x3 + a2x

2 + a1x+ 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 thereforeend up with a reduced quartic which it is convenient to write in the followingway

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 originalquartic. It is not true that we can always do this, but by means of anothermiracle we can transform the equation into one with the same roots wherewe can. Let t be a new variable whose value will be determined later. Wemay 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 perfectsquare. 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 tofind a specific value of t say t1. We then get

(x2 + t1)2 = (a+ 2t1)

(x+

b

2(a+ 2t1)

)2

.

It follows that the roots of the original quartic are the roots of the followingtwo quadratics

(x2 + t1)−√a+ 2t1

(x+

b

2a+ 4t1

)= 0

and

(x2 + t1) +√a+ 2t1

(x+

b

2a+ 4t1

)= 0.


Example 5.6.3. Solve the quartic

x4 = 1− 4x.

We shall find a value of t below

(x2 + t)2 = t4 + 2x2t+ t2 = 2x2t− 4x+ (1 + t2)

which makes the righthand side a perfect square. This requires us to find aroot 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 twoquadratics

(x2 + 1)−√

2(x− 1) = 0 and x2 + 1 +√

2(x− 1) = 0.

The roots of our original quartic are therefore

1± i√√

8 + 1√2

and−1±

√√8− 1√

2.

5.6.3 Symmetries and particles

Although quadratic equations had been solved in antiquity, it was not untilthe 16th century that cubics and quartics were first solved. This great leapforward in the development of algebra was centred on a group of Italianmathematicians — Scipione del Ferro (1465–1525), Niccolo Tartaglia (1500–1557), Girolamo Cardano (1501–1576), Ludovico Ferrari (1522–1562), RafaelBombelli (1526–1572) — whose antics are worthy of an opera or Shakespearecomedy but the importance of their work cannot be overemphasized. Buttwo points arise. First, the solution of quadratics, cubics and quartics seemto rely on mathematical miracles. Second, we appear to see a pattern: tosolve cubics we need to solve an associated quadratic and to solve quarticswe need to solve an associated cubic. These two points were investigated bya number of mathematicians in great depth: in particular, Lagrange (1736–1813), Ruffini (1765–1822) and Abel (1802–1829). The expectation was high


that quintics should be solvable by using quartics in a way that continuedthe pattern. Then came the great surprize. Ruffini and Abel proved thatthe pattern does not continue and that one cannot always describe the rootsof a quintic in radical form — there are, of course, five roots — the pointis that these roots cannot in general be written down using an algebraicformula. The question is why and the answer to this question also explainsthe 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 himelsewhere, some of it accurate, instead I shall focus on his mathematics.By building on the work of Lagrange, he explained the miracles above andmuch more. His approach was new: to determine whether the roots of apolynomial could be expressed in algebraic terms as radical expressions, hestudied the symmetries of the polynomial. Just what this means is explainedin a subject known as Galois theory after its founder. Crucially, this is nota mere extrapolation of existing algebraic manipulation, instead it involvesworking 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 understandthe tricks that were used to solve differential equations. It was in this waythat symmetry came to play a fundamental role in physics. If you hear aparticle physicist talking about symmetries, they are paying an unconscioustribute to Galois’ bold work in studying the nature of the roots of polynomialequations.

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