Oxford University Mathematical Institute. Complex Numbers

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1 Version of 30 September 2017 Oxford University Mathematical Institute Complex Numbers Notes by Peter M. Neumann (Queen s College) 1. Introduction Students coming to Oxford to study mathematics arrive with a wide range of background knowledge and training. In respect of complex numbers some will know much and have acquired useful confidence, others will be almost beginners. But it is likely that, however much or little you have already learned, there will be less unanimity of language and perception than is desirable in a large group of students who are studying mathematics together to a deep level. The purpose of the two lectures, and therefore the purpose of these lecture notes, is to serve as a reminder to those who have met the topic before, to introduce the ideas to those new to the topic, and perhaps most importantly to establish a common language and notation. 2. Complex numbers For us a complex number is an entity of the form a + bi. Here a, b are real numbers and i = 1. The real number system R which should be familiar to us all consists of rational numbers p/q and the irrational numbers like 2 or e that fit between them to make a system that many of us picture using the number line It is the ingredient i of a complex number that took mathematicians of the 16 th to 19 th centuries much time and mental readjustment to come to terms with. We learn quite early in our lives that minus times minus is plus. As a consequence, the square of any real number is non-negative, so the equation x 2 = 1 has no real roots. Complex numbers result from the imaginative idea of simply inventing an entity i that is a root of this equation, that is, for which i 2 = 1. Indeed, the notation i is said to have been invented in recognition that this must be an imaginary number having nothing to do with quantity and little to do with reality. Note. Sometimes we ll use the form a + bi, sometimes a + ib. There is no mathematical difference. In some contexts there is a mild psychological difference. As is well known, in the Real world, there are quadratic equations such as x = 0 and x 2 x + 1 = 0 that have no roots. Once we have complex numbers, however, all quadratic equations with real coefficients have roots: If a 0 and ax 2 + bx + c = 0 then x = b ± b 2 4ac. If b 2 4ac, the so-called 2a discriminant, is non-negative then the roots are real numbers; if b 2 4ac < 0 then b 2 4ac = 4ac b 2 i. Thus if b 2 4ac then the equation has real roots and if b 2 4ac < 0 then roots exist as complex, non-real numbers. 1

2 Exercise 2.1 Which of the following quadratic equations have real roots, which do not? 3x 2 + 2x 1 = 0; 2x 2 6x + 9 = 0; 4x 2 + 7x 9 = 0; 4x 2 9x + 5 = Arithmetic of complex numbers The collection of all complex numbers is written C (notation that is pretty much standard in mathematics everywhere in the world). Addition and multiplication are basic binary operations on C (in the precise sense of the technical term binary operation that is to be explained in other lecture courses). They are what you would expect: (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) (c + di) = (ac bd) + (ad + bc)i. That rule for multiplication comes simply from the calculation (a + bi) (c + di) = ac + adi + bci + bdi 2 and exploiting the fact (or assumption?) that i 2 = 1. We ll come to division later. Exercise 3.1 Put each of the following complex numbers into standard form a + bi: (1 + 2i)(3 i); (2 + i)(1 2i); (1 + i) 4 ; (1 3i) The Argand diagram The complex number a+bi may be identified with the point (a, b) of the euclidean plane R 2. For this reason, some people call a + bi the cartesian form of a complex number. Just as we speak of the real line when we represent real numbers geometrically on the number line, so we speak of the complex plane (also known as the Argand diagram ) when we picture C as a 2-dimensional geometrical object. If z C and z = a + bi then a is known as the real part Rez (or Rez or Rz notation varies from book to book and from lecturer to lecturer but its meaning should always be clear from the context) and b is known as the imaginary part Imz (or Imz or I z ). It is important that R C in the sense that R = {z C Imz = 0}. In fact this is not just a containment as set, it is containment as arithmetical structure. For, (a + 0i) + (c + 0i) = (a + c) + 0i and (a + 0i) (c + 0i) = (ac) + 0i, so the arithmetic of real numbers identified as a special kind of complex numbers is precisely the same as the arithmetic of R. We often express this by writing R C rather than R C. With this identification R becomes the x-axis in the complex plane. The y -axis consists of numbers of the form bi. These are generally known as pure imaginary complex numbers. Clearly, any complex number is uniquely the sum of a real number and a pure imaginary number. Addition of complex numbers may be represented geometrically in the Argand diagram by the so-called parallelogram law, as shown in the diagram. There is also a geometrical description of multiplication, but it is more complicated and will be discussed later. 2

3 Exercise 4.1. Show that there is no binary relation < on C for which the following conditions are satisfied: for any z C exactly one of z < 0, z = 0, 0 < z holds; if z, w C, 0 < z and 0 < w then 0 < z + w; if z C and z 0 then 0 < z 2. Note. It is a consequence that there is no order-relation on C analogous to that on R. Thus inequalities can only be valid between real numbers. Formulae such as i < 2i and 1 + i < 2 + i make no sense. 5. Conjugate and modulus of a complex number For z = a + bi C the complex conjugate z (for which you might sometimes see the notation z, though very rarely in the Oxford mathematics courses) is defined by z := a bi. Interpreted geometrically complex conjugation is reflection of the Argand diagram in the real axis. Here is an easy, but disproportionately useful and important fact about it: Theorem C1: If z, w C then z + w = z + w and z w = z w. Because this is a very important fact, it is important that you yourself have internalised the reasons why it is true. That is why the proof is left as an exercise if it gives you trouble then please consult your Tutors or write me a note. Exercise 5.1. Write out a proof of this theorem. For z = a + bi C the absolute value or modulus (these terms are used interchangeably) z is defined by z := a 2 + b 2. Since a 2 + b 2 0 this real number has a non-negative square root and this is what is meant by the symbol. Thus z 0, and z = 0 if and only if z = 0. Geometrically, z is the distance of z from 0 in the complex plane. Three details about the modulus that follow directly from the definition are worth noting: (1) z = z ; (2) if x is real then x = x 2, that is, x = x if x 0, x = x if x < 0, which accords with the definition of absolute value for real numbers (in their own right); (3) if z = a + bi then, since a 2 + b 2 a 2, z a, that is, Rez z ; similarly, of course, Imz z. Theorem C2: (1) If z C then z = z z. (2) If z, w C then z w = z w. For the proof of (1) note that if z = a + bi then z z = (a + bi)(a bi) = a 2 i 2 b 2 = a 2 + b 2, so that z = z z by definition of the modulus. Now to prove (2) calculate as follows: z w 2 = (z w)(z w) = (z w)(z w) [by Theorem C1] = (z z)(ww) = z 2 w 2, and the desired result comes by taking square roots. Suggestion: Get into the habit of working with complex numbers as single entities. There is rarely any need to write them in terms of their real and imaginary parts. Our proof of (2) illustrates this. 3

4 Theorem C3 [The Triangle Inequality]: If z, w C then z + w z + w. Proof. Geometrically, this is the fact that the length of any side of a triangle is always at most the sum of the lengths of the other two sides. In this case the relevant triangle has its vertices at 0, z and z + w in the complex plane. The side from 0 to z has length z, the side from z to z + w has length w, and the side from 0 to z + w has length z + w. It is perhaps worth seeing an alternative, more algebraic proof. z + w 2 = (z + w)(z + w) = (z + w)(z + w) = zz + wz + z w + ww. Now wz = z w and so wz +z w = 2Re(z w). By the third of the notes that follow the definition of absolute value, Re(z w) z w. We also have z w = z w = z w. Therefore z + w 2 = z 2 + 2Re(z w) + w 2 z z w + w 2. Thus z + w 2 ( z + w ) 2, and taking positive square roots we find that z + w z + w. Exercise 5.2. Let z, w C and suppose that z 0. Show that z + w = z + w if and only if there is a real number λ 0 such that w = λz. 6. Division of complex numbers We have just seen that if z C then z z = z 2 and that if z 0 then z 2 > 0. It follows that if z 0 then 1 z = z z 2, that is, 1 a + bi = a a 2 + b 2 b a 2 + b 2 i. Therefore if z, w C and z 0 then w z = w z z z = wz. Thus C is very similar to Q or z 2 R in that it has addition, multiplication and division by non-zero numbers. These operations satisfy the usual rules of arithmetic rules that are familiar, but only to those who have taken the time to stop and think what they are. In language to be introduced in other courses, the complex number system C is a field (just as Q and R are fields). Exercise 6.1. Work out 7 2i i, i and a few other such complex number calculations 1 i until you are confident that you fully understand division in C. Exercise 6.2. form a + bi: Let z 1 := 1 + i and z 2 := 2 3i. Put each of the following into standard z 1 + z 2 ; z 1 z 2 ; z 1 z 2 ; z 1 /z 2 ; z 1 z 2 ; z 1 z The argument of a complex number In plane geometry and its applications it is often useful to switch between cartesian and polar coordinates. In the Argand diagram the polar coordinates of a complex number z are (r, α) where r is the distance from 0 and α is the angle from the positive x-axis (the positive real line) anti-clockwise to the ray joining 0 to z. It should be clear from the definition of absolute value that r = z. The angle α is known as the argument arg z. In principle it is many-valued. For any integer k the angle α + 2kπ would serve just as well. Many mathematicians would express this by saying that arg z is only determined modulo 2π. 4

5 Two important details about the argument are worth noting: that arg 0 is not defined, and that arg z = arg z. Sometimes one specifies that π < arg z π to make the argument single-valued. This single-valued version of arg z is known as its principal value. Another convention that some authors use is that the principal value should satisfy 0 arg z < 2π, but this version does not combine well with the equation arg z = arg z. Switching back and forth between polar and cartesian coordinates requires trigonometry. If z = a + bi C \ {0} and r = z, α = arg z then a = r cos α, b = r sin α. So in terms of modulus and argument, z = r(cos α + i sin α). Going the other way, if z = a + bi 0 and r := z, α := arg z then r = a 2 + b 2 (of course this is how z was defined) and, since b/a = r sin α/r cos α = tan α, we find α as tan 1 (b/a) where, however, the inverse tangent function has (as always) to be interpreted with great care. Exercise 7.1. Find the modulus and argument of each of the following complex numbers: 1 + 3i; (2 + i)(3 i); (1 + i) 5. Exercise 7.2. On separate Argand diagrams sketch each of the following subsets of C: A := {z z < 1}; B := {z Rez = 3}; C := {z 1 4 π < arg z < 1 4 π}; D := {z arg(z i) = 1 2π}; E := {z z 3 4i = 5}; F := {z z 1 = z i }. We have seen in Theorem C2 that if z, w C then z w = z w. How is arg(z w) related to arg z and arg w? Well, if α := arg z and β := arg w, so that z = z (cos α + i sin α), w = w (cos β + i sin β), then z w = z w (cos α + i sin α)(cos β + i sin β). Using the trigonometrical addition formulae for sine and cosine we see that (cos α + i sin α)(cos β + i sin β) = (cos α cos β sin α sin β) + i(sin α cos β + cos α sin β) = cos(α + β) + i sin(α + β). Thus we have proved the following companion to Theorem C2(2): Theorem C4: If z, w C then arg(z w) = arg z + arg w. Note. It is important here that arg z is not restricted to the principal value. Even if π < arg z π and π < arg w π, it could well happen that arg z + arg w > π or arg z + arg w < π. For fundamental facts like this one, it is essential that the argument is understood to be determined only modulo 2π. The theorem indicates that the argument function behaves somewhat like a logarithmic function. In fact, it is an important constituent of the complex logarithmic function but that is material for a more advanced study than can be offered in this short course. Theorem C4 yields a geometric interpretation of multiplication as promised above, at the end of 4. As a transformation of the Argand diagram, the operation multiply by w leaves the point 0 unmoved, scales distances from 0 by a factor w, and rotates the plane anticlockwise 5

6 about 0 through an angle arg w. It is worth remembering that this anticlockwise convention for rotations is standard not just in the context of the complex plane, but also in geometry, in mechanics in most of mathematics. There is more to the convention, though: if α > 0 then an anticlockwise rotation through angle α is a clockwise rotation through angle α. Exercise 7.3. For each of the following complex numbers w, what transformation of the Argand diagram does multiplication by w represent? 8. De Moivre s Theorem i; (1 + i); (1 i); (3 + 4i). If θ R then cos 2 θ + sin 2 θ = 1. Therefore complex numbers of the form cos θ + i sin θ are of absolute value 1. Conversely, we have seen that if z = 1 and α := arg z then z = cos α + i sin α. Define S 1 := {z C z = 1}, [standard notation] the unit circle in the complex plane. Described a little differently, S 1 = {cos θ+i sin θ θ R }. We know, since absolute values are multiplicative, that if z, w S 1 then also z w S 1, that is, S 1 is closed under the multiplication of complex numbers; moreover, if z S 1 then z 1 = z S 1. In terms that some of you will already have met and all of you will meet before long, this means that S 1 is a group. It is a very important group known as the circle group. Theorem C5 [De Moivre s Theorem]. If z S 1, θ := arg z and n Z then arg(z n ) = nθ. Equivalently, (cos θ + i sin θ) n = cos nθ + i sin nθ. For n 0 this comes quickly by mathematical induction: it is certainly true if n = 0 (since z 0 = 1 and arg 1 = 0) or if n = 1; if it is true when n = m then arg(z m+1 ) = arg(z m )+arg z = m. arg z + arg z (by inductive assumption) and so arg(z m+1 ) = (m + 1) arg z, that is, it is then true when n = m + 1. If n < 0, say n = N where N > 0, then z n = (z 1 ) N = z N, and so arg(z n ) = N arg z = N( arg z) = N arg z = n arg z. Thus the assertion is true for all integers n. Note. Both for De Moivre s Theorem and for its proof it is important that we do not have to construe an argument θ as being an angle in the range ( π, π]. Even if θ lies in this range, nθ need not. De Moivre s Theorem has many uses. One of them is to express the cosines and sines of multiples of an angle θ as polynomials in cos θ and sin θ. For example: cos 3θ + i sin 3θ = (cos θ + i sin θ) 3 = cos 3 θ + 3i cos 2 θ sin θ + 3i 2 cos θ sin 2 θ + i 3 sin 3 θ = (cos 3 θ 3 cos θ sin 2 θ) + i(3 cos 2 θ sin θ sin 3 θ), and, by equating real and imaginary parts, we find that and cos 3θ = cos 3 θ 3 cos θ sin 2 θ = cos 3 θ 3 cos θ(1 cos 2 θ) = 4 cos 3 θ 3 cos θ sin 3θ = 3 cos 2 θ sin θ sin 3 θ = 3(1 sin 2 θ) sin θ sin 3 θ = 4 sin 3 θ + 3 sin θ. 6

7 Exercise 8.1. Show that if θ R then cos 5θ = 16 cos 5 θ 20 cos 3 θ + 5 cos θ; sin 5θ = (16 cos 4 θ 12 cos 2 θ + 1) sin θ. Exercise 8.2. Show that if m is a positive integer then (1 + i) 4m = ( 1) m 2 2m. 9. Roots of unity If z C, n N and z n = 1 then z is known as a root of unity an n th root of unity, to be precise. For such numbers, since z n = z n = 1 and z > 0, we have z = 1. Thus complex roots of unity lie in S 1. Theorem C6. Let z C. Then z is an n th root of unity if and only if z = 1 and there exists an integer k such that arg z = 2kπ/n. Proof. Suppose first that z = 1 and arg z = 2kπ/n for some integer k. Then z n = 1 and, by De Moivre s Theorem, arg(z n ) = n 2kπ/n = 2kπ. Thus z n has the same modulus and the same argument as 1, and therefore z n = 1. Conversely, if z n = 1 then, as we have already seen, z = 1, and moreover, if α := arg z then by De Moivre s Theorem again, nα = arg 1 = 2kπ for some integer k, and therefore arg z = 2kπ/n. Corollary: For any positive integer n the number of n th roots of 1 is n. For, by Theorem C5, the n th roots of 1 are the complex numbers cos(2kπ/n)+i sin(2kπ/n) for 0 k n 1, and there are precisely n of these. Examples: The square roots of 1 in C are just its real square roots 1 and 1; the fourth roots of 1 in C are 1, i, 1, i. Note. If z is an n th root of unity and z m 1 for each natural number m in the range 0 < m < n then z is said to be a primitive n th root of unity. Theorem C7. Let z C and let n be an integer 2. If z is an n th root of unity and z 1 then z n 1 + z n z 2 + z + 1 = 0. This equation holds, in particular, for primitive n th roots of unity. For, let z be an n th root of unity. Since 0 = z n 1 = (z 1)(z n z + 1) and z 1 by assumption, it must be the case that z n z + 1 = 0. Example. Let ω be a primitive cube root of 1. Then ω 2 + ω + 1 = 0. Solving this quadratic equation in the usual way we find that ω = 1 2 ( 1 ± 3) = 1 2 ( 1 ± i 3). I find it comforting that this is consistent with the facts that { ω = cos(2π/3) + i sin(2π/3) or ω = cos(4π/3) + i sin(4π/3), and { cos(2π/3) = cos(4π/3) = 1 2 sin(2π/3) = sin(4π/3) =

8 Exercise 9.1. Write down the primitive 6 th roots of unity and the primitive 8 th roots of unity in standard form a + bi. Exercise 9.2. Is 3 + 4i a root of unity? Is i a root of unity? Exercise 9.3 [Harder]. of unity. Find all pairs (a, b) of rational numbers for which a + bi is a root Exercise 9.4. Let φ := cos(2π/5) + i sin(2π/5), a primitive 5 th root of 1. Define α := φ + φ 4, β := φ 2 + φ 3. (i) Show that α and β are real numbers. (ii) Show that α + β = 1 and αβ = 1 (so that α, β are the roots of x 2 + x 1 = 0). (iii) Deduce that cos(2π/5) = 1 4 ( 5 1). Exercise 9.5. Let p be a prime number let ω be a primitive p th root of unity and let m p 1 { be a positive integer. Show that ω m k 0 if p does not divide m, = p if p divides m. k=0 10. A formula discovered by Euler We observed in 7 above that Theorem C4 indicates that the argument behaves somewhat like a logarithm. A famous formula published by Leonhard Euler in 1748, e iθ = cos θ + i sin θ, explains, or at least is consistent with, the logarithmic nature of the argument. In this notation Theorem C4 says that e iα e iβ = e i(α+β), just as we would expect of the exponential function. The exponential function may be introduced in many different ways. An important approach is to say that e x is that function f(x) which is its own derivative and for which f(0) = 1. That such a function exists is not obvious, however. An alternative is to use a power series, defining e x to be the infinite sum (once a sense can be attached to the idea of adding infinitely many numbers together) 1 + x 1! + x2 2! + + xn n! +. Substituting x = iθ and separating the series into real and imaginary parts we find two power series, recognisable as those for cos θ and for sin θ respectively, so that e iθ = cos θ + i sin θ. Some of you may have seen this derivation before you arrived at Oxford. It begs rather a lot of questions, though. In what sense can we add infinitely many numbers together? Is it sensible to take a formula that is valid for real numbers and expect it to hold for complex numbers? Certainly that is not always a good idea for example, x 2 1 is a formula that is true for all real numbers x, and yet it fails trivially in C since by definition i 2 = 1. One of the great successes of 19 th century mathematics was the development of answers to awkward questions and criticisms such as these. By the end of the Prelim Analysis courses you will have acquired an understanding of what might, if mathematicians had been unlucky, have gone wrong with the reasoning above and how we can be sure that it leads in fact to perfectly valid results. As a matter of fact, the reasoning sketched above is a little different from Euler s. In some ways he was even more inventive. Starting from the equation (sin z) 2 + (cos z) 2 = 1, 8

9 he took what was at the time the amazing step of introducing 1 and factorising it as (cos z + 1 sin z)(cos z 1 sin z) = 1. This historical matter is of great interest to some of us but is not relevant to the Prelim course, so although I ll say a few more words about it I ll put them into an appendix. Let s take Euler s formula for granted. Then for a complex number z we have z = re iθ where r := z and θ := arg z. Although this formula expresses nothing more than the polar form of z, it does so in a notation that is particularly suggestive and particularly easy to use in calculations. Here is one: Theorem C8. Let z C and let n be a positive integer. Then z has n th roots in C. In fact, if z 0 then there are precisely n distinct complex numbers w such that w n = z. The reason is this. If z = 0 the assertion is trivial, so now suppose that z 0. Write z = re iα where r := z and α := arg z. Then r > 0 so we know (do we? yes, methods for a proof occur quite early in Analysis I) that there is a unique positive real number s (written n r) such that s n = r. Now let β := α/n and let w := se iβ. Then w n = s n (e iβ ) n = s n e i nβ = re iα = z. This proves the existence of one n th root of z. If φ is any one of the n th roots of 1, of which we know (from 9) that there are precisely n, then (φw) n = φ n w n = w n = z : therefore z has at least n distinct n th roots. On the other hand, if u n = z then (u/w) n = u n /w n = z/z = 1, so u/w is an n th root of unity and therefore u = φw, where φ is one of the n th roots of 1: therefore z has precisely n distinct n th roots. Exercise For n := 2, 3, 4, 5 locate the n th roots of 16i on an Argand diagram. 11. Polynomial equations Early in our acquaintance with algebra we learn that if ax 2 + bx + c = 0 and a 0 then x = b ± D 2a where D := b 2 4ac. When we were young, and when mathematics was several centuries more primitive (or innocent?) than it is now, the number D (known as the discriminant of the equation) was expected to be positive. Once a square root i of 1 had entered the scene every quadratic equation with real coefficients turned out to have two roots in C. Well, not quite. Only one root if D = 0. But in that case the equation may be rewritten as a(x + b/2a) 2 = 0 and its single root b/2a may be deemed to be a root of multiplicity 2, so that in this sense it again has two roots. Those were real quadratic equations. We now have more numbers available to serve as coefficients. The algebraic manipulations involved in completing the square to find the classic formula for the roots of a quadratic equation work just as well in C as they do in R. Moreover, by Theorem C7 every non-zero complex number D has precisely 2 distinct square roots. Thus: Theorem C9. Every quadratic equation with complex coefficients has either 2 distinct roots in C or a single root with multiplicity 2 Exercise Find the square roots of i. Now solve the equation z 2 (2 + 2i)z + (7 22i) = 0. By Theorem C7, if c 0 then the polynomial equation x n = c has precisely n roots in C. And if c = 0 then we think of the equation as having 0 as a root of multiplicity n. In this 9

10 sense the equation always has n roots, counting multiplicities. Consider the problem of solving the general polynomial equation of degree n, a 0 x n + a 1 x n a n 2 x 2 + a n 1 x + a n = 0. It is assumed that a 0 0 since otherwise the equation would be of lower degree. It is also now assumed that a i C for 0 i n. Dividing through by a 0 does not alter the roots and so we treat equations f(x) = 0 where f(x) is a so-called monic polynomial, f(x) = x n + c 1 x n c n 2 x 2 + c n 1 x + c n with complex coefficients and leading coefficient equal to 1. The Factor Theorem (an elementary application of the Division Algorithm for polynomials with complex coefficients) states that if u C and f(u) = 0 then there is a polynomial g(x) such that f(x) = (x u)g(x). Applying the rules for multiplying polynomials we see that g(x) is monic and of degree n 1. It follows easily that if f(x) = 0 has distinct roots u 1, u 2,..., u m in C then there exists a monic polynomial g m (x) of degree n m that has complex coefficients and for which f(x) = (x u 1 )(x u 2 ) (x u m )g m (x). Consequently n m and the equation f(x) = 0 of degree n has at most n roots in C. So far so good. Nothing deeper than (advanced) High School mathematics here. This fact, that a polynomial equation of degree n can have at most n roots, was seen, if through a glass darkly, by Thomas Harriot in about 1605, by Albert Girard (1629), by René Descartes (1637), by Isaac Newton (probably about 1667 but not published until 1707), and by many other mathematicians in clearer and clearer form as the years went by. The belief that a polynomial equation of degree n does have n complex roots if multiplicities are taken into account came fifty years later. It gradually grew from about the middle of the 17 th century. Its modern form could not properly be formulated until complex numbers were introduced in the mid-18 th century and became reasonably widely accepted early in the 19 th century. Nevertheless, it was understood quite a long time before that as the assertion (A): any polynomial with real coefficients may be expressed as a product of linear and quadratic factors (with real coefficients). In this form there were many attempted proofs, none of them sufficiently convincing. The following result, however, with a remarkable number of reasonably convincing proofs C. F. Gauss, for example, published his first in 1799 and his fourth in 1849, fifty years later) is now a little over 200 years old: Theorem C10. When multiplicities are taken into account every complex polynomial of degree n has exactly n roots. That is, if f(x) is a monic polynomial x n +c 1 x n 1 + +c n 1 x+c n with coefficients in C then there exist u 1, u 2,..., u n C (not necessarily distinct) such that f(x) = (x u 1 )(x u 2 ) (x u n ). Although this is called The Fundamental Theorem of Algebra its natural placement within mathematics lies in Analysis or Topology. Its proofs require ideas that lie a long way beyond school mathematics, and somewhat beyond first-year university courses. A For detail and sources see for example Jacqueline Stedall, Mathematics emerging: a Sourcebook , OUP Oxford 2008, Chapter 12, or Jacqueline Stedall, From Cardano s great art to Lagrange s reflections: filling a gap in the history of algebra. European Mathematical Society, Zürich

11 very attractive proof will emerge from theorems proved in the Oxford second-year Complex Analysis course. You may come across other proofs in courses on Topology. For your first year at Oxford, however, we want you to take it on trust. That is, we want you to appreciate what it says and believe it (provisionally, until you are in a position to understand a proof), but use it only when you cannot make progress without it. The key to the connection between assertion (A) and the Fundamental Theorem of algebra is the following elementary fact. Theorem C11. Let f(x) be a polynomial with real coefficients. If u C and f(u) = 0 then also f(u) = 0. The reason is this. Write f(x) = a 0 x n + a 1 x n a n 2 x 2 + a n 1 x + a n where a m R for 0 m n. Using Theorem C1 and induction on the number of summands we find that f(u) = a 0 u n + a 1 u n a n 2 u 2 + a n 1 u + a n, By assumption, however, a m = a m for 0 m n and so f(u) = f(u). Thus f(u) = 0 if and only if f(u) = 0. What Theorem C11 means is that the non-real roots of a polynomial with real coefficients occur in complex conjugate pairs. Now if u C and u / R, then (x u)(x u) is the real quadratic polynomial x 2 2Re(u)x+ u 2. It follows from the Fundamental Theorem of Algebra that assertion (A) is true. As was mentioned above, this fact was appreciated (if not proved) in ever clearer form from about the middle of the 17 th century. By some time in the 18 th century (I do not know quite when) it was also realised that Theorem C10 is actually equivalent to assertion (A) even although the former appears at first sight to be much stronger. Exercise Let (A) be the assertion that any polynomial with real coefficients may be expressed as a product of linear and quadratic factors (with real coefficients). Let (B) be the assertion that any polynomial with complex coefficients may be expressed as a product of linear factors (with complex coefficients). Write down straightforward proofs that if (A) is true then (B) must also be true, and that if (B) is true then (A) must also be true. Exercise Show that σ 8 = 1 if and only if σ is a primitive 16 th root of 1. Find the roots of the equation x 8 = 1, and use them to write x as the product of four quadratic factors with real coefficients. Acknowledgements: I am grateful to Professor Francis Kirwan and her predecessors as lecturers on Complex Numbers for many ideas, for the majority of the questions in the exercise sheet, and for the suggestion that some historical comments should be included. I thank my wife Sylvia for her critical reading and careful proofreading of early drafts of these notes and of the accompanying exercise material. 11

12 Appendix: Notes on Euler s formula (1748) Note 1. Leonhard Euler (b in Basel, Switzerland; d in St Petersburg, Russia) was a wonderfully inventive and prolific mathematician, and a great teacher. There is a commonly held belief that the equation e i π = 1 is something that Euler was proud of as he might well have been, except that he does not seem to have noticed it. The general formula e i θ = cos θ + i sin θ was a discovery that he published in 1748 but he never focussed on the special case θ = π. ( Note 2. Euler s argument starts from the description of e z as 1 + z ) i where i is an i infinite magnitude in present times this is written e z = lim (1 + z ) n. From this he moved n n neatly to power series for e z, cos z and sin z, but his argument is nevertheless not quite what I have sketched in Section 10 above. These images come from the bottom of p. 103 and the top of p. 104 of Euler s Introductio in analysin infinitorum (Lausanne, 1748). Although the language is Latin and the typography is different from modern usage, at the end of 138 the formulae e +v 1 = cos v + 1. sin v and e v 1 = cos v 1. sin v can easily be recognised. ΠMN: Queen s: 30.ix