Chapter 5 Complex Numbers

Home , Complex multiplication, Cube root

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 rˆolein 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. Com- plex 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 1 i 3. − . Multiply out and simplify to get i. √2 − 128 CHAPTER 5. COMPLEX NUMBERS

The final operation is division. We have to show that when a + ib = 0 the reciprocal 6 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 ∈ z¯ = a bi. − The numberz ¯ is called the complex conjugate of z. Why is this operation useful? 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 numbers z, the number zz¯ 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 = 0. The ﬁrst step is to multiply top and bottom by the complex conjugate of6 a + bi. We therefore get a bi a bi 1 − = − = (a bi) . (a + bi)(a bi) a2 + b2 a2 + b2 − − Examples 5.1.2. Carry out the following calculations.

1+i 1. i . The complex conjugate of i is i. Multiply top and bottom of the i+1 − fraction to get − = 1 i. 1 − i 2. 1 i . The complex conjugate of 1 i is 1 + i. Multiply top and bottom − i(1+i) i 1 − of the fraction to get 2 = −2 . 4+3i 3. 7 i . The complex conjugate of 7 i is 7 + i. Multiply top and bottom − (4+3i)(7+i) −1+i of the fraction to get 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

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 = √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 bdp)2 + (ad + bc)2 = (ac)2 + (bd)2 + (ad)2 + (pbc)2 = (a2 + b2)(c2 + d2). − Thus the result follows. 1Plural: 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 y2)+2xyi = a+bi, and so equating real and imaginary parts, we have to solve− the following two equations x2 y2 = 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 . 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

a √a2 + b2 a + √a2 + b2 x2 = + = . 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 y2 = 3 and 2xy = 4. − 5.1. COMPLEX NUMBER ARITHMETIC 131

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 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

4z2 + 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 ( 13 16i) z2 + iz + − − = 0. 4 Next, we observe that

i 2 1 z + = z2 + iz . 2 − 4 Thus i 2 1 z2 + iz = z + + . 2 4 Our equation therefore becomes

i 2 1 13 z + + + 4i = 0. 2 4 − 4 − 132 CHAPTER 5. COMPLEX NUMBERS

We therefore have i 2 z + = 3 + 4i. 2 Taking square roots of both sides using a previous calculation, we have that i z + = 2 + i or 2 i. 2 − − i 3i It follows that z = 2 + 2 or 2 2 . Now check that these roots really do work. − − 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. 2+3i (d) 4+i . 1 3 (e) i + 1+i . 3+4i 3 4i (f) 3 4i 4+4− i . − − 2. Find the square roots of each of the following complex numbers and check 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)2 x2 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 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. 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 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

n n 1 anx + an 1x − + ... + a1x + a0 − where ai are complex numbers, called the coeﬃcients, is called a polynomial. We assume a = 0. The degree of this polynomial is n. We abbreviate this n 6 to deg. If an = 1 the polynomial is said to be monic. The term a0 is called n the constant term and the term anx is called the leading term. Polynomials can 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 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).

m n Proof. Let f(x) have leading term amx and let g(x) have leading term bnx . m+n Then the leading term of f(x)g(x) is ambnx . Now ambn = 0 and so the degree of f(x)g(x) is m + n, as required. 6 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.

2 7 3x 2 x 2x2 + 4x + 5 6x4 +− 5x3 + 4x2 + 3x + 2 6x4 + 12x3 + 15x2 7x3 11x2 + 3x + 2 − 3 − 2 35 7x 14x 2 x − 2 −41 − 3x + 2 x + 2 136 CHAPTER 5. COMPLEX NUMBERS

The procedure is repeated one more time with the new polynomial

2 7 3 3x 2 x + 2 quotient 2x2 + 4x + 5 6x4 +− 5x3 + 4x2 + 3x + 2 6x4 + 12x3 + 15x2 7x3 11x2 + 3x + 2 − 3 − 2 35 7x 14x 2 x − 2 −41 − 3x + 2 x + 2 2 12 15 3x + 2 x + 2 29 x 11 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 7 3 29 11 6x4 + 5x3 + 4x2 + 3x + 2 = 2x2 + 4x + 5 3x2 x + + 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)f (x) by Proposition 5.2.7 and the − 1 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 re- flects 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 non- constant 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 n n 1 f(x) = anx + an 1x − + ... + a1x + a0 − where the ai are real numbers. Let z be a complex root. Then

n n 1 0 = anz + an 1z − + ... + a1z + a0. − Take the complex conjugate of both side and use the properties of the complex conjugate to get

n n 1 0 = anz¯ + an 1z¯ − + ... + a1z¯ + a0 − and soz ¯ 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 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 this polynomial is (z z¯)2. You can check that if z is complex and non-real then z z¯ 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 quadratic. 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

n n 1 f(x) = anx + an 1x − + ... + a1x + a0 − r be a polynomial with integer coeﬃcients. If s 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.

r Proof. Substituting s into f(x) we have, by assumption, that

r n r n 1 r 0 = an( ) + an 1( ) − + ... + a1( ) + a0. s − s s Multiply through by sn to get

n n 1 n 1 n 0 = anr + an 1sr − + ... + s − r + a0s . − We now make two observations. First, r a sn. I claim that r and sn are | 0 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 n are coprime. It follows that r a0. Second, s anr . By a similar argument to the previous case s a . | | | n 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)2 x3 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 143

5.3.1 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. Posi- tive 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 sin θ cos θ

0◦ 0 1 1 √3 30◦ 2 2 45 1 1 ◦ √2 √2 √3 1 60◦ 2 2 90◦ 1 0

5.3.2 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 zn = 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 andi3 = 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 − for free.

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 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 a motivating example. Example 5.3.3. We ﬁnd the three cube roots of 1. There are two ways of writing these roots: trigonometric form and radical form. By radical form 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 2π 2π by 3 and ω2 = cos 240◦ + i sin 240◦ which is twice 3 . If we put ω = ω1 2 then in fact ω2 = ω . This is the trigonometric form of the roots. ω

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 ω2 = 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.

2π 16 cos = 1 + √17 + 34 2√17 17 − − q + 68 + 12√17 16 34 + 2√17 2(1 √17) 34 2√17 − − − − s q q 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 2, ω√3 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 complex number. Put θ θ u = √n r cos + i sin , n n the obvious nth root, and put 2π 2π ω = cos + 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

n 1 u, uω, . . . , uω − .

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. Consider the function ex. All you need to know about this function is that it is equal to its derivative and e0 = 1. We would like to write

x 2 3 e = a0 + a1x + a2x + a3x + ... 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

x 2 e = a1 + 2a2x + 3a3x + ...

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 = 2a + 3 2 a x + ... 2 · · 3 1 Now put x = 0 and we get that a2 = 2 . Continuing in this way we quickly 1 spot the pattern for the values of the coeﬃcient an. We ﬁnd that an = n! 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

2 3 4 ex = 1 + x + x + x + x + .... • 2! 3! 4! 3 5 7 sin x = x x + x x + .... • − 3! 5! − 7! 2 4 6 cos x = 1 x + x x + .... • − 2! 4! − 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 1 1 eiθ = 1 + (iθ) + (iθ)2 + (iθ)3 + ... 2! 3! that is 1 1 1 eiθ = 1 + iθ θ2 θ3i + 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 = 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

1 ix ix (a) sin x = (e e− ). 2i − 1 ix ix (b) cos x = 2 (e + e− ).

4 1 Hence show that cos x = 8 [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 = (a , a ) and b = (b , b ), define a + b = (a + b , a + b ). • 1 2 1 2 1 1 2 2 If a = (a , a ) define a = ( a , a ). • 1 2 − − 1 − 2 If a = (a , a ) and b = (b , b ), define • 1 2 1 2 ab = (a b a b , a b + a b ). 1 1 − 2 2 1 2 2 1 2 1 x x 1 x x Compare (a) and (b) below with sinh x = (e e− ) and cosh x = (e + e− ). 2 − 2 152 CHAPTER 5. COMPLEX NUMBERS

If a = (a , a ) = 0 deﬁne • 1 2 6 1 a1 a2 a− = ( , − ). 2 2 2 2 a1 + a2 a1 + a2 It is now a long exercise to checkp that all thep 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. 5.6. RADICAL SOLUTIONS 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 + 4v2 + 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 Radical solutions

By the fundamental theorem of algebra, we know that any non-constant polynomial has roots. Such roots can be explicitly calculated using root- ﬁnding 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 3 2 f(x) = a3x + a2x + a1x + a0 where a3 = 0. I shall assume all coefficients are real though the theory works in general.6 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 2 a3 the term in x is zero. Put x = y 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 complex 2π 2π variables. Let ω = cos 3 + i sin 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 roots u + v, uω + vω2, uω2 + vω. Now we can solve x3 + px + q = 0 if we can ﬁnd u and v such that

p = 3uv, q = u3 v3. − − − 5.6. RADICAL SOLUTIONS 155

Now if we cube the ﬁrst equation, we get the following two equations p − = u3v3, q = u3 + v3. 27 − If we regard u3 and v3 as the unknowns we know their sum and we know their product. This means that u3 and v3 are the roots of the quadratic equation

p3 x2 + qx = 0. − 27 We therefore have that

1 27q2 + 4p3 u3 = q + 2 − r 27 ! and 1 27q2 + 4p3 v3 = q . 2 − − r 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± 3 u = 1 + √28. We must then choose v to be q

3 u = 1 √28. − q 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 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 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. RADICAL SOLUTIONS 157

5.6.2 Quartic equations Let 4 3 2 f(x) = a4x + a3x + a2x + 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 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 b 2 (x2 + t )2 = (a + 2t ) x + . 1 1 2(a + 2t ) 1 It follows that the roots of the original quartic are the roots of the following two quadratics b (x2 + t ) √a + 2t x + = 0 1 − 1 2a + 4t 1 and b (x2 + t ) + √a + 2t x + = 0. 1 1 2a + 4t 1 158 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 + 2x2t + t2 = 2x2t 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 quadratics

(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 1 √8 1 ± and − ± − . p√2 p√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 5.6. RADICAL SOLUTIONS 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. 160 CHAPTER 5. COMPLEX NUMBERS