Analysis Mtha5001y (2015–16)

Home , Absolute convergence, Function series

ANALYSIS MTHA5001Y (2015–16)

David Asper´o University of East Anglia

Contents

1 The Complex Plane 3 1.1 Polar coordinates ...... 8

2 Metric spaces and continuity 9 2.1 Continuity ...... 11

3 Open sets, topologies, uniform continuity, and convergence 13 3.1 Small interlude on linear orders ...... 21

4 Sequences 22 4.1 Cauchy sequences ...... 26

5 Series 27 5.1 Absoluteconvergence...... 29

6 Sequences and series of functions 30 6.1 Seriesoffunctions...... 32 6.2 Power series ...... 33

1 7 Holomorphic functions 37 7.1 The Cauchy–Riemann equations ...... 41 7.2 Harmonicfunctions...... 45 7.3 Somepopularfunctions...... 47

8 Functions as Transformations 52

9 The reciprocal function 53

10 Fractional linear transformations 55 10.1 The extended plane ...... 56 10.2 M¨obiustransformations ...... 56

11 Conformal mappings 59

12 Curves, paths, and connectedness 63

OVERVIEW

1. AUTUMN SEMESTER. We will lay the basics of complex analysis. The main focus will be the notions of series of complex functions and of holomorphic function. The following are the main topics to be covered.

The complex plane; the modulus as a measure of distance. • Basic topological notions. • Continuity. • Sequences of functions. Uniform and point–wise convergence for sequences of • functions. Power series representing real and complex functions. Radius of convergence. • Di↵erentiability of complex functions. Holomorphic functions. • Cauchy–Riemann equations. • Elementary functions. • Fractional linear transformations and conformal functions. • 2. SPRING SEMESTER. We will focus on the theory of complex integration along paths. This theory will contain the main highlights of the module. The following is a list of topics likely to be covered then (changes may occur).

2 More topology of the complex plane. • Review of di↵erentiability of complex functions, holomorphic functions, the • Cauchy–Riemann equations and elementary functions deﬁned by power series. Paths, contours, connectedness and simple-connectedness. • Integration along a path, the Estimation Theorem, integration of power series, • the Fundamental Theorem. Cauchy’s Theorem, Cauchy’s Integral Formulae, Taylor’s Theorem, Liouville’s • Theorem, the Identity Theorem, Laurent expansions. Singularities, residues and Cauchy’s residue theorem, techniques for ﬁnding • residues, summation of series and other applications.

1 The Complex Plane

This material is familiar but we shall take this opportunity to review it and set the basis for the rest of the course. Our starting point will be the problem of ﬁnding solutions of polynomial equations with real coecients, i.e., solutions of an expression of the form

n anx + ...+ a1x + a0 =0 where n is some integer and ai R for all i. 2 Linear equations (when n =1)ofcourseareeasyandhavealwaysexactlyonesolution when a =0. 1 6 Quadratic equations are a bit trickier: The solutions of

x2 + px + q =0

2 2 are given by x = p + p q and x = p p q,providedtheseexpressions 0 2 4 1 2 4 p2 make sense. In other words,q these solutions will existq (in R)ifandonlyif q 0. 4 The simplest quadratic equation without a real solution is

x2 +1=0

Is this the end of the story? It won’t be for us. For various reasons mathematicians have found it desirable to work with numbers which, “if they existed,” would satisfy equations like x2 +1=0.1

1It would be perhaps more accurate to say that mathematicians were naturally driven to work with ﬁelds containing all real numbers and square roots of negative real numbers.

3 We will stipulate the existence of a new number, i,theimaginaryunit,2 which satisﬁes

i2 = 1 Of course, i is not in R.Wewouldliketobeabletousethisnumberincombination with the usual real numbers. In other words, we would like to have a number system C which

(1) contains all the real numbers together with this new number i.

This number system should come endowed with new arithmetical operations of addition, ,andmultiplication, , such that (2) when restricted to the real numbers, and are exactly the usual operations, let us call them + and ,onR. · Moreover,

(3) and should have all the usual properties that we are used to from working with the real numbers (they should be associative, commutative, and so on).

Putting (1), (2) and (3) together we have, in particular, that if a, b are reals, then a (i b) should be a number in our system, and that for all reals a1, a2, b1, b2 the following should hold:

(a1 (i b1)) (a2 (i b2)) = ((a1 + a2) (i b1)) (i b2) =(a + a ) (i (b + b )) 1 2 1 2 (a1 (i b1)) (a2 (i b2)) = a1a2 (((i i) b1b2) (i (a1b2 + a2b1))) =(a a b b) (i (ab + a b )) 1 2 1 2 1 2 2 1 where the last equation should hold since i i = 1. By now we have quite a lot of information on how our intended number system should behave. Of course, from now on we will write + to represent both of + and ,and to represent both of and ,evenifthesearecompletelydi↵erentoperations.Wewill · also adopt the customary· abbreviations in use when doing arithmetic with the reals; in particular, we will remove parentheses in certain cases (as for example when we write a + b c, which doesn’t really make sense, in order to denote a +(b c)). We will also tend to· suppress the symbol (and thus will tend to write ab instead· of a b), and will write a2 to denote a a,etc.Thisway,wemaywritetheaboveequationsasfollows:· · · (a1 + ib1)+(a2 + ib2)= (a1 + a2)+ib1 + ib2 =(a1 + a2)+i(b1 + b2)

2This name is very misleading. i is not any more ‘imaginary’ than, say, ⇡, p2 or, for that matter, 1.

4 2 (a1 + ib1) (a2 + ib2)=a1a2 +(i b1b2 + i(a1b2 + a2b1)) · =(a a b b )+i(a b + a b ) 1 2 1 2 1 2 2 1 Our next move is to stipulate that our new numbers, which we are going to call ‘complex numbers’, are exactly the numbers of the above form (in other words, that they are expressions of the form a + ib,wherea and b are reals and i is some new object), and that the operations satisfy exactly the above requirements. Noting that we can unambiguously identify an expression like a + ib,wherea, b R,withtheordered 2 pair (a, b) R2 prompts us to make the following deﬁnition. 2

Deﬁnition 1.1 The complex plane, C, is R2 endowed with the following operations + and : ·

1. (a1,b1)+(a2,b2)=(a1 + a2,b1 + b2) 2. (a ,b ) (a ,b )=(a a b b ,a b + a b ) 1 1 · 2 2 1 2 1 2 1 2 2 1 where all expressions ai + aj, a1a2 b1b2, aibj refer of course to the usual addition, subtraction and multiplication in the reals.

We will of course identify a complex number (a, b)inC (which is nothing but the real plane R2)withtheexpressiona + ib. Addition and multiplication of complex numbers then satisfy the following.

(a1 + ib1)+(a2 + ib2)=(a1 + a2)+i(b1 + b2)(1)

(a + ib )(a + ib )=a a b b + i(a b + a b )(2) 1 1 2 2 1 2 1 2 1 2 2 1 where a1, a2, b1 and b2 are reals, and where the expressions a1 + a2, a1a2 and so on of course refer to the usual operations on the reals. Throughout the notes, if I write something like a + ib Iwillusuallyunderstand implicitly that a and b are reals. On the other hand, in an expression like u + iv, u and v will be typically complex numbers but not necessarily reals. If necessary, I will make this precise in the relevant context, though. I will also often use letters like z, w to denote complex numbers. We identify x R with (x, 0) (or, what amounts to the same thing, with x + i0), so 2 R C. With this identiﬁcation, R is also called the real axis. Also, the line ia : a R is called✓ the imaginary axis. You can naturally visualize the real axis and the{ imaginary2 } axis as being orthogonal to each other and meeting at exactly 0 (0 is also called the origin).

5 Notation 1.2 Given z = a + ib C with a and b reals, we call a the real part of z, and denote it by Re(z), and call b the2 imaginary part of z, and denote it by Im(z).

For example, Re(i)=Re( i)=0,Im(i)=1,andIm( i)= 1. It is easy to check – which we are not doing here – that (C, +, )isacommutative ring with identity,whichmeansthatitisanicelybehavedalgebraicstructure,inthe· sense that the usual arithmetical computations that we are used to from working with the reals can be performed in it. In particular, given any number a + ib there is another number, namely a ib,suchthat (a + ib)+( a ib)=0 (in other words, a ib is the additive inverse of a + ib). Note that a ib is obtained by reﬂecting a +ib with respect to the origin. The existence of additive inverse means that C comes also with a natural operation of subtraction that behaves exactly in the way you would expect. Also, given a + ib =0,3 we have that 6 1 (a + ib) (a ib)=1 (3) · a2 + b2 (in other words, 1 (a ib), which of course we may write also as a2+b2 a ib , a2 + b2 is the multiplicative inverse of a + ib). The existence of these multiplicative inverses means that (C, +, )is,notonlyacommutativeringwithidentity,butinfactaﬁeld.In · other words, (C, +, )comesalsoequippedwithdivision.4 ·

Notation 1.3 Let z = a + ib C, with a and b reals. 2 1. The complex number a ib is called the conjugate of z and is denoted by z. 2. The real number pa2 + b2 is called the modulus of z and is denoted by z . | | Note that the conjugate of z is obtained by reﬂecting z it along the real axis. Also, the modulus of z is a measure of “how far” z is from the origin (we will make this more precise in a while).

3And of course a + ib = 0 if and only if at least one of a, b is not 0. 6 4However, unlike the reals, (C, +, ) is not an ordered ﬁeld, i.e., there is no order relation on C · compatible with the operations. In fact, we can naturally picture C as a plane but not as a line. We will see a proof of this later on.

6 From equation 3 it follows that if z = 0 is a complex number, then its multiplicative 6 1 inverse, which of course we can denote z ,isgivenby 1 z = (4) z z 2 | | 1 Thus, z is obtained by reflecting z along the real axis, and then multiplying the 1 resulting complex number by the real number z 2 . In particular it follows that if z is 1 | | 1 very far from 0, then z is very close to 0 and, vice versa, if z is very close to 0, then z is 1 very far from 0. Later on we will take a closer look at the function sending z to z . In general, dividing complex numbers needs a trick, which relies on the fact that (a + ib)(a ib)=a2 iab + iba i2b2 = a2 + b2.Fromthiswegetthat (a + ib) (a + ib)(c id) (ac + bd)+i(bc ad) = = . (c + id) (c + id)(c id) c2 + d2 It is remarkable that the addition of i to R and closure under multiplication and addition lets us not only solve the equation x2 +1=0(whosesolutionsarei and i), but in fact every polynomial equation: Theorem 1.4 (Fundamental theorem of algebra) Any equation n anx + ...+ a1x + a0 =0, where n is some integer and ai C for all i, has some solution in C. 2 Aconsequenceoftheabovetheoremisthateverypolynomialp(x)withcomplex coecients can be factored as p(x)=(x ↵ )n1 ... (x ↵ )nm 1 · · m for ↵i C and integers ni. 2 Theorem 1.4, despite its pompous name, is not that fundamental in algebra, but it does say something remarkable; that (C, +, )isanalgebraically closed field.Afieldis · algebraically closed if every polynomial equation has some solution in it. Q and R are certainly not algebraically closed. C is by no means the only algebraically closed field. In fact the algebraic closure Q⇤ of Q is an algebraically closed field contained in C and much smaller than C.5 5 5 3 p2, i, p7+p11 + i, and so on, belong all to Q⇤. On the other hand, many interesting numbers, such as ⇡ or e, do not belong to Q⇤.ThenumbersinQ⇤ are called algebraic numbers, and the numbers in C which are not in Q⇤ are called transcendental numbers. Surprisingly, it is not known whether or not e + ⇡ is algebraic and whether or not e⇡ is algebraic (although it is expected that none of them is). It is not even known whether or not e + ⇡ is rational, and whether or not e⇡ is rational (!). On the other hand it is easy to see that at least one of e + ⇡, e⇡ is transcendental. The proof is as follows: Consider the polynomial p(x)=(x e)(x ⇡)=x2 2(e + ⇡)x + e⇡.Ife + ⇡ is algebraic, then so is 2(e + ⇡)sinceQ⇤ is a field and therefore closed under the operations of arithmetic. Hence, if e⇡ were also algebraic, then p(x) would have all its coecients in Q⇤.ButQ⇤ is algebraically closed. Therefore, all of its roots, namely e and ⇡, would be in Q⇤. But it is known, by deep theorems in analysis, that e and ⇡ are in fact both transcendental.

7 There are proofs of of Theorem 1.4 which should be accessible to you in a little while, and it would be a good idea to look at them later this semester.6 In any case a proof will be given in the second semester.

Exercise 1.5 Prove that the following is true for all z, w C. 2 1 z¯ Re z =(z +¯z)/2, Im z =(z z¯)/(2i), z 2 = zz,¯ = , | | z z 2 | | z w =¯z w,¯ zw =¯zw,¯ z/w =¯z/w,¯ z + w z + w . ± ± | || | | |

1.1 Polar coordinates

We can represent the number z = a + ib on the plane using its coordinates (a, b), but we can also use polar coordinates. Namely we can represent z uniquely as a pair z , ✓ where z is the distance from z to the origin, and ✓ is the angle between the positive| | part of| the| x-axis and the line connecting z with the origin. In other words, and using elementary trigonometry, the number z just described is the number z (cos(✓)+i sin(✓)). | |

Notation:Iwilltemporarilyusethenotationcis(✓)todenotethenumbercos(✓)+ i sin(✓).7

Using the above notation, the number z Iamtalkingaboutis z cis(✓). | | The angle ✓ is certainly not unique; in fact, the map ✓ (cos ✓, sin ✓)fromR into 7! the unit circle in R2 is 2⇡-periodic. On the other hand, if z =0,thenthereisaunique ✓ ( ⇡,⇡]suchthatz = z cis(✓). This ✓ is called the principal6 argument of z,butwe will2 sometimes refer to it simply| | as the argument of z.8 We will denote this number also by arg(z). Recall now the formulas relating the cosine and sine of an angle ↵ + to those of ↵ and

cos(↵ + )=cos↵ cos sin ↵ sin (5)

sin(↵ + ) = sin ↵ cos +sin cos ↵ (6)

The following proposition is an immediate consequence of the above equations.

6See for example “The fundamental theorem of algebra: an elementary and direct proof”, by Oswaldo Oliveira. 7Later on we will prove that there is a certain ‘exponentiation function’ ez, deﬁned for all z C, and 2 such that ei✓ =cis(✓) for every ✓ R. Here e denotes the familiar Euler’s constant from real analysis. 8Of course, for z = 0, any ✓ is2 such that z = z cis(✓). | |

8 Proposition 1.6 Given complex numbers z1 = r1 cis(✓1) and z2 = r2 cis(✓2),

z z = r r cis(✓ + ✓ ) 1 · 2 1 2 1 2 The following are some immediate consequences of the above proposition, some of them quite useful or illuminating.

For all z , z ,arg(z z )=arg(z )+arg(z ). • 1 2 1 2 1 2 For every z C, z z¯ = z 2. • 2 · | | Given any number z, iz is the result of rotating z anti-clockwise around the origin • ⇡ by 2 . For all reals r 0, ✓,andeveryintegern 1, (r cis(✓))n = rn cis(n✓). In particular, • (cos(✓)+i sin(✓))n =cos(n✓)+i sin(n✓)(7)

which is the so–called de Moivre’s formula for the case ✓ R. 2 Given any z =0andanyintegern 1, z has exactly nn-th roots. If z = r cis(✓), • these roots are6 ✓ +2⇡k z = pn r cis( ) k n for k =0, 1,...n 1(notethattheyarealldi↵erentandthatz can have at most nn-th roots). In other words, the polynomial xn z can be written xn z = (x z ) k k

Exercise 1.7 Given b C, b =0, show that the line with origin 0 and direction b is 2 6 given by z C Im( z )=0 . { 2 | b }

2 Metric spaces and continuity

We will use the formal notion of distance (or metric) in order, among other things, to make sense of the notion of continuity of functions, which of course is one of the main objects of study in analysis.

9 We all have an informal idea of the general notion of distance between two points. Moreover, we are all familiar with certain speciﬁc examples, coming from mathematics, of “distances”. Take for example the real numbers with the “distance” given by a b , where x represents the absolute value of x:Thedistancebetweentworealsa and| b is| the absolute| | value of a b. Another very classical example of “distance” is the Euclidean distance (or metric) in R2,givenby

d((a ,b ), (a ,b )) = (a a )2 +(b b )2 1 1 2 2 1 2 1 2 p What about the complex numbers? The modulus z of a complex number z can be seen as a measure of the distance from z to 0. This intuition| | provides us with the right tool to deﬁne yet another notion of “distance”, this time between complex numbers: Given z1, z2 in C, the distance between z1 and z2 is given by ⇢(z ,z )= z z 1 2 | 1 2|

If z1 = a1 + ib1 and z2 = a2 + ib2,theaboveformulagives

⇢(z ,z )= (a a )2 +(b b )2 1 2 1 2 1 2 Perhaps not surprisingly, this is just thep Euclidean metric on C after identifying C with R2 (by identifying a + ib with the pair (a, b)). The following deﬁnition of metric tries to isolate the main ingredients in the intuitive idea of distance and generalises well–known properties of the Euclidean metric on R2.

Deﬁnition 2.1 Let X be a set and d : X X R a function. d is a metric if and only if the following holds for all x, y, z in⇥X. !

(1) d(x, y) 0 (2) d(x, y)=0if and only if x = y.

(3) d(x, y)=d(y, x)

(4) d(x, z) d(x, y)+d(y, z) (triangle inequality)  A pair (X, d) is a metric space if d is a metric on X.

It is important to note that metrics are functions with values in R,andnotin(say) C.OneimportantreasonforthisisthatinR there is a natural notion of order, whereas in C there isn’t. And we certainly want to be able to compare distances.9 Here are some examples of metric spaces (check that this is true):

9We want to be able to say that the distance between these two points is bigger or smaller than the distance between those other two points.

10 (R,d), where d(x, y)= x y and denotes absolute value. • | | || R2 with the usual (Euclidean) metric. • For every set X and for all x, y X,letd(x, y)=0ifx = y and let d(x, y)=1if • x = y.Thisd is called the discrete2 metric on X.Then(X, d)isametricspace. 6 1 1 1 R with the metric l given by l ((a1,b1), (a2,b2)) = a1 a2 + b1 b2 . l is called • | | | | the Manhattan (or taxicab) metric,and(R,l1)isametricspace.

Note for example that if d is the Euclidean metric, then l1((0, 0), (1, 1)) = 2 whereas d((0, 0), (1, 1)) = p2.

Here is a more sophisticated example of metric space: Let a

b d(f,g)= f(x) g(x) dx | | Za It turns out – and it is not dicult to prove – that (X, d)isametricspace(itis instructive to draw a picture to get an idea of what this metric is like).10

2.1 Continuity

Consider the following functions f, g, h and i deﬁned on R and with range contained in R.Forallx R: 2 f(x)=x2 • g(x)= 1ifx 0andg(x)=x2 if 0

Continuity of a (real) function at a point x0 is sometimes meant to encapsulate the informal idea of a function being ‘smooth’ around x0,inthesensethatonecandraw its graph, at least for values of the domain “close to x0” without lifting the chalk from 10Things like this will certainly not be in the exam, but it is nevertheless a good idea to think about them.

11 the board or the pencil from the paper. A description of continuity at x0 which doesn’t mention chalk or pencils and will be easier to formalise is the following: A function f is continuous at x0 if points that are close to x0 are mapped to points that are close to f(x0)or,moreprecisely,ifforeverypointx (in a suciently small neighbourhood of x0), the distance between f(x)andf(x0)isnot“toobig”compared to the distance between x and x0.ThismeansthatifIhandyouapositiverealnumber✏,then,nomatterhow small ✏ is, you should be able to find a positive real number so that for every x in the domain of f,ifx is at a distance from x0 less than ,thenx is mapped to a point at adistanceoff(x0)lessthan✏. Note that now that we are equipped with the notion of metric, we can make perfect (mathematical) sense of the above description. Our ocial definition of continuity will express exactly the situation described above. Now to the ocial definition.

Definition 2.2 Let (X, d), (Y,⇢) be metric spaces, let f : X Y be a function and ! let x0 X. f is continuous at x0 i↵for every ✏ R, ✏>0, there is some R, >0, such that2 for all x X,IFd(x, x ) <, THEN 2⇢(f(x),f(x )) <✏. 2 2 0 0 We say that f is continuous if f is continuos at every x X. 0 2 At this point it is perhaps worth remarking that perhaps we didn’t make such a great job formalising our initial intuition of continuity by means of the above definition: If you look at the function in the fourth example – which is in fact continuous at x =1 according to the Definition 2.2 with respect to the usual metric on R11 –, you will realise that it is by no means ‘smooth’ around x =1.Soperhapstheformalideaofcontinuityis not so well–suited after all to express the intuitive idea of continuity. In fact, the stronger notion of di↵erentiability will work better for that. On the other hand, continuity, even if it is a too general concept for some purposes, is also an extremely powerful notion which you will come across all the time, one of the reasons being that it is applicable in contexts where the notion of being di↵erentiable doesn’t even make sense. There are even more dramatic examples showing how badly the general notion of continuity captures the idea of ‘smoothness’ in general: Let a be a real such that 0 < 3 1 a<1, b an odd integer, and suppose ab > 1+ 2 ⇡ (for example a = 2 and b =1000). Consider now the following function f, defined by K. Weierstrass in 1872:

f(x)= an cos(bn⇡x) n 0 X It turns out that, for this choice of a, b,thefunctionf is continuous everywhere but is not di↵erentiable in any point (!)

11Try to prove that.

12 Example 2.3 The function g given at the beginning of this section is not continuous at 1 x =0(where of course we endow R with the usual metric). To see this take ✏ = 2 > 0 and note that, for every >0 there are y R, y✏. | | | | | | Example 2.4 If (X, d) and (Y,⇢) are metric spaces and d is the discrete metric, then every function f : X Y is continuous. In fact, given any x X and any ✏>0, 1 ! 2 check that = 2 does the job. [Verify this.]

3 Open sets, topologies, uniform continuity, and convergence

The notion of continuity can be deﬁned in a yet more general way, not involving " and .Tounderstandthis,letustalkaboutopensets. Abasicnotioninametricspaceisthenotionofopen ball.Let(X, d)beametric space and a X. An open ball around a is any set of the form 2 Bd(a)= x X : d(x, a) <" , " { 2 } where ✏ is a positive real number. Such an open ball is said to be centred in x and to be of radius ".Intuitively,inametricspaceasetO is open if ‘it does not have an edge’, namely for every x O there is an open ball centred in x which is entirely contained in O.Inotherwords:2

Deﬁnition 3.1 A subset O of a metric space (X, d) is open if for any x O there is ">0 such that Bd(x) O. 2 " ✓ d We often write B✏(x)forB" (x)whend is understood. Examples of open sets in R,withtheusualmetric,areopenintervalsandunionsof open intervals. In particular, R itself is open. If X is any set and d is the discrete metric on X,thenany subset of X is open.

Observation 3.2 In C with the usual metric (i.e., the Euclidean metric), open balls are open disks. On the other hand, if we look at C with the metric l1, then an open ball is the interior of a square.12 Also, note that every open disk is the union of some (inﬁnite) collection of interiors of squares, and that the interior of every square is the union of an (inﬁnite) collection of open disks. In other words, every open ball with respect to the Euclidean metric is a union of open balls with respect to l1, and every open ball with respect to l1 is a union of open balls with respect to the Euclidean metric.

12For example, the open l1–ball centred in 0 of radius 1 is the interior of the square with vertices (1, 0), (1, 1), ( 1, 0), ( 1, 1).

13 The deﬁnition of continuity can be given in terms of open sets:

Definition 3.3 Suppose that (X, d) and (Y,⇢) are metric spaces and f : X Y is a ! function. f is continuous at x0 X i↵for every open set O in Y with f(x0) O, there is an open set U in X containing2 x such that f(U) O. 2 0 ✓ We say that f is continuous if it is continuous at every x X. 0 2 This definition is useful because it gives us a geometric intuition of continuous functions (say, functions f : C C), even when we are not able to graph such functions. ! So now we have two definitions of continuity, which doesn’t look like a great idea.

Proposition 3.4 Deﬁnitions 2.2 and 3.3 are equivalent.

It will be useful and natural at this point to deﬁne the notions of topology and topological space. The notion of topological space is ubiquitous in mathematics. For example, many facts in analysis are instances of much more general topological facts. In general, topology will provide us with a very useful language to do analysis.13

Deﬁnition 3.5 Let X be a set and let ⌧ be a collection of subsets of X. We say that ⌧ is a topology on X if it satisﬁes the following:

(1) ⌧ and X ⌧. ;2 2 (2) For every n N and for all U1,...,Un ⌧, U1 ... Un ⌧ (⌧ is closed under ﬁnite intersections).2 2 \ \ 2

(3) For every family Ui i I ⌧, i I Ui ⌧ (⌧ is closed under arbitrary unions).14 { | 2 }✓ 2 2 S (X,⌧) is a topological space if ⌧ is a topology on X.

Our most prominent example of topological will be the following:

Example 3.6 If (X, d) is a metric space, then

⌧ = U X U open with respect to the metric d d { ✓ | } is a topology on X. [Verify that this is true.]

13Also, even if you don’t like generality and care mostly only about, say, R or C, you should bare in mind that topology is an extremely rich source of notions and proof methods that, when applied to R and C, yield a very interesting and non-trivial theory (Lebesgue measure, category, etc.) that simply wouldn’t exist if we didn’t have topology in the background. 14 By i I Ui we mean of course the set x : there is some i I such that x Ui . 2 { 2 2 } S 14 Deﬁnition 3.7 Given a topological space (X,⌧), elements of ⌧ are also called open sets. Also, a subset Y of X is closed if and only if X Y is open. \ Remark 3.8 Note that and X are always open and closed. ; Also, note that when (X,⌧)is,forexample,C with the usual topology (i.e., the topology induced by the Euclidean metric), then most subsets of X are neither open not closed! One set which is neither open nor closed is Q Q,viewedasasubsetofC. Another set ⇥ which is neither open nor closed is z C : z 1 0 . { 2 | | }\{ } AtopologicalspaceX is said to be Hausdor↵ i↵for all a, b X,ifa = b,thenthere are open sets U, V such that a U, b V and U V = . 2 6 2 2 \ ; Proposition 3.9 Every metric space is Hausdor↵.

Proof. Let (X, d)beametricspace,a, b X, a = b.Let✏ = d(a, b) > 0. Then 2 6 U = B ✏ (a)andV = B ✏ (b)areopensetscontaininga, b,resp.,suchthatU V = . 2 2 ✏ ✏ \ ; To see this, suppose c U V .Butthend(a, b) d(a, c)+d(c, b) < 2 + 2 = ✏,i.e., d(a, b) <✏,whichisacontradiction.2 \ 

Remark 3.10 Di↵erent metrics can induce the same topology. For example consider the metrics l1 and the Euclidean metric on C (let us call it d). These metrics are di↵erent (for example, d(0, 1+i)=p2 whereas l1(0, 1+i)=2). On the other hand, by 1 Observation 3.2, if ⌧l1 and ⌧d are the topologies on C induced by l and d, respectively, then ⌧l1 = ⌧d, since any open set in either topology ⌧ is a union of open sets in the other topology ⌧ 0, and therefore belongs to ⌧ 0.

Let us make yet another deﬁnition of continuous function, this time for topological spaces in general.

Deﬁnition 3.11 Let (X,⌧ ) and (Y,⌧ ) be topological spaces, let f : X Y be a func- 1 2 ! tion, and let x0 X. We say that f is continuous at x0 if for every open set V ⌧2, if f(x ) V , then2 there is an open set U ⌧ such that x U and f[U] V . 2 0 2 2 1 0 2 ✓

We say that f is continuous if and only if f is continuous at every x X. 0 2 Fortunately this deﬁnition coincides with our previous two deﬁnitions in the case of metric spaces:

Proposition 3.12 Let (X, d) and (Y,⇢) be metric spaces, let f : X Y be a function, ! and let x0 X. Let ⌧d and ⌧⇢ be the topologies on X and Y induced by d and ⇢, respectively.2 The following are equivalent:

15 (1) f is continuous at x0 with respect to the metrics d and ⇢ (i.e., in the sense of Deﬁnition 3.3).

(2) f is continuous at x0 with respect to the topologies ⌧d and ⌧⇢ (i.e., in the sense of Deﬁnition 3.11).

Proof. For this proof, given , ✏ R, z X,andw Y ,let 2 2 2 Bd(z)= x X d(x, z) < { 2 | } and B⇢(w)= y Y ⇢(y, w) <✏ ✏ { 2 | }

Let us show first that (1) implies (2): Suppose f is continuous at x0 with respect to the metrics d and ⇢.LetV be an open set in ⌧ such that f(x ) V .Bythedefinition ⇢ 0 2 of topology induced by ⇢ and since f(x0) V ,thereissome¯y Y and some positive ⇢ ⇢ 2 2 ✏ R such that f(x0) B✏ (¯y)andB✏ (¯y) V . Note that ✏0 := ✏ ⇢(f(x0), y¯) > 0since 2 2⇢ ⇢ ✓ ⇢ ⇢(f(x0), y¯) <✏.ButB (f(x0)) B (¯y)sinceforeveryy B (f(x0)), ✏0 ✓ ✏ 2 ✏0 ⇢(y, y¯) ⇢(y, f(x )) + ⇢(f(x ), y¯) < (✏ ⇢(f(x ), y¯)) + ⇢(f(x ), y¯)=✏,  0 0 0 0 where the first inequality is true by the triangle inequality for ⇢ [draw a picture]. Since f is continuous at x0 with respect to the metrics d and ⇢,wehavethatthereissome d ⇢ d positive R such that f[B (x0)] B (f(x0)). But B (x0)isanopensetin⌧d and 2 ✓ ✏0 contains x0,sowearedone.

(2) implies (1): Suppose f is continuous with respect to the topologies ⌧d and ⌧⇢.Let d ⇢ ✏ R be positive. We want to ﬁnd a positive R such that f[B (x0)] B✏ (f(x0)). 2 ⇢ 2 ✓ For this, note that B✏ (f(x0)) is an open set in ⌧⇢ containing f(x0). Hence, since f is continuous at x0 in the topological sense, there is some open set U in ⌧d containing x0 ⇢ such that f[U] B✏ (f(x0)). Since U is open in ⌧d and x0 U,thereissomeopenball d ✓ d d 2 B (¯x)suchthatx0 B (¯x)andB (¯x) U.Let0 = d(¯x, x0)annotethat0 > 0. d 2 d ✓ d But then B (x0) B (¯x)sinceforeveryx B (x0), 0 ✓ 2 0 d(x, x¯) d(x, x )+d(x , x¯) < ( d(x , x¯)) + d(x , x¯)=,  0 0 0 0 where the ﬁrst inequality is true by the triangle inequality for d [draw another picture]. This gives us what we want, since

d d ⇢ f[B (x0)] f[B (¯x)] f[U] B (f(x0)) 0 ✓ ✓ ✓ ✏ ⇤

There is a connection between the continuity of real functions and the continuity of complex functions. Let us start with an example:

16 Example 3.13 Let f(z)=z2 5. Then f is a complex function which can also be expressed as f(x + yi)=x2 +2ixy y2 5=(x2 y2 5) + i(2xy). In general, every complex function f can be expressed in terms of two real functions, in the form f = u + iv, u : C R and v : C R.Intheaboveexample, u(x, y)=x2 y2 5andv(x, y)=2!xy.Thesearecalledthe! components of f. u is the real component and v is the imaginary component. The theory of continuity carries over from R2 to C (as we have seen, R2 and C are the same object from a topological point of view) to obtain that a complex function f is continuous i↵each of its components is continuous. In fact:

Proposition 3.14 Let f : C C be given by ! f(x + iy)=u(x, y)+iv(x, y) where x, y, u(x, y) and v(x, y) are all in R. Let a + ib C. Then the following are equivalent: 2

(1) f is continuous at a + ib.

(2) u : R R R is continuous at (a, b) and v : R R R is continuous at (a, b) ⇥ ! ⇥ ! (where R R is given the usual Euclidean metric and R is given the usual absolute value metric).⇥

This makes it easy to check the continuity of complex functions. Other concepts related to continuity carry over to C as well. An important one is the deﬁnition of uniform continuity:

Definition 3.15 Suppose that (X, d) and (Y,⇢) are metric spaces, f : X Y , and A X. We say that f is uniformly continuous on A if for every ">0 there! is >0 such✓ that for all x A and all x X, if d(x, x ) <, then ⇢(f(x),f(x )) <". 0 2 2 0 0 The di↵erence between continuity and uniform continuity is that in the definition of uniform continuity the value of depends only on ",notonthechoiceofx0. Also, uniform continuity is defined using subsets of the space, rather than single points x0. Try to think of a function that is continuous but not uniformly continuous (hint: make the domain an open interval and see how the epsilons and deltas change as you go close to the edge of the interval). You may recall from the theory of the functions of real variable that any continuous function defined on a closed interval is actually uniformly continuous. The analogue of that in the theory of complex numbers is that any continuous function defined on a closed disk is uniformly continuous.

17 Terminology 3.16 Let (X, d) be a metric space, let r R, r>0, and let x0 X. The closed ball centred in x with radius r is B (x )= x2 X d(x, x ) r . 2 r 0 { 2 | 0  } Note:If(X, d)isametricspace,theneveryclosedballin(X, d)isclosedinthe topological sense for the topology induced by d.[Check.] The following is a natural notion of boundedness for metric spaces.

Deﬁnition 3.17 A subset of a metric space is bounded if and only if it is contained in a closed ball.

For example any interval in R is bounded, while (0, )isnot(withtheusualmetric). 1

Theorem 3.18 Suppose n N and that A is a closed and bounded subset of Rn with respect to the usual (Euclidean)2 metric. Suppose (Y,⇢) is a metric space and f : A Y is continuous. Then f is uniformly continuous on A. !

Warning:Theorem3.18isnot true for metric spaces in general (instead of Rn): Consider for example the function f : Q R sending x Q to 1 if p2

Theorem 3.19 Suppose (X, d) and (Y,⇢) are metric spaces and A X is compact. If f : A Y is continuous, then f is uniformly continuous on A. ✓ !

Theorem 3.20 (Heine–Borel theorem) Let A be a subset of Rn, for some n N. Then the following are equivalent. 2

(1) A is compact.

(2) A is closed and bounded.

In the two theorems above, ‘compact’ refers to the following notion.

18 Definition 3.21 Given a topological space X and A X, A is compact if and only if for any open cover of A, i.e., any collection =✓ U i I of open sets such U U { i | 2 } that A i I Ui, there is a finite subcollection of which still covers A (i.e., there are ✓ 2 U i1,...,im, for some m N, such that A Ui1 ... Uim ). S 2 ✓ [ [ All statements in Proposition 3.22 follows directly from the definitions (try to prove them).

Proposition 3.22 1. Suppose X is a Hausdor↵topological space. Then every compact subset of X is closed.

2. Every closed subset of a compact set is itself compact.

3. If X and Y are topological spaces, A X is compact and f : X Y is a continuous function, then f[A] is a compact✓ subset of Y . !

Proof. (of Theorem 3.19): Let ✏ R,0<✏.Forevery¯x A there is some x¯ such d ⇢ 2 d 2 that f[Bx¯ (¯x)] B✏/2(f(¯x)) (where of course Br (x)denotestheopenballcentrediny ✓ ⇢ of radius r with respect to the metric d and Br (y) denotes the open ball centred in y of radius r with respect to the metric ⇢). Now, = B (¯x) x¯ A is an open cover of U { x¯/2 | 2 } A. Hence, by compactness of A there is a ﬁnite number of pointsx ¯1,...,x¯m in A such that A B /2(¯x1) ... B /2(¯xm) ✓ x¯1 [ [ x¯m Let such that 0 < /2foralli =1,...m,forexample  x¯i =min /2,..., /2 { x¯1 x¯m }

(note that >0since = x¯1 /2,...,x¯m /2 is ﬁnite). Let x, x0 A be such that { d } 2 d(x, x0) <.Leti be such that x B /2(¯xi). Then 2 x¯i

d(x0, x¯ ) d(x0,x)+d(x, x¯ ) <+ /2 /2+ /2= i  i x¯i  x¯i x¯i x¯i and therefore ⇢(f(x0),f(¯xi)) <✏/2. But then

⇢(f(x),f(x0)) ⇢(f(x),f(¯x )) + ⇢((f(¯x ),f(x0)) <✏/2+✏/2=✏  i i This ﬁnishes the proof. ⇤

You can easily ﬁnd proofs of Theorem 3.20 if you are interested. I am not proving this fundamental theorem here, but will nevertheless give a proof sketch to illustrate the crucial use of the completeness of R,inthesensethatyouhaveseenintheﬁrstyear analysis module (in contrast, Q is certainly not complete, which is of course exploited in the counterexample that I have presented right after Theorem 3.18).

19 Proof sketch of Theorem 3.20:ByProposition3.22,ifA is compact then it is closed. It is not dicult to see that it is bounded as well. For the other direction I will only do the case n =1.Sinceeveryclosedsubsetof a compact set is compact by Proposition 3.22, it is enough to prove that every closed interval [a, b]ofR is closed. Let be an open cover of [a, b]. We want to see that there U is a ﬁnite subcollection 0 which also covers [a, b]. Since every member of is a union of open intervals, weU ✓ mayU – and will – assume that every member of is actuallyU an open interval. U Let X = x [a, b] [a, x]iscoveredbyaﬁnite 0 { 2 | U ✓U} Note that a X,soX = ,andthatifx X and y

1. x1 < x¯, 2. y X and z [a, y)impliesz X,and 2 2 2 3.x ¯ is the least upper bound of X, we have that x X.ButthenthereareU ,...,U such that [a, x ] U ... U . 1 2 1 m 2U 1 ✓ 1 [ [ m Then U1,...,Um (y, y0) is a ﬁnite subcollection of that covers [a, x2]andso x X{. But then}x ¯[is{ not an} upper bound of X sincex

Deﬁnition 3.23 Suppose that f is a real function. Then we say

lim f(x)=a x x0 ! if for any ">0 there is >0 such that for all x, 0 < x x < = f(x) a <". | 0| )| |

20 Like in the definition of continuity, this is really a specific case of a definition that can be made in any metric space:

Deﬁnition 3.24 Suppose that (X, d) and (Y,⇢) are metric spaces, f : X Y is a function, and x X. We say that ! 0 2 lim f(x)=a x x0 ! if for any ">0 there is >0 such that for all x, 0

Proposition 3.25 Let f, g : C C be functions, let a, b C and let z0 C. Then ! 2 2

1. If limz z0 f(z)=a and limz z0 g(z)=b, then limz z0 (f(z)+g(z)) = a + b. ! ! !

2. If limz z0 f(z)=a and limz z0 g(z)=b, then limz z0 (f(z)g(z)) = ab. ! ! ! f(z) a 3. If limz z0 f(z)=a, limz z0 g(z)=b and b =0, then limz z0 = . ! ! 6 ! g(z) b

The proofs of (1)–(3) are exactly as in the real case. Note that we are not asking that f and g be continuous. The reason why (1)–(3) are true is that addition, multiplication and division are continuous functions from C C to C. [Try to see this.] ⇥

3.1 Small interlude on linear orders

The following observation ﬁts well in this section.

Recall that a linear order on a set X is a binary relation <⇤ on X such that

1. for all x, y in X, x<⇤ y,ory<⇤ x,orx = y,

2. for no x, y it holds that x<⇤ y and y<⇤ x,and

3. for all x, y, z in X,ifx<⇤ y and y<⇤ z,thenx<⇤ z.

Alinearlyorderedsetcanthusbevisualisedasaline.

21 Example 3.26 The relation < on R deﬁned by x

(4) For all x, y in R, if 0

(5) For all x in R, 0

So the natural order on R “behaves well” with respect to multiplication (it also behaves well with respect to addition). Let us see that there is, on the other hand, no linear order on the complex plane which behaves well with respect to the arithmetical operations.

Proposition 3.27 There is no linear order <⇤ on C with the following properties.

(a) For all x, y in C, if 0 <⇤ x and 0 <⇤ y, then 0 <⇤ xy.

(b) For all x in C, 0 <⇤ x if and only if x<⇤ 0.

Proof. Suppose first that 0 <⇤ i.Then0<⇤ ii = 1by(a),whichimplies0<⇤ ( 1)( 1) = 1 again by (a). But this implies 1 <⇤ 0by(b).Butthenwehave0<⇤ 1 and 1 <⇤ 0whichcontradicts(2)inthedefinitionoflinearorder.By(4)inthe definition of linear order we thus have that i<⇤ 0andtherefore0<⇤ i by (b). But then 0 <⇤ ( i)( i)= 1againby(a),andagainwereachacontradictionasinthe previous case. This contradiction finishes the proof. ⇤

4 Sequences

For us, sequences will be functions with domain N; in other words, things of the form 15 (an)n N.Wewillsometimeswritealso an n1 0. Most of the time we will be looking at 2 { } sequences of complex numbers, i.e., sequences in which each an is in C. Because we have been careful to deﬁne our concepts in the generality of metric spaces, almost everything we say will apply to any metric spaces. In particular they will apply to C and R.Thereforeweshallnotdistinguisheverytimeifourobjectsarecomplexor real.

15 I start counting from 0, but some people start counting from 1. They would write an n1 1.This is basically a matter of notation and makes absolutely no di↵erence in anything we are doing{ } here.

22 Deﬁnition 4.1 A sequence (an)n N is said to converge to the point a if 2 ( ">0)( N)( n N) a a <". 8 9 8 | n | In this case we say that a is the limit of (an)n N and we write limn an = a or limn an = 2 !1 a.

Note: The expressions limn an = a and limz z0 f(z)standforcompletelydi↵erent !1 ! things. To begin with, the ﬁrst expression refers to a sequence (an)n,whereasthesecond expression refers to a function f. Some examples of sequences:

1 (i) The sequence ( )1 n converges to 0, n  n (ii) The sequence (( 1) )n N =(1, 1, 1, 1,...)doesnotconverge. 2 n (iii) The sequence (i )n N does not converge either. 2 1 Let us verify (iii): Set " = 3 .Supposethereissomea and some N N such that for every n N, a a <"= 1 .Inparticular, a a < 1 .Butthen,2 | n | 3 | N | 3 1 1 2 2= a a a a + a a < + = , | N N+2|| N | | N+2| 3 3 3 which is a contradiction. The ﬁrst equality follows since iN iN+2 = iN (1 i2) = iN 1 ( 1) =1 2=2. | | | | | |·| | · ⇤ The use of the word the in the notion of the limit is justiﬁed by the following:

Theorem 4.2 A sequence may converge to at most one limit.

Proof. Suppose for contradiction that an n1 0 converges to both a and b and that a = b. Therefore a b > 0, so let " = a {b /4,} still > 0. By the deﬁnition of convergence6 | | | | there are N,M > 0suchthatforalln N we have an a <"and for all n M we have a b <". Taking n>max(N,M)andusingthetriangleinequalityweobtain:| | | n | a b a a + a b <"+ " =2" =2(a b /4) < a b , | || n| | n | | | | | acontradiction.

Advertisement 4.3 Taking limits of sequences of functions is an extremely useful way of generating new functions, like sin(z), cos(z), ez, and so on, from (sequences of) previously deﬁned functions. Note that with the resources we have seen so far (i.e., the usual arithmetical operations on C) and using just composition of functions, the only functions that you would ‘naturally’ expect to ﬁnd in analysis that we can actually generate at the moment are polynomials and, more generally, ‘rational’ functions (i.e., p(x) functions of the form q(x) , where p(x) and q(x) are polynomials). It could be argued that the study of limits of sequences functions is in fact the most fruitful notion in analysis.

23 The notion of the limit is di↵erent from the following notion of a limit point.

Deﬁnition 4.4 A point a is alimitpointofthe sequence (an)n i↵, for every ">0, the set n N : an a <" { 2 | | } is inﬁnite.

For example, the sequence 1,-1,1,-1, ... has two limit points, namely 1 and -1. There- fore a limit point is not in any sense unique. The connection between the limit and a limit point is given by the following.

Fact 4.5 (1) A convergent sequence has exactly one limit point, namely the limit of the sequence. (2) A point is a limit point of a sequence i↵it is the limit of a subsequence of the sequence.

The proof of part (1) is easy. The proof of part (2) has been done in class and is easy too.

We say that a sequence (xn)n is bounded if its range xn : n N is bounded. { 2 } One of the most beautiful early theorems in analysis is the following theorem on the existence of limit points.

Theorem 4.6 (Bolzano–Weierstrass) Suppose that (an)n is a bounded sequence. Then it has a limit point.

Proof. Let us see two proofs of this theorem. The ﬁrst proof will be more general and streamlined. The second proof will be probably more visual.

Proof 1: We note the following general fact about closed sets:

Fact:ForanytopologicalspaceX,thecollectionofclosedsubsetsofX is closed under arbitrary intersections.

This fact follows immediately from the fact that being a closed set means being the complement of an open set, together with the fact that the collection of open sets is closed under arbitrary unions (by deﬁnition). Indeed, if Fi is closed for each i in some index set I,thenX Fi is open for each i and so i I (X Fi)isopen.But \ 2 \ i I Fi = X ( i I (X Fi)) and is therefore closed. ⇤ 2 \ 2 \ S T Now, from theS above fact it follows that given any set A there is a minimal (under )closedsetF that contains A. This minimal closed set is of course ✓ F F X closed,A F { | ✓ ✓ } \ 24 It is called the closure of A and is denoted by A.

Since a 1 is contained in a bounded closed set F , a 1 F ,andsoinparticular { n}0 { n}0 ✓ a 1 is a closed and bounded set. { n}0 Suppose towards a contradiction that a 1 does not have any limit point. { n}0

Claim 4.7 a 1 = a 1. { n}0 { n}0

Proof. Obviously a 1 a 1. Now suppose a a 1 a 1.Suppose">0 { n}0 ✓ { n}0 2 { n}0 \{ n}0 were such that B (a) a 1 = a ,...a were ﬁnite. Let " \{ n}0 { n1 nm }

"0 =min ", a a ,... a a > 0 { | n1 | | nm |} and note that B" (a) an 01 = .ThenC B✏ (a)wouldbeclosedand an 01 C B✏ (a) 0 \{ } ; \ 0 { } ✓ \ 0 since a 1 is the minimal closed set that includes a 1,soa/ a 1. Contradiction. { n}0 { n}0 2 { n}0 ⇤

Now, since an n N has no limit points, arguing as in the above proof we ﬁnd for { } 2 every n some "n > 0suchthatB"n (an) ai i N = an .Then, \{ } 2 { }

= B"n (an) n N U { | 2 } is an open cover of an n N.Since an n N = an n N is compact by the Heine–Borel { } 2 { } 2 { } 2 theorem, there is a ﬁnite set of indices, n1,...,nm,suchthat

ai i N B"n (an1 ) ... B"n (anm ) { } 2 ✓ 1 [ [ m

This is a contradiction since an / B"n (an1 ) ... B"nm (anm )foranyn = n1,...,nm. ⇤ 2 1 [ [ 6

Proof 2:Letusprovethisonlyintherealcase.Wemayclearlyassumethat ai i N 2 is infinite, or otherwise we are done. { } 0 0 Suppose an n N I := [↵0,0], for ↵0 <0.SplitI into two disjoint closed 1 { } 21 ✓ 1 0 ↵0 ↵0+0 1 intervals I0 and I1 of the same length, I0 =[↵0,↵0 + 2 ]=[↵0, 2 ]andI1 = 0 ↵0 ↵0+0 1 [↵ + , ]=[ , ]. Then there is some r(0) 0, 1 such that I has infinite 0 2 0 2 0 2{ } r(0) intersection with ai i N (since ai i N is infinite and since the union of two finite sets 1 { 1} 2 { } 2 is finite). Set I = Ir(0) =[↵1,1]. n In general, fix n and suppose I =[↵n,n] has been defined and has infinite n intersection with ai i N. Now we repeat the same construction: We split I into { } 2 n+1 n+1 n+1 ↵n+n two disjoint closed intervals, I0 and I1 ,ofthesamelength,I0 =[↵n, 2 ] and In+1 =[↵n+n , ]. Then there is some r(n) 0, 1 such that In+1 has infi- 1 2 n 2{ } r(n) nite intersection with ai i N for the same reason as in the previous paragraph. Set n+1 n+1 { } 2 I = Ir(n) =[↵n+1,n+1].

25 Let now↵ ¯ be the least upper bound of the sequence ↵n n.Thisupperboundexists ¯ { } by the completeness of R. Let also be the greatest lower bound of n n.Thislower { } bound exists again by the completeness of R.Clearly¯↵ ¯,andinfact¯↵ = ¯ since 0 ↵0  | | limn n ↵n =limn 2n = 0. Now it is easy to check that↵ ¯ is a limit point !1 | | !1 n of ai i N.Infact,givenanyopenballB centred in↵ ¯, B contains I for some (in fact, { } 2 n inﬁnitely many) n.ButeverysuchI contains inﬁnitely many members of ai n N. ⇤ { } 2 The proof of the following proposition is easy and – if you think about it – boils down to the fact the usual topology on C is generated by the set of ‘open squares’, i.e., the sets of the form

c + id : c R,d R, c a <✏, d b <✏ , { 2 2 | | | | } for a, b R and ✏>0.16 2

Proposition 4.8 For every n let zn = an + ibn, an, bn R. Then (zn)n converges in C 2 if and only if both sequences (an)n and (bn)n converge in R. In that case, limn zn = !1 limn an + i limn bn. !1 !1

The following proposition is easy too. Its proof is as in the real case.

Proposition 4.9 Let (zn)n and (wn)n two convergent complex sequences and let c C. Then 2

1. (zn + wn)n converges and limn zn + wn =limn zn +limn wn, !1 !1 !1

2. (znwn)n converges and limn znwn =limn zn limn wn, and !1 !1 · !1

3. (czn)n converges and limn czn = c limn zn. !1 · !1

4.1 Cauchy sequences

We shall prove Cauchy’s criterion for convergence. For that we need a deﬁnition.

Deﬁnition 4.10 A sequence (an)n N is a Cauchy sequence if 2 ( ">0)( N)( n, m > N) a a <". 8 9 8 | n m|

Theorem 4.11 A sequence (an)n N converges i↵it is a Cauchy sequence. 2 16In other words, the usual topology on C is the product topology on R2 (where R is given the usual topology). Given topological spaces (X,⌧1), (Y,⌧2), the product topology on X Y is the collection of all unions of sets of the form U V ,whereU ⌧ and V ⌧ . ⇥ ⇥ 2 1 2 2

26 Proof. In the forward direction, suppose that lim an = a.Let">0begiven.Then there is N such that for all n N we have an a <"/2. Therefore if n, m N we have that a a a a + a a <"|/2+"/|2=",sothesequenceisCauchy. | n m|| n | | m| In the other direction, let N be such that for all m, n N we have a a < 1, so in | n m| particular an aN < 1. Then let r 1 be large enough so that all a1,...aN 1 are within | | r from aN ,forexampler =max a1 aN ,... aN 1 aN .Thenallan are within r from {| | | |} aN ,andhencethesequence an 01 is bounded. By the Bolzano–Weierstrass theorem the sequence has a limit point,{ call} it a.

We claim that a is in fact the limit of an 01.Forthis,notethatgiven">0wecan ﬁnd N such that a a <"/2foralln,{ m} N.Ontheotherhand,weknowthat | n m| B"/2(a) contains inﬁnitely many points, so let n0 N be such that an0 B"/2(a). Then for m n we have that a a a a +a a <"/2+"/2=2 ".Thisshows 0 | m || m n0 | | n0 | that a is the limit of the sequence. ⇤

This theorem will be very useful when studying inﬁnite series.

5 Series

The notation for an inﬁnite series of complex numbers is n 0 zn.Weoftenrefertothe sequence (sn)n N of partial sums, i.e., sn = z0 + ...+ zn.Theseriesissaidtoconverge 2 P if the sequence of partial sums converges. In that case, lim sn is called the sum of the series. Depending on the context, we denote the sum of this series also by n1 0 zn (it is then of course understood that this limit exists). The precise way in which we are using this notation should always be clear from the context. Here is the ocialP deﬁnition.

Deﬁnition 5.1 A series (of complex numbers) is an expression of the form n 0 zn, where zn C for all n. Given, m, the m-th partial sum of n 0 zn is sm = 0 n m zn. 2 P  We say that n 0 zn converges i↵ limn sn = a exists.P In that case, weP call a the !1 sum of the series n 0 zn. P P If a series does not converge, we say that it diverges.

Remark 5.2 The notion of series tries to make sense of the idea of ‘adding infinitely many numbers’ (whenever this is possible). There are other similar notions, for example the notion of infinite product: Given a sequence (an)n, an is defined as limn pn, n !1 where pn = a1 ... an for all n, provided this limit exists, of course. For example, · · Q

2 p2 2+p2 2+ 2+p2 = ... ⇡ 2 · p 2 · q p2 ·

27 and ⇡ 2 2 4 4 6 6 4n2 = ...= 2 1 · 3 · 3 · 5 · 5 · 7 · 4n2 1 n 1 Y In this module we will not deal with inﬁnite products, though, but only with series.

Since any complex number can be written in the form a+ib for reals a, b,everyﬁnite n n n sum j=0 zj of complex numbers can be written in the form j=0 aj + i j=0 bj where n n aj and bj are real. This allows us to use the notation j=1P j=1 P P

P P 1 1 1 zn = an + i bn n=0 n=0 n=0 X X X and to derive the following two propositions using Propositions 4.8 and 4.9, respectively.

Proposition 5.3 A complex series n zn converges if and only, letting zn = an + ibn with an and bn in R for all n, both n an and n bn converge as series of real numbers. In that case, P P P zn =( an)+i( bn) n n n X X X

Proposition 5.4 Let zn and un be complex series and let c C. n n 2 P P 1. If both n zn and n un converge, then n(zn + un) converges too. In that case, P P P (zn + un)=( zn)+( un) n n n X X X

2. If n zn converges, then n czn converges too. In that case, P P czn = c( zn) n n X X

Remark 5.5 It is not true in general that if n zn and n wn are convergent series, 1 then (zn wn)=( zn)( wn). The following is a counterexample: Letting zn = n n · n n P P 2 for all n n we have that n 1 zn =1. Of course 1=1 1. But 1 > n 1 un where P P P · 1 un = zn zn for all n. To see this inequality, note that n un = n 1 22n ; in other words, · P P un = zn if and only if n is even, and un =0if n is odd. In particular, all terms occurring P P in n zn are positive, and n un misses some of them (in fact, inﬁnitely many of them). From this it readily follows that n un < n zn. (Hint to see this: Use the fact that P P 1 n an + n bn = n(an + bn) for suitable an, bn.) As a matter of fact, n un = 3 since 1 n P P n un = n 1( 4 ) (for this, see later when we talk about geometric series). P P P P P P

28 By the Cauchy condition, if n zn converges then in particular for all ">0thereis N such that for all n N we have sn+1 sn <",so zn+1 <"since zn+1 = sn+1 sn. P | | | | This implies that limn zn =0.Thisisanecessary,butnotasucientcondition,for !1 the convergence of a series as the following example shows.

Example 5.6 n 1 1/n diverges to inﬁnity even though limn 1/n =0. !1 The CauchyP condition for sequences can be restated as

Theorem 5.7 A sequence (an)n converges if and only if limm,n an am =0. !1 | |

In the above, limm,n an am =0means:Foreveryreal✏>0thereissome !1 | | N N such that for all n>m N, an am <✏. 2 | | Applying this to series we obtain the following useful test.

Theorem 5.8 A series n zn converges if and only if

P lim (zm + zm+1 + ...+ zn)=0 m,n !1 Proof. If s is the n-th partial sum of z ,thens s = z + ...+ z .Sincewe n n n n m m+1 n have limm,n sn sm =0ifandonlyiflimm,n (sn sm)=0,thisgivesuswhatwe !1 !1 want. | | P

5.1 Absolute convergence

Consider a complex series n zn and the associated real series n zn .Byapplying repeatedly the triangle inequality we obtain that for any n, m, | | P P z + z + ...+ z z + z + ...+ z , | m m+1 n|| m| | m+1| | n| so if the latter quantity tends to 0, so does the former. Hence, if z converges, then n | n| so does zn.Wehave n P DeﬁnitionP 5.9 A series z converges absolutely if z converges. n n n | n| It follows that absoluteP convergence implies convergence.P The opposite is not true, n as witnessed by the series n 1( 1) 1/n,whichconvergesbutnotabsolutelysince n n 1 ( 1) 1/n = n 1 1/n does not converge. | | P P Absolute convergenceP provides us with an easy way to test convergence of complex series, by looking at a real series associated to them. The main valuable property of absolute convergence is the following fact which we shall not prove.

29 Fact 5.10 The terms of an absolutely convergent series can be rearranged in any way; more precisely, the result of rearranging the terms of any absolutely convergent series in any way is another absolutely convergent series with the same sum as the original series.

This fact is not true of series that are convergent but not absolutely convergent. For example, given any r R,itispossibletoﬁndapermutation⇡ of the indices 2 n of the convergent sequence n n zn given by zn =( 1) 1/n in such a way that the corresponding series n 1 z⇡(n) converges precisely to r. P P 6 Sequences and series of functions

Let me show you this ad again:

Advertisement 6.1 Taking limits of sequences of functions is an extremely useful way of generating new functions, like sin(z), cos(z), ez, and so on, from (sequences of) previously deﬁned functions. Note that with the resources we have seen so far (i.e., the usual arithmetical operations on C) and using just composition of functions, the only functions that you would ‘naturally’ expect to ﬁnd in analysis that we can actually generate at the moment are polynomials and, more generally, ‘rational’ functions (i.e., p(x) functions of the form q(x) , where p(x) and q(x) are polynomials). It could be argued that the study of limits of sequences functions is in fact the most fruitful notion in analysis.

Let us start to look at sequences functions. The members of these sequences will be actual functions rather than numbers. Then we will consider the corresponding notion of series of functions.

Suppose that we have a sequence (fn)n of functions, fn : E C,whereE C is ! ✓ given, and that for every x E the sequence of numbers (fn(x))n converges. Then we may deﬁne the limit function2f on E by letting

f(x)=limfn(x) n for all x E.Wesaythatthesequence(f ) converges (point–wise) to f.17 2 n n

Example 6.2 Let fn(x)=x/n for n 1. Then the sequence (fn)n 1 converges point- wise to the function f(x)=0, the constant 0 function.

For practical uses (examples will be seen) when speaking of convergence of a sequence of functions we would like to be able to say that certain good properties of the sequence of functions are preserved by the limit of the sequence. For example, if each function fn is continuous, we would like to say that the limit function is continuous as well. This is not true in general: 17Of course the same notion of point–wise convergence and the corresponding notion of limit function can be deﬁned in the general context of metric spaces.

30 Example 6.3 For every n, let f (x)=xn, for x [0, 1]. Each of these functions is a n 2 polynomial, so in particular continuous. The sequence (fn)n converges to the function f deﬁned on on [0, 1] sending every t<1 to 0 and sending 1 to 1, which is certainly not continuous.

To avoid this situation we shall in fact use a di↵erent, stronger concept of convergence. This is the concept of uniform convergence.

Deﬁnition 6.4 Given a set E C, a sequence (fn)n of functions, fn : E C, converges uniformly to a function✓f if for any ">0 there is N such that for all!x E, if n N, then f (x) f(x) <". 2 | n | The di↵erence with point–wise convergence is that we have one N which works for all x. We have the following:

Theorem 6.5 Suppose that a sequence (fn)n of continuous functions converges uniformly to a function f. Then f is continuous.

Proof. Let E be the common domain of all these functions. Let ">0 be given. Because of the uniform convergence there is N such that for all x E,ifn N,then 2 fn(x) f(x) <"/3. Let x0 E be any point. Because fN is continuous at x0 there is | >0 such that| f (x) f (x2 ) <"/3whenever x x <.Then | N N 0 | | 0| f(x) f(x ) f(x) f (x) + f (x) f (x ) + f (x ) f(x ) <". | 0 || N | | N N 0 | | N 0 0 |

The above theorem is not true in general for convergent sequences of continuous functions:

Example 6.6 The sequence of function (fn)n we have seen in Example 6.3 converges to the function f there but certainly does not converge uniformly.

Exercise 6.7 Prove or disprove the following: Let E C and suppose fn : E C ✓ ! is uniformly continuous for all n. Suppose (fn)n converges uniformly to f. Then f is uniformly continuous.

31 6.1 Series of functions

Given now a sequence (fn)n of functions, we can make sense of the series n fn.We say that this series converges to the function f i↵the corresponding sequence of partial P sum functions gn(x)= k n fk(x)converges,asasequenceoffunctions,tof.Inthis  case we may call f the sum, or the limit, of fn. And of course we say that fn P n converges uniformly i↵(gn)n converges uniformly. P P We can apply the Cauchy criterion to uniform convergence of series. Namely:

Proposition 6.8 A series n fn(x) of functions converges uniformly on E i↵for every ">0 there is N such that for all x E we have that for all m n N, P 2 f (x)+f (x)+...+ f (x) <" | n n+1 m | Proof : Exercise. (Hint: Use that s (x) s (x)=f (x)+f (x)+...+ f (x), m n n n+1 m where sn stands for the n-the partial sum.) Using the above proposition we can find an easy test for the convergence of a series of functions. For this, we first need to define the following relation between series:

Deﬁnition 6.9 Given a series n fn(x) of functions and a series n an of positive real numbers, we say that an is a majorant for fn(x) if there is a constant M such n P n P that for some N, for all n N and all x we have fn(x) Man. We say that M is a coecient of majorisationP . P | |

Then we have the following.

Theorem 6.10 (Weierstrass M test) Suppose that (an)n is a sequence of positive real numbers such that n an is a majorant for n fn(x) and such that n an converges. Then fn(x) converges uniformly. n P P P P Proof. We apply the Cauchy criterion to n fn(x). Let M be the coecient of majori- sation. Let ">0begiven.Because an converges there is N1 > 0suchthatforall n P n N1 and all m we have that an + an+1 + ...+ an+m <"/M.LetN2 > 0besuchthat P for all n N2 and all x we have fn(x) Man.LetN =max N1,N2 .Thenforall n N,alln and all x we have | | { } f (x)+f (x)+...+ f (x) + f (x) + ...+ f (x) | n n+1 n+m | | n | | n+m | Ma + ...+ Ma = M(a + ...a ) M("/M)=". n n+m n n+m  ⇤⇤

Be careful: The above is just a useful test. There are uniformly continuous series of functions which do not satisfy this test. The following is a “cheap” example.

32 Example 6.11 Consider the series n 1 fn, where each fn is the constant function ( 1)n+1 with value . This series converges uniformly to the constant function with value n P ( 1)n+1 the sum of the series n 1 n , which is ln(2). However, n 1 fn does not have any 1 majorant n 0 an which converges since otherwise n 1 n would converge too. P P As weP can see, it is important to know which seriesP of positive real numbers converge, so I will remind you of some of the tests for this, without proving them.

Theorem 6.12 Suppose that an,bn are positive real numbers for all n. (1) (Comparison Test) If a b for all large enough n, then convergence of b n  n n n implies convergence of an. n P an+1 (2) (Ratio test) Suppose limPn = r. !1 an

1. If r<1, then n an converges. 2. If r>1, then a diverges. Pn n 3. If r =1, thenP the test doesn’t give any information; the series may or may not converge. (3) (Ratio Comparison Test) If for all large enough n, a /a b /b , then con- n+1 n  n+1 n vergence of bn implies convergence of an. The ﬁrst of theP tests is a special case of theP majorant test.

6.2 Power series

A power series is a function series of the form

1 a (z a)n n n=0 X for some constants a ,a ,a ... and a. { 0 1 2 } n An important example of such a series is a geometric series n1=0 z .Thisexample is easy to analyse, namely: N n 1 zN+1 P For z =1, n=0 z = 1 z by the familiar formula for the sum of a ﬁnite geometric progression6 (which is easy to prove by induction on N). If z < 1wehave N+1 P | | limN z =0,so !1 1 zN+1 1 lim = N !1 1 z 1 z n 1 1 Therefore the geometric series n=0 z converges to 1 z exactly when z < 1. If z 1 it clearly diverges as its terms become unbounded (so they certainly do| | not tend| to| 0). P The geometric series will be used quite often, sometimes in di↵erent forms. Let us see an example:

33 (z 1 i)3 Example 6.13 Suppose somebody asks us to write 2+i z as the sum of a power series that converges if and only if z 1 i < 1. | |

One way to deal with this request would be to remember that the geometric series n n 0 z converges if and only if z < 1. Hence, letting w = z 1 i, | | P wn = (z 1 i)n n 0 n 0 X X converges if and only if z 1 i = w < 1, which is what we want about our series. n | | n | | Also, if n 0 w = n 0(z 1 i) converges, then it converges to P P 1 1 1 = = 1 w 1 (z 1 i) 2+i z n It follows that if n 0(z 1 i) converges, then P (z 1 i)3 (z 1 i)3 (z 1 i)n = (z 1 i)n = 2+i z n 0 n 3 X X In conclusion, an answer is (z 1 i)n n 3 X n This is of course the power series n 0 an(z a) where an =0if n 2 and an =1if n 2, and where a =1+i.  P The geometric series coupled with the Weierstrass M test allows us to study power series in general. We ﬁrst recall the notions of lim sup and lim inf.

Deﬁnition 6.14 Suppose that (an)n is a sequence of positive numbers. Then lim supn an = a if a is the sup of the set of all limit points of (an)n, i.e. all the limits of subsequences of a . We deﬁne lim inf a as the inf of the set of all limit points of (a ) . { n}n n n n n Note:

(1) If limn an exists, then lim supn an =liminfn an =limn an.

(2) If b>lim supn an and an n is bounded, then all but ﬁnitely many points of an n are less than

34 Theorem 6.15 (Hadamard) Let a0,a1, C be given and deﬁne the number R by ···2 1 R = . lim sup a 1/n n | n| Then the following holds.

(1) For any a C, 2 n 1. the power series n1=0 an(z a) converges absolutely for all z such that z a R(so in particular the series n | | 1 a (z a) is divergent). n=0 n P Proof. For any positive r˜ 1/R =limsupan ,so an < 1/r˜ for{| all n|su}ciently large.2 For such n, a < 1/r˜n,so | | | | | n| sup a (z a)n = a rn < (r/r˜)n < 1 | n | | n| z Br(a) 2 So (r/r˜)n < is a majorant for a (z a)n.Since (r/r˜)n < (as r/r˜ < 1), 1 n 1 | | we get the absolute uniform convergence on Dr(a)bytheWeierstrassMtest. P P P If z a = r>R,take˜r (R, r). Then there exist n arbitrarily large such that 1/n| | n2 n n an 1/r˜.Then,an(z a) (r/r˜) for inﬁnitely many n.Butlimn (r/r˜) = . !1 Hence,| | (a (z a)n) |does not converge| and therefore a (z a)n does not converge1 n n n n either. ⇤⇤ P Thus we see that the set of points where a power series converges consists of a disk z a

Remark 6.16 R = is allowed, and we informally allow things like “ 1 =0” and “ 1 = ” [check the proof1 and see that this actually makes sense]. 1 0 1 Remark 6.17 On the circle of convergence, many di↵erent behaviours are possible. zn diverges for all z =1. zn/n diverges for z =1, else converges, but not absolutely (this follows| from| the fact that the partial sums of zn are bounded for z =1 andP 1/n 0). zn/n2 convergesP absolutely on z 1. Sierpi´nskigave a (complicated)6 example of# a function which diverges at every point| | of the unitP circle except z =1. P

35 (z a)n Example 6.18 Given a C and ⇢ R, ⇢>0, prove that the series n 0 nn con- 2 2 verges uniformly on z C : z ⇢ . { 2 | | } P

(z a)n Answer: By Hadamard’s formula, n 0 nn has radius of convergence 1 P 1 = = 1 1 1 limsup n limsupn 1 n| nn | n (z a)n By Hadamard’s theorem we therefore know that n 0 nn converges uniformly on the closed disk z C : z a ⇢ for every ⇢>0. Now, given any ⇢>0 let ⇢⇤ be such that { 2 | | } P z C : z ⇢ z C : z a ⇢⇤ { 2 | | }✓{ 2 | | } (z a)n Then, since n 0 nn converges uniformly on z C : z a ⇢⇤ , it converges { 2 | | } uniformly also on z C : z ⇢ (by the deﬁnition of uniform convergence it is trivially trueP that if{ a2 sequence| | of functions} converges uniformly on some set B and A B, then that sequence of functions converges uniformly on A as well). ✓ Sometimes the radius of convergence of a power series can be computed in an easier way:

n Proposition 6.19 Let n 0 an(z a) be a power series and R its radius of convergence. Suppose P a lim n = r n |an+1 | exists. Then R = r.

Proof. We prove r R ﬁrst. Let r =lim an .For z a =˜r0suchthat˜rr and assume a (z a)n converges. We will reach a contradiction. If |a (z | a)n n n P P

36 n n converges, then limn an(z a) =0,soinparticulartheset an(z a) n N is {| || 2 } bounded by some constant K>0. Now for x R with r 1, n | anx | r so a xn diverges. This contradiction ﬁnishes the veriﬁcation that r R. | n| ⇤⇤ P Note that r = is allowed in the above proposition. 1

7 Holomorphic functions

Let us start looking at one of the most important concepts in complex analysis, that of a holomorphic or analytic function.

Deﬁnition 7.1 Let E C be open and let f : E C be a function. ✓ ! 1. We say that f is di↵erentiable (or derivable) at z i↵

f(z + h) f(z) lim h 0 ! h exists. In this case we write f(z + h) f(z) f 0(z)=lim h 0 ! h 2. We say that f is holomorphic at z if and only if there is an open set O E such that z O and such that f is di↵erentiable at w for every w O. ✓ 2 2 3. We say that f is holomorphic (or analytic) on E i↵ f is holomorphic at every z E. 2

Note:IfE C is open and f : E C is a function, then f is holomorphic in E i↵ f is di↵erentiable✓ at every point in E!.

Remark 7.2 One reason we ask, in this definition, that f be defined on some open set containing z is that we want to rule out pathologies like a function being defined in exactly one point and being “di↵erentiable” in that point for vacuous reasons.

37 We can easily see that for any z0 in the domain of an analytic function f we have

f(z) f(z0) f 0(z0)= lim z z0 z z ! 0 Also, by applying the basic rules concerning sums and products of limits we can see that the sum and product of every two analytic functions is again analytic, and so is the quotient (whenever deﬁned). Moreover, if f and g are derivable at z,then (f + g)0(z)=f 0(z)+g0(z), (fg)0(z)=f 0(z)g(z)+f(z)g0(z), etc. The proofs of these facts are identical to the proofs in the real case. Some familiar functions are easily seen to be analytic, such as the function f(z)=zn for every n 0. Using the above paragraph we can also see that all the familiar formulas n n 1 for taking derivatives still hold, so for example the derivative of z is nz ,etc. We shall talk much more about holomorphic functions in the following lectures, where amainresultweshallproveisthateveryholomorphicfunctionhasderivativesof every order.Thisisnot true for di↵erentiable real functions in general.

Example 7.3 Let f : R R be the function that sends x 0 to 0 and sends x 0 to 2 !  x . f is di↵erentiable everywhere, f 0(x)=0if x 0, and f 0(x)=2x is x 0. However, f 0(h)f 0(0) f 0(h) f 0 is not di↵erentiable at x =0(i.e., limh 0 =limh 0 does not exist). ! h ! h

Example 7.4 We can elaborate on the above example. Given an integer N 1, let f(x)=0if x 0 and let f(x)=xN+1 if x 0. Then it is easy to check as in the above example that fis derivable N times at x =0 but not N +1times.

Example 7.5 For an example with a di↵erent look, let N 1 be an integer and let g : R R be given by g(x)=xN x , where x is the absolute value of x. Then g is derivable! at x =0exactly N times. | | | |

The following fact will be used in the proof of Theorem 7.7.

1/n Fact 7.6 limn n =1. !1

1/n 1/n Proof. Note that n>1 implies n > 1, and so in particular n =1+n for some n > 0whenevern>1. So n(n 1) n(n 1) n =(1+ )n =1+n + 2 + ...>1+ 2 n n 2 n 2 n It follows that 2 n 1 2 n < n(n 1) = n 2

38 2 2 so 0 <n < .Therefore,sincelimn =0,wehavethatlimn n =0,and n !1 n !1 therefore q q 1/n lim n =lim(1 + n)=1 n n !1 !1 ⇤

We will eventually see that holomorphic functions can be represented as sums of a power series. This theorem underlines the importance of power series. Towards this result we can already derive a very interesting theorem:

n Theorem 7.7 Let anz be a power series with radius of convergence R. For z

n 2. f 0(z) is the sum of the series obtained from anz by deriving each term, i.e., n f 0(z)= n 0(n +1)an+1z . P n n 3. The seriesP n 0(n +1)an+1z has the same radius of convergence as anz . P P n Proof. By deriving each term of the series we mean that the derivative of anz is n 1 n nanz ,sotheterm-wisederivativeof n1=0 anz is

1 P n 1 n nanz = (n +1)an+1z n=1 n 0 X X First let us calculate the radius of convergence of this series. 1/n 1/n 1/n 1/n Since limn n =1,wehavethatlimsupn n an =limsupn an ,givingthe same radiae of convergence for both series by Hadamard’s| | formula. | | Let us now represent

1 1 n n 1 k f(z)= anz =(a0 + a1z + ...+ an 1z )+ akz n=0 X Xk=n and call the ﬁrst summand sn(z)andthesecondRn(z). Let us also write f1(z)= n 1 limn sn0 (z), which is by deﬁnition 1 nanz .Wehavetoshowthatf 0(z)=f1(z) !1 n=1 for all z with z

1 k 1 k 2 k 2 k 1 ak(z + z z0 + ...+ zz0 + z0 ) Xk=n

39 [check that this is true]. Let us fix ⇢ such that z <⇢0begiven.WecanfindN0 such that for all n N0 we have 1 P P 1 Rn(z) Rn(z0) k 1 k ak ⇢ <"/3. | z z0 | | | Xk=n Similarly, there is N1 such that for all n N1 we have sn0 (z0) f1(z0) <"/3. By the definition of limits and derivatives there is>0suchthatif0| < z z| <,then | 0| sn(z) sn(z0) s0 (z ) <"/3 | z z n 0 | 0 Therefore we showed that f 0(z0)existsandisequaltof1(z0). ⇤ Note that we can repeat the same reasoning for higher derivatives and get 2 f 00(z)=2a2 +6a3z +12a3z + ... and in general (k +1)! (k +2)! f (k)(z)=k!a + a z + a z2 + ... k 1! k+1 2! k+2 Note, by the way, that the radiae of converge do not change (this follows of course from applying the above theorem repeatedly). In particular, if f(z)isholomorphicandgiven by a power series, then it has derivatives of any given order, and these derivatives are also given by power series. (k) Substituting z =0inthelastequationgivesthatf (0) = ak k!andthereforethe original power series expansion becomes · 1 f (n)(0) f(z)= zn n! n=0 X It is an amazing theorem due to Taylor that in fact every holomorphic function can be represented as a sum of a convergent power series:

Theorem 7.8 If f(z) is holomorphic in the region E containing z0, then for any open disk centred at z0 and contained in E we have that for any z in the disk, 1 f (n)(z ) f(z)= 0 (z z )n n! 0 n=0 X Proving this theorem is a bit beyond our means at the moment but we shall return to it in the second semester.

40 7.1 The Cauchy–Riemann equations

As we already observed, every complex function f(z)canalsobeconsideredintermsof two real functions of two real variables x, y,wheref becomes

f(x, y)=u(x, y)+iv(x, y)

Example 7.9 Let f(z)=z2. Writing z = x + iy we obtain that f(z)=x2 +2ixy y2, which can be written as f(z)=(x2 y2)+2ixy. Writing u(x, y)=x2 y2 and v(x, y)= 2xy we obtain f(x, y)=u(x, y)+iv(x, y).

It turns out that the di↵erentiability (i.e. analyticity) of f is connected with the di↵erentiability of each u, v,butnotinastraightforwardmanner.Weshallderivea formula which gives this connection. Recall from our discussion of metric spaces that a set E C is said to be open if ✓ for any x E there is ">0suchthatB"(x) E.Letf : E C be a function with 2 ✓ ! E C open. Abusing the notation slightly as in Example 7.9 we shall write f(x, y)as an⇢ alternative for f(x + iy).

If f 0(z)existsforsomez = x + iy E,then 2 f(z + h) f(z) f(x + h, y) f(x, y) @f f 0(z)= lim =lim = (x + iy), h R,h 0 h R,h 0 2 ! h 2 ! h @x˜ by the deﬁnition of partial derivatives in R2,and f(z + ih) f(z) f(x, y + h) f(x, y) @f f 0(z)= lim =lim i = i (x + iy). h R,h 0 h R,h 0 2 ! ih 2 ! h @y˜ Thus complex-di↵erentiability of f at z implies not only that the partial derivatives of f exist there, but also that they satisfy the Cauchy–Riemann equations @f @f = i . @x @y If f = u + iv, then this equation is equivalent to the system @u @v @v @u = , = . @x @y @x @y

Thus,

Theorem 7.10 If f = u + iv is holomorphic in E, where u : E R and v : E R, then ! !

41 1. the Cauchy–Riemann equations are satisﬁed there, and

2. for every x + iy E, 2 @u @v f 0(z)= (x + iy)+i (x + iy) @x˜ @x˜

Example 7.11 (1) Let f(z)=z2. We expect this to be an analytic function and that f 0(z)=2z everywhere. Let us verify the Cauchy-Riemann equations. We have already 2 2 found out that u(x, y)=x y and v(x, y)=2xy. We have ux(x, y)=2x = vy(x, y) and u (x, y)= 2y = v (x, y). y x (2) Let f(z)=ez = exeiy = ex cos y + iex sin y. Then u(x, y)=ex cos y and v(x, y)= ex sin y. We can easily check that the Cauchy–Riemann equations are satisﬁed in the whole plane.18

Since the Cauchy–Riemann equations are necessary for the existence of the derivative, they can also be used to locate points where a given function does not have one.

Example 7.12 Let f(z)= z 2. Now we have f(x+iy)=x2+y2 so u(x, y)=x2+y2 and | | v(x, y)=0. Hence ux(x, y)=2x and vy(x, y)=0while uy(x, y)=2y and vx(x, y)=0. Hence the Cauchy-Riemann equations are not satisﬁed unless x = y =0and so f 0(z) cannot exist unless z =0.19

We know that for a complex function f deﬁned on an open set E the di↵erentiability of f at z implies not only that the partial derivatives of f exist there, but also that they satisfy the Cauchy–Riemann equations @u @v @v @u = , = . @x @y @x @y

The fact that f is holomorphic in E implies that the partial derivatives of f exist and are continuous on E.20 Moreover:

Theorem 7.13 Let E be an open subset of C and let f : E C be a function. If the partial derivatives of f exist and are continuous on E, and! the Cauchy-Riemann equations are satisﬁed there, then f is holomorphic in E.

18Where we anticipate that f(z)=ez is the function given by u + iv,whereu(x, y)=ex cos(y) and v(x, y)=ex sin(y) for z = x + iy (x, y R). Later on we will see this deﬁnition again and we will study this function with more detail. 2 19 We are not saying that f 0(z)existsifz = 0, but rather that in order for f 0(z)toexist,z would have to be 0 (which we obtain from the fact that the Cauchy–Riemann equations are not satisﬁed at any z = 0). In fact, it is not dicult to see that f 0(0) does not exist after all. 20Remember,6 we haven’t seen a proof of this yet.

42 To prove this theorem we shall recall the following classical result from real analysis:

Theorem 7.14 (Mean Value Theorem) Suppose that a, b are real numbers such that a

Proof of Theorem 7.13. Let z0 = x0 + iy0 E.Wemustshowthatf 0(z0)exists, that is, that the limit 2 f(z + h) f(z ) lim 0 0 h 0 ! h exists. Let r be small enough that the disk of radius r around z0 is contained in E and choose h = s + it with 0 < s + it

f(z + h) f(z ) f˜(x + s, y + t) f˜(x ,y ) 0 0 = 0 0 0 0 h s + it

f˜(x + s, y + t) f˜(x ,y + t) s f˜(x ,y + t) f˜(x ,y ) t = 0 0 0 0 + 0 0 0 0 , s · s + it t · s + it ˜ where we have set f(x, y):=f(x + iy). Next, we deﬁne functions u1 :[x0,x0 + s] R, ! v1 :[x0,x0 + s] R, u2 :[y0,y0 + t] R and v2 :[y0,y0 + t] R by ! ! !

u1(x):=u(x, y0 + t),

v1(x):=v(x, y0 + t),

u2(y):=u(x0,y),

v2(y):=v(x0,y), where u(x, y):=Ref˜(x, y)andv(x, y):=Imf˜(x, y). Then it follows that

f˜(x + s, y + t) f˜(x ,y + t) 0 0 0 0 = s u (x + s) u (x ) v (x + s) v (x ) 1 0 1 0 + i 1 0 1 0 s s

43 and f˜(x ,y + t) f˜(x ,y ) 0 0 0 0 = t u (y + t) u (y ) v (y + t) v (y ) 2 0 2 0 + i 2 0 2 0 . t t Now we apply the mean-value theorem, Theorem 11.7, getting

u1(x0 + s) u1(x0) @u = u0 (x + s⇤)= (x + s⇤,y + t), s 1 0 @x 0 0

v1(x0 + s) v1(x0) @v = v0 (x + s⇤⇤)= (x + s⇤⇤,y + t), s 1 0 @x 0 0

u2(y0 + t) u2(y0) @u = u0 (y + t⇤)= (x ,y + t⇤) t 2 0 @y 0 0 and v2(y0 + t) v2(y0) @v = v0 (y + t⇤⇤)= (x ,y + t⇤⇤) t 2 0 @y 0 0 for suitable real numbers s⇤ (0,s), s⇤⇤ (0,s), t⇤ (0,t), t⇤⇤ (0,t). Combining everything done so far, we obtain2 2 2 2

f(z0 + h) f(z0) @u @v s = (x + s⇤,y + t)+i (x + s⇤⇤,y + t) + h @x 0 0 @x 0 0 · s + it ✓ ◆ @u @v t (x ,y + t⇤)+i (x ,y + t⇤⇤) . @y 0 0 @y 0 0 · s + it ✓ ◆ Note that @u @u @v @v (x + s⇤,y + t) (x ,y ) 0, (x + s⇤⇤,y + t) (x ,y ) 0, @x 0 0 @x 0 0 ! @x 0 0 @x 0 0 ! @u @u @v @v (x ,y + t⇤) (x ,y ) 0, (x ,y + t⇤⇤) (x ,y ) 0 @y 0 0 @y 0 0 ! @y 0 0 @y 0 0 ! as h 0sinces⇤,s⇤⇤,t⇤,t⇤⇤ 0ash 0andthepartialderivativesarecontinuous. Moreover! s/(s + it) stays bounded! (by! 1). Thus | | f(z0 + h) f(z0) @u @v s lim h 0 =lim (x0,y0)+i (x0,y0)+R(h) + ! h h 0 @x @x · s + it ! ✓ ◆ @u @v t lim (x0,y0)+i (x0,y0)+R¯(h) h 0 @y @y · s + it ! ✓ ◆ for functions R and R¯ such that limh 0 R(h)=limh 0 R¯(z) = 0. Hence this limit is ! ! equal to @f˜ s @f˜ t (x ,y ) + (x ,y ) @x 0 0 s + it @y 0 0 s + it

44 @f˜ @f˜ Now if the Cauchy–Riemann equations hold at (x0,y0), i.e., if @y (x0,y0)=i @x(x0,y0), then the last expression is precisely equal to @f/˜ @x(x0,y0). We still need to take care of the case when at least one of s and t is 0. This I leave as an (easy) exercise for you [using the fact that the Cauchy–Riemann equations are satisﬁed]. ˜ All in all we have that f 0(z0)existsandisequalto@f/@x(x0,y0). ⇤

We shall also use notation ux, vy etc. in place of the partial derivative notation. The Cauchy–Riemann equations can be used to prove various theorems involving analytic functions. We shall give an example. To state it, we deﬁne:

Deﬁnition 7.15 A function f is locally constant on E if E can be written as a union of disjoint open sets, on each of which the function is constant.

If a function f has derivative equal to 0, this does not imply that f is constant. For example, if f is deﬁned on B1(0) by f(z)=2andonB1/2(10) by f(z)=3,thenthe derivative of f is 0 everywhere f is deﬁned, but f is certainly not constant. It is however locally constant. In fact the following is true.

Fact 7.16 If f : E C C is a holomorphic function, E open, then f 0 is the 0 function on E if and✓ only if!f is locally constant.

We will come back to this fact later on.

Theorem 7.17 Suppose that f is a complex function on an open set E such that both f and f¯ are holomorphic. Then f is locally constant on E.

Proof. Let f = u + iv where u, v are real. Then f¯ = u iv.BytheCauchy–Riemann equations applied to f we have u = v and u = v . Applying the Cauchy–Riemann x y y x equations to f¯ we get ux = vy and uy = vx.Thisgivesthatallofthesepartial derivatives must be identically 0, so the function f is locally constant by Fact 7.16.

7.2 Harmonic functions

Arealfunctionoftworealvariablesu(x, y)issaidtobeharmonic if there is another real function of two real variables v(x, y)suchthatthecomplexfunctionf = u + iv is analytic. In such a case we call the function v(x, y)theharmonic conjugate of u. Harmonic functions show up often in various areas of mathematics and physics. Note:Ifu : R2 R is harmonic, then u has continuous partial derivatives of every ! order. This is true since there is v : R2 R such that f = u+iv is holomorphic, together !

45 with the results on holomorphic functions that we will see in the second semester, and together with the Cauchy–Riemann theorem we have just seen. How do we recognise a harmonic function?

Example 7.18 Let u = x2 y2 + xy. We want to ﬁnd a function v such u + iv is analytic. Using the Cauchy Riemann equations we have @v @u = =2x + y @y @x @v @u = =2y x @x @y Integrating the ﬁrst of the last two equations and substituting the result in the second we obtain successfully y2 v =2xy + + h(x) 2

2y + h0(x)=2y x Here h(x) is the ‘constant of integration’, and is a function of x (because we integrated a partial derivative with respect to y). From the last relation we ﬁnd

h0(x)= x and after integration x2 h(x)= + c 2 where c is a constant of integration. Putting all these results together we have

y2 x2 v =2xy + + c 2 2

Therefore we have succeeded in ﬁnding a function v as desired and it follows that u is harmonic.

Example 7.19 Let u = x2 + y2. We want to ﬁnd a function v such u + iv is analytic. Using the Cauchy Riemann equations we have @v @u = =2x @y @x @v @u = = 2y @x @y

46 Integrating the ﬁrst of the last two equations and substituting the result in the second we obtain successfully v =2xy + h(x)

2y + h0(x)= 2y Again, here h(x) is the ‘constant of integration’ and is a function of x (because we integrated a partial derivative with respect to y). From the last relation we ﬁnd

h0(x)= 4y

This is a contradiction because h0(x) does NOT depend on y and cannot therefore be equal to 4y. Therefore we cannot ﬁnd v such that u + iv is analytic and u is not a harmonic function.

If we only want to know if a given function is harmonic, without ﬁnding its harmonic conjugate, we can use the so called Laplace equation:

Theorem 7.20 (Laplace) A function u(x, y) is harmonic i↵ uxx and uyy are deﬁned everywhere and satisfy the Laplace equation

uxx + uyy =0.

Proof of half of the theorem:Weshallprovethattheequationisanecessary condition: Suppose that u is harmonic, hence there is a function v such that u and v together satisfy the Cauchy–Riemann equations. Hence ux = vy,andthereforeuxx = vyx. On the other hand, u = v so u = v .Weknowthatv and v are continuous y x yy xy x y and di↵erentiable (as u + iv is analytic), so vxy = vyx by a theorem in real analysis. Therefore u + u = v v =0. xx yy yx xy ⇤

Note:Thereareharmonicfunctionsu, v such that u + iv is not holomorphic. An example is obtained by setting u(x, y)=x and v(x, y)= y.Thenthefunctionsending x + iy to x iy (for x, y R), which of course is the conjugation function, cannot be analytic. This follows immediately2 from the fact that his function is f¯,wheref is the identity, together with the fact that the identity is analytic and not locally constant and that, as we have seen, if g is an analytic function deﬁned on an open set andg ¯ is analytic, then g is locally constant.

7.3 Some popular functions

In this section we look at some complex functions f with the following properties:

f is popular / ubiquitous in mathematics. •

47 The restriction of f to R,callitf ⇤,isevenmorepopular. • f,unlikef ⇤, cannot be defined on all of C by appealing to ‘natural’ geometric ideas • or arithmetical notions (such as in the definition of real trigonometric functions or real exponential functions), but instead is defined explicitly as the sum of a certain power series.

This power series is precisely the extension to C of the Maclaurin expansion of f ⇤. • We will be using the following lemma.

n n Lemma 7.21 Let f(z)= n 0 anz , where n 0 anz has radius of convergence R (recall that R might be ). Then the following holds. 1 P P n (i) n 0 anz has radius of convergence R. P n (ii) f(¯z)= n 0 anz for z

Proof.

(i) We have that 1 1 = = R, lim sup a lim sup a n | n| n | n| n so by Hadamard’s formula, n 0 anz has radius of convergence R. (ii) We have that P

n k k f(¯z)= an(¯z) = lim ak(¯z) =lim ak(¯z) , n n n 0 !1 k n !1 k n X X X where the last equality holds since the conjugation function is continuous, and

k k n lim ak(¯z) =lim a¯kz = a¯nz n n !1 k n !1 k n n 0 X X X (iii) Immediate from Hadamard’s theorem together with (i) and (ii).

The following lemma, which we have already mentioned, can be established by exactly the same argument as for real di↵erentiable functions.

Lemma 7.22 Let E C be open and let f, g : E C be analytic functions. Then ✓ ! fg is analytic and its derivative is given by (fg)0 = f 0g + fg0.

48 Deﬁnition 7.23 A function f : C C is entire if and only if f is analytic on all of ! C.

We are ready for our ﬁrst function.

z zn Deﬁnition 7.24 e = n 0 n! for all z C. 2 P Fact 7.25 1. ez is an entire function.

z z 2. For all z, (e )0 = e . 3. e0 =1.

4. ez+w = ezew for all z, w C. 2 5. ez =0for all z C. 6 2 6. Let z = x + iy with x, y R. Then ez = ex. In particular, eiy =1for all y R. 2 | | | | 2

1 1/n! Proof. (1): Letting an = n! for all n,sincelimn 1/(n+1)! =limn n+1 = ,byProposition n 1 6.19 we know that the radius of convergence of the series z is inﬁnity and hence, n n n! by Hadamard’s theorem, z converges everywhere. n n! P z z n n 1 n 1 (2): We know that (eP)0 is obtained as (e )0 = n 1 n! z = n 1 z (n 1)! = zn z n 0 n! ,andthisisagaine . P P 0 00 0n 0 21 P (3): e = 0! + n 1 n! =1+ n 1 0=1+0,whichistruesince0 =1. z c z (4): Fix c CP,andletf(z)=Pe e .By(2)togetherwithLemma7.22,f 0(z)= z c z z c z2 e e e e = 0. Hence there is a constant Kc such that f(z)=Kc for all z. Also: c 0 c c c z c z f(c)=e e = e by (3), so Kc = e .Thene = e e for all z, c. Hence, letting w+z z w+z z z w c = w + z, e = e e = e e . z z z (5): Since e e =1,wehavethate =0. 6 (6): ez 2 = ezez = ezez¯ (by Lemma 7.21), and also ezez¯ = ez+¯z = e2x =(ex)2. Hence, ez = e|x =| ex since ez and ex are positive reals. The second assertion follows of course | | | | | | from taking x =0. ⇤

Let us extend now the usual trigonometric functions to the complex plane.

Definition 7.26 For all z C, 2 21Since 00 = 1 does not seem to be obvious, let me point out the following: If n and m are integers, then nm is the cardinality of the set of all function from a fixed set of m things to a fixed set of n things; hence 00 = 1 since there is exactly one function from the empty set into the empty set, which is precisely the empty set. On the other hand, 0n =0ifn>0 since there is no function from a nonempty 00 set into the empty set. It follows that 0! = 1 since 0! = 1 by definition.

49 n z2n 1. cos(z)= n 0( 1) (2n)! P n z2n+1 2. sin(z)= n 0( 1) (2n+1)! P Fact 7.27 1. cos(z) and sin(z) are entire functions.

2. cos( z)=cos(z) and sin( z)= sin(z) for all z.

3. cos(z)0 = sin(z) and sin(z)0 =cos(z) for all z. 4. eiz =cos(z)+i sin(z) for all z C. In particular, ei✓ =cos(✓)+i sin(✓) for all 2 ✓ R (Euler’s identity). 2 1 iz iz 1 iz iz 5. cos(z)= (e + e ) and sin(z)= (e e ) for all z C. 2 2i 2 6. cos2(z)+sin2(z)=1for all z C. 2 Proof.

1. As in Fact 7.25 (1).

2. For all z,

( 1)2nz2n z2n cos( z)= ( 1)n = ( 1)n =cos(z) (2n)! (2n)! n 0 n 0 X X and ( 1)2n+1z2n+1 z2n+1 sin(z)= ( 1)n = ( 1)n =sin(z) (2n +1)! (2n +1)! n 0 n 0 X X n z2n 3. We know that for all z,(sin(z))0 is obtained as the sum of n 0( 1) 2n+1(2n+1)! = n z2n n 0( 1) 2n (2n)! ,whichiscos(z)bydefinition.Similarly,weknowthatforallz, P (cos(z)) is obtained as the sum of P 0 z2n 1 z2n+1 z2n+1 ( 1)n2n = ( 1)n+1 = ( 1)n , (2n)! (2n +1)! (2n +1)! n 1 n 0 n 0 X X X n z2n+1 and n 0( 1) (2n+1)! is sin(z)bydefinition. iz P iz (iz)k 4. e is, by definition on e , equal to the limit of the partial sums k n k! .Each of these sums can be written as  P ( 1)kz2k ( 1)kz2k+1 + i , (2k)! (2k +1)! 2k n 2k+1 n X X

50 and so ( 1)kz2k ( 1)kz2k+1 eiz =lim + i n (2k)! (2k +1)! !1 2k n 2k+1 n ! X X On the other hand, cos(z)+i sin(z)is,bydeﬁnitionofcos(z)andsin(z), the limit of the partial sums

z2k z2k+1 ( 1)k + i ( 1)k (2k)! (2k +1)! k n k n X X In other words,

z2k z2k+1 cos(z)+i sin(z)= lim ( 1)k + i ( 1)k n (2k)! (2k +1)! !1 k n k n ! X X These two limits are the same, and therefore eiz =cos(z)+i sin(z)forallz. If x is real, cos(x), as we have deﬁned it here, is just the usual trigonometric function you have seen in the past since

x2n cos(x)= ( 1)n (2n)! n 0 X is the Mclaurin expansion of cos(x)around0.Similarly,sin(x) is the usual trigonometric function you have seen. Therefore, for every real ✓,

ei✓ =cos(✓)+i sin(✓),

where cos(✓)andsin(✓)denotetheusualtrigonometricfunctions.

5. By (2) and (4) we know that

1 iz z 1 (e + e )= (cos(z)+i sin(z)+cos( z)+i sin( z)) = cos(z) 2 2 and

1 iz z 1 (e e )= (cos(z)+i sin(z) cos( z) i sin( z)) = sin(z) 2i 2i

1 iz iz 1 iz iz 6. We know that cos(z)= (e + e )andsin(z)= (e e ). Therefore 2 2i

z 2 1 2iz 2iz 1 2iz 2iz cos (z)+sin (z)= (e + e +2) (e + e 2) = 1 4 4

51 Corollary 7.28 For all x, y R, ex+iy = ex(cos(y)+i sin(y)). 2

Corollary 7.29 The range of ez is C 0 . \{ }

Proof. Let z C, z =0.Thenz = r(cos(✓)+i sin(✓)) for some reals r, ✓, r>0. But then eln(r)+i✓ =2eln(r)(cos(6 ✓)+i sin(✓)) = r(cos(✓)+i sin(✓)) = z (where ln(r)denotesthe natural logarithm of r).

So the range of the function sending z to ez misses exactly one point in C.The following theorem says that if an entire function misses at least two points in C then it misses all points except for one.

Theorem 7.30 (Little Picard’s theorem) Suppose f : C C is an entire function and ! there are z0 = z1 in C such that range(f) C z0,z1 . Then f is constant. 6 ✓ \{ } The real trigonometric functions cos and sin are of course periodic with period 2⇡ as you know. On the other hand, the real exponential function certainly is not. Compare this with the following.

Fact 7.31 The complex exponential function ez is periodic with period 2⇡i.

Proof. Given any z C, z = x + iy with x, y R, 2 2 ez+2⇡i = ex(cos(y +2⇡)+i sin(y +2⇡)) = ex(cos(y)+i sin(y)) = ez

8 Functions as Transformations

Unlike the case for real functions, functions f : C C are usually dicult to visualise. ! The reason is that a function as above is really a subset of R2 R2, that is, essentially ⇥ a subset of R4, and we cannot visualise 4–dimensional space (I presume). Instead of visualising the whole function we can collect partial information about it by looking at the way in which it transforms certain regions of the plane. Linear transformations w = Az + B, A, B C. We can break the study of the linear transformation into two steps. First we consider2 the transformation of the type w = z + B.Thinkingofz = x + iy and B = a + ib we obtain w =(x + a)+i(y + b). In terms of vectors we obtain w~ = ~z + B~ .Geometrically,thisrepresentsthetranslationby B~ . Notice that this transformation maps a shape onto a congruent shape.

52 Now we can consider the transformation w = Az,whereA is a complex number. It is easiest to write z in its polar form, z = ⇢eit and let A = rei✓ (where ⇢, r, t, ✓ R, r, ⇢ 0). Therefore we get that Az = r⇢ei(t+✓). 2 This means that the transformation given by w = Az is an expansion by a factor of r followed by an anti-clockwise rotation of ✓ radians around the origin. This transformation preserves shape (i.e. sends a triangle to a similar triangle, squares to squares, etc.), but it does not preserve the area. Now we can view the general linear transformation f(z)=Az + B as being obtained in two steps from the transformations we mentioned above. In other words, f(z)= (f1 f0)(z)wheref0(z)=Az and f1(z)=z + B.Thatis,inordertoobtainw from z we can ﬁrst expand by a factor of r and rotate anticlockwise by ✓ radians around the origin to obtain w0,andthentranslatew0 by B to obtain w. Now we are ready to tackle problems of the following sort:

Example 8.1 Find an analytic transformation that maps the square with vertices ( 1/2, 1/2) onto a square with area 9. ± ± To solve this problem we note that the area of the given square is 1, so we need to transform it into a square with area 9 times larger. Therefore we need to expand the size of the edge 3 times. It follows that multiplying by 3 will work, i.e., f(z)=3z.Of course other functions would do as well, for example f(z)=3z + B for any ﬁxed B C. 2

9 The reciprocal function

Consider the transformation w =1/z. This function is not deﬁned for z =0anditis clearly an involution.22 Also, we already mentioned – and it is straightforward to verify – that the theory of derivation for real functions goes through for holomorphic functions to the point of proving that if f and g,deﬁnedonsomedomainE,areholomorphicinE,andg(z) =0 f f 0g g0f 6 for all z E,then g is holomorphic in D and its derivative is given by g2 .Itfollows 2 1 that 1/z is holomorphic in C 0 and that its derivative is z2 . Using this together \{ } 1 with Lemma 7.22, we see that if n 1isaninteger,thenthefunction zn is holomorphic n in C 0 and its derivative is n+1 . \{ } z 1/z has a nice representation as the sum of a power series in the vicinity of 1: Let w =1 z.Then 1/z = wn = (1 z)n = (( 1)(z 1))n = ( 1)n(z 1)n n 0 n 0 n 0 n 0 X X X X 22A function f is an involution if f f is the identity.

53 for all z such that w = 1 z < 1. | | | | We can see the geometric e↵ect of this function by breaking it into two steps. Namely 1 write w(z)=1/z =¯z/( z 2)andconsideru(z)= z.Thenw(z)=¯u(z). The e↵ect of | | z 2 u on z is a multiplication by a real number, so any| point| will be mapped to a point of the same argument but a possibly di↵erent modulus. The e↵ect of w on u(z)isareﬂection around the x axis. From this we immediately see that 1/z ﬁxes the unit circle S2, maps everything outside S2 inside S2 and, conversely, maps everything inside S2 outside S2.

Let L be a vertical line L cutting the real axis at some point x0.Wesawthatif x0 > 1, then 1/z maps L to C 0 ,whereC is a circle, symmetric around the real axis, with diameter less than 1 containing\{ } 0. And since it is an involution it of course maps C 0 to L.Similarly,wesawthatifx0 < 1, then 1/z maps L to C 0 ,where this time\{C is} a circle, symmetric around the real axis, with diameter bigger\{ than} 1 and containing 0. And of course it again maps C 0 to L.Ifx0 = 1, then the circle C has of course diameter 1. Also, 1/z maps a line through\{ } the origin to a line passing through the origin (for example it maps the line x = y to the line x = y). Both the function w = Bz + C and the reciprocal function 1/z have the property that they map the circles and lines to circles and / or lines. For w = Bz + C this is easy to see. Let us show that this is true also for 1/z: Every line and every circle (in R2) can be expressed by an equation of the form

a(x2 + y2)+bx + cy + d =0 (8) with a =0forcircles,anda =0forlines.Conversely,givenanychoiceofrealsa, 6 b, c, d,thesetofpoints(x, y) R2 satisfying (8) is a circle or a line depending on whether or not a =0.Fortheconverse,thecrucialpointisthatthecoe2 cientofboth the x2 term and the y2 is the same (this is the reason why this set of points is not a non-circular ellipse). Now, letting w =1/z = u + iv for z =0,u, v R,weget x y 2 2 6 2 u = and v = ,andtherefore0=d(u + v )+bu cv + a if and only if x2+y2 x2+y2 x2 y2 bx cy 0=d( (x2+y2)2 + (x2+y2)2 )+ x2+y2 + x2+y2 + a if and only if

d(x2 + y2)(x2 + y2) a(x2 + y2)+bx + cy + = a(x2 + y2)+bx + cy + d =0 (x2 + y2)2

Note that in going from the ﬁrst to the last expressing we are “reversing the role” of the a, b, c, d’s. From these equations, by taking all possible relevant combinations for the parameters a and b,weimmediatelygetthat1/z maps

acircle(a =0)notpassingthroughtheorigin(d =0)toacirclenotpassing • through the6 origin, 6

54 acircle(a =0)passingthroughtheorigin(d =0)toalinepassingthroughthe • origin, 6 aline(a =0)notpassingthroughtheorigin(d =0)toacirclepassingthrough • the origin, and ﬁnally 6 a line (a =0)passingthroughtheorigin(d =0)toalinepassingthroughthe • origin.

10 Fractional linear transformations

We have seen the geometric e↵ect of the general linear function w = Bz + C and the reciprocal function 1/z.Thesetwofunctionsarespecialcasesoffractionallinear transformations. Deﬁnition 10.1 A function f is a fractional linear transformations if there are constants a, b, c, d in C such that c and d are not both 0 and such that f is given by az + b f(z)= . cz + d We will sometimes refer to fractional linear transformations as FLT’s. It is straightforward to see that the composition of two fractional linear transformation is a fractional linear transformation. In fact, it is easy to check that if a z + b f(z)= 1 1 c1z + d1 and a w + b g(w)= 2 2 , c2w + d2 a b a b i.e., if f and g are given by, respectively, the matrices 1 1 and 2 2 ,then c1 d1 c2 d2 az+b ✓ ◆ ✓ ◆ (g f)(z)= cz+d for a = a1a2 +c1b2, b = b1a2 +d1b2, c = a1c2 +c1d2,andd = b1c2 +d1d2, i.e., the coecients correspond to the product matrix a b a b a a + c b b a + d b 2 2 1 1 = 1 2 1 2 1 2 1 2 c d c d a c + c d b c + d d ✓ 2 2 ◆✓ 1 1 ◆ ✓ 1 2 1 2 1 2 1 2 ◆ We will focus mostly on c = 0 because otherwise this is just a linear transformation, which we have already studied.6 In this case we can write a bc ad 1 f(z)= + . (9) c c cz + d Indeed, the expression on the right–hand side is equal to bc ad + acz + ad az + b = c2z + dc cz + d

55 10.1 The extended plane

It turns out that it is easier to study FLT’s if we discuss the extended plane –alsoknown as the Riemann sphere –, which here we will denote by C⇤.Thisisthecomplexplane with one extra point, .Wecanimaginethisplanealsoasaspherewherethenorth pole corresponds to 1. 1 Let us assume that this sphere is the unit sphere S3 in R3;inotherwordsS3 = (x, y, z) R3 x2 + y2 + z2 =1.23 There is then a natural continuous bijection { 2 | } f : S3 (0, 0, 1) C given by \{ }! x + iy F ((x, y, z)) = 1 z Its inverse is G : C S3 (0, 0, 1) given by ! \{ } 2Re(z) 2Im(z) z 2 1 G(z)=( , , | | ), 1+ z 2 1+ z 2 1+ z 2 | | | | | | which is a continuous function wrapping C around this sphere– without the north pole. ˜ 3 The mapping F can be extended in a unique way to a continuous bijection f : S C⇤ by sending of course the north pole (0, 0, 1) to and specifying that an open set! in S3 1 containing (0, 0, 1) is mapped to an open set in C containing and vice versa. Thus, atypicalopensetU = F˜[B]containing –whereB is an open1 disk in S3 centred in (0, 0, 1) – is the complement of a closed disk1 centred in 0. This closed disk will have big radius if B’s radius is small and will have small radius if B’s radius is big.24 We can see that F˜ maps lines from the complex plane to circles passing through the north pole, and circles from the complex plane to circles on the sphere which do not pass through the north pole. We will soon encounter continuous complex functions deﬁned on all of C with the possible exception of one point. Given such a function f we will use 1/z in order to extend f to a unique continuous function f˜ from the extended plane into itself. We will do this in the following way: Suppose limz f(z)=w0,wherew0 C or maybe w0 = . ˜ !1 2 1 1 Then we set f( )=w0.Likewise,iff is not deﬁned at z0 C and limz z0 =0,we 1 2 ! f(z) will let f˜(z )= . 0 1

10.2 Möbius transformations az + b Definition 10.2 A fractional linear transformation given by w = is a Möbius cz + d transformation if and only if ad bc =0. 6 23Many other choices of spheres would work equally well in order to implement this idea. 24 It is not dicult to see that C⇤, with this topology, is a compact space. (This is immediately true 3 3 since C⇤ is a continuous image of S , S is compact, and the continuous image of a compact space is compact by Proposition 3.22.) C⇤ is the so–called one–point compactification of C.

56 az + b Note that if f(z)= is a fractional linear transformation with c =0,thenf is cz + d 6 either a M¨obius transformation or a constant function. Again, suppose c = 0. When cleared of fractions, (9) becomes Azw+Bz+Cw+D =0 which is linear in both6 z and w.Thisisalsocalledbilinear.Solvingtheoriginalequation for z we ﬁnd that

dw + b z = (10) cw a which shows that each point in the w plane (except a/c)istheimageofexactlyonepoint in the z-plane (except d/c). We can extend the domain of definition to the extended plane by letting f( d/c)= and f( )=a/c.Theextendedfunctionisabijectionof the extended plane onto itself1 and its1 inverse is given by (10). The above shows of course that every Möbius transformation is injective (i.e., sends di↵erent points to di↵erent points). The following fact is easy to verify given the fact that FLT’s are determined by 2 2matriceswithcomplexentriesandthattheiscomposition is given by the corresponding⇥ product matrix, together with the fact that every Möbius transformation has an inverse and this inverse is also a Möbiustransformation. (The above considerations cover the case when c =0.) 6 Fact 10.3 The Möbiustransformations form a group under composition.

az + b In general, given a fractional linear transformation f(z)= ,wewilloften cz + d identify f with the unique continuous extension f˜ of f to a function from the extended plane into itself. This function is computed as follows:

1. If c =0,thenf is not defined at z = d/c.Inthatcase,f˜( d/c)= . Note 6 1 that, if ad bc =0(i.e.,f is a Möbiustransformation), then limz d/c f(z)= 6 ! 1 since limz d/c cz + d =0andlimz d/c az + b = ad/c + b =0. ! ! 6 b az + b a + z 2. f˜( )=limz =limz ,z=0 ,whichisa/c if c =0,andis if !1 !1 6 d 1 cz + d c + z 6 1 c =0andz =0. 6 Let us see next that a fractional linear transformation which is not the identity can have at most 2 fixed points.

Proposition 10.4 Let f : C⇤ C⇤ be a fractional linear transformation. If f has at least 3 ﬁxed points, then it is in! fact the identity.

57 az + b Proof. Let f be given f(z)= . cz + d Note: is a fixed point of f if and only if c =0anda =0. 1 6 az + b We have that z C is a fixed point of f if and only if = z i↵ az + b = cz2 + dz 2 cz + d i↵0 = cz2 (a d)z b. If is a fixed point (i.e., c =0),thenthisisalinearequation,sowithatmost1 solution,1 with no solutions (if a d =0andb =0),oroftheform0=0(ifa d =0and b =0).Inthislastcasef is the identity, and6 in the other cases it has at most 1 + 1 = 2 fixed points ( together with at most 1 fixed point in C). 1 If is not a fixed point, then we get a quadratic equation which, by the same analysis1 as above, has either at most 2 solutions or is of the form 0 = 0, in which case a d =0=c = b and f is again the identity. ⇤

Proposition 10.5 Let z1, z2, z3 be distinct points in C. Then there is a unique M¨obius transformation f sending z1 to 0, z2 to 1, and z3 to , respectively. Moreover, this transformation is given by 1 (z z )(z z ) f(z)= 1 2 3 (z z )(z z ) 3 2 1

Proof. By plugging in values it is easy to verify that f is a function sending z1 to 0, z2 to 1, and z to . To see that it is indeed a Möbius transformation, we compute ad bc 3 1 for the corresponding coecients, that is, z3(z2 z3)(z2 z1)+z1(z2 z3)(z2 z1)= (z z )(z z )(z z ) =0asz = z for all i= j. 2 3 2 1 3 1 6 i 6 j 6 Uniqueness: Suppose f and g are two Möbiustransformation as in the conclusion. 1 Then g f is a Möbiustransformation fixing z , z , z ,andthereforeithastobethe 1 2 3 identity by Proposition 10.4. Hence f = g. ⇤

Corollary 10.6 Let z1, z2, z3 and w1, w2, w3 be points in C such that zi = zj and w = w for all i = j. Then there is a unique M¨obiustransformation sending 6 z to w , i 6 j 6 1 1 z2 to w2 and z3 to w3.

Proof. Let f be the unique Möbius transformation such that f(z1)=0,f(z2)=1 and f(z3)= and let g be the unique Möbius transformation such that g(w1)=0, 1 1 g(w )=1andg(w )= .Thenh = g f is a Möbiustransformation as desired. 2 3 1 The proof that is is unique is the same as in the proof of Proposition 10.5. ⇤

In fact, it is easy to see that the unique M¨obius transformation in the above corollary is given by (z z )(z z ) (w w )(w w ) 1 2 3 = 1 2 3 (z z )(z z ) (w w )(w w ) 3 2 1 3 2 1

58 1 Indeed, note that, in the above proof, we have that for all z, w, g (f(z)) = w i↵ (z z )(z z ) (w w )(w w ) g(w)=f(z), which gives the equality 1 2 3 = 1 2 3 . (z z )(z z ) (w w )(w w ) 3 2 1 3 2 1 It will be sometimes useful to have the following piece of notation at hand.

Notation:Givencomplexnumbersz, z1, z2, z3, (z z )(z z ) (z,z ,z ,z ):= 1 2 3 1 2 3 (z z )(z z ) 3 2 1 is called the cross ratio of z, z1, z2, z3.

Example 10.7 Let us find a Möbiustransformation mapping 0 to i, 1 to 2, and 1 to 4. For this, we compute (z 0)(1 ( 1)) 2z (z,0, 1, 1) = = (z ( 1))(1 0) z +1 and (w i)(2 4) 2(w i) (w, i, 2, 4) = = , (w 4)(2 i) (w 4)(2 i) so we are looking for a Möbiustransformation given by 2(w i) 2z = (w 4)(2 i) z +1 In other words, (16 6i)z +2i w = h(z)= (6 2i)z +2 11 Conformal mappings

Mappings that preserve (the magnitude and orientation of) angles are said to be conformal.Thereareatleasttwonotionsintheprevioussentencewhichhaven’tbeendeﬁned yet, so let’s do that now. To start with, let us say that a curve is a continuous function :[a, b] C,where [a, b]isaclosedrealinterval,andletussaythat is smooth if it is di↵erentiable! and has a continuous derivative.

Of course, the derivative of a curve with parameter interval [↵,]atapointt0 has the expected meaning: (t) (t0) (t0)0 =lim t t0,t [↵,] t t ! 2 0 Note t is a real variable whereas this limit is a complex number.

59 Deﬁnition 11.1 Let :[a, b] C be a smooth curve such that 0(a) =0. The tangent ! 6 to at z = (a) is the half line given by z0+t0(a) t R,t>0 (We require 0(a) =0 so this is a well–deﬁned half line.) { | 2 } 6

Deﬁnition 11.2 Let 1 :[a1,b1] C and 2 :[a2,b2] C be smooth curves and ! ! suppose z = (a )= (a ), 0 (a ) =0and 0 (a ) =0. Then the angle between and 1 1 2 2 1 1 6 2 2 6 1 at z is ✓ =arg(0 (a )) arg(0 (a )). 2 1 1 2 2 Given two distinct nonzero points z1, z2 in C, the angle between z1 and z2 is arg(z1) arg(z2).

We saw at the beginning of the course that arg(z/w)=arg(z) arg(w)forallz, w C, w = 0. Hence, in the above deﬁnition we have that 2 6

10 (a1) arg(10 (a1)) arg(20 (a2)) = arg( ) 20 (a2) and that arg(z ) arg(z )=arg(z /z ) 1 2 1 2 Also, note that for all z1 and z2 as in the above deﬁnition, the angle between z1 and z2 is precisely the angle between 1 and 2 at 0, where 1 and 2 are given by, respectively, i arg(z1) i arg(z2) 0 1(t)=te and 2(t)=te (t R ). 2

Deﬁnition 11.3 Let f : E C C, where E is open. We say that f is a conformal ✓ ! function if f has continuous derivatives in E and for every two smooth curves 1 : [a ,b ] E, :[a ,b ] E, if (a )= (a ), 0 (a ) =0and 0 (a ) =0, then 1 1 ! 2 2 2 ! 1 1 2 2 1 1 6 2 2 6 arg(0 (a )) arg(0 (a )) = arg((f )0(a ) (f )0(a )). 1 1 2 2 1 1 2 2 Note that a conformal function is holomorphic by deﬁnition. The following form of the chain rule can be proved by the usual argument.

Lemma 11.4 Let E C, let :[a, b] E be a curve, let f : E C, and suppose ✓ ! ! 0(a) and f((a))0 exist. Then (f )0(a)=f 0((a))0(a). The following proposition characterises conformal mappings.

Proposition 11.5 If f : E C is holomorphic and f 0(z) =0for all z E, then f is conformal. ! 6 2

60 Proof. Let z E and let :[a ,b ] E, :[a ,b ] E be smooth curves with 2 1 1 1 ! 2 2 2 ! (f 1)0(a1) f 0(z)10 (a1) z = 1(a1)=2(a2), 10 (a1) =0and20 (a2) =0.Then = = 6 6 (f 2)0(a2) f 0(z)20 (a2) (a ) 10 1 ,wheretheﬁrstequalityholdsbyLemma11.4. 20 (a2)

Note: f(z)=z2 is not conformal since it does not preserve angles at 0, but it is conformal on C 0 .Indeed,f 0(0) = 0 and f 0(z) =0forz =0. \{ } 6 6 There is a converse to Proposition 11.5.

Proposition 11.6 Suppose f : E C is conformal (E C open). Then f is holo- ! ✓ morphic and f 0(z) =0for all z E. 6 2 Möbius transformations are among the most useful examples of conformal functions. az + b Indeed, let f(z)= ,witha, b, c, d C.Then,forallz dom(f)=C d/c , cz + d 2 2 \{ } a(cz + d) c(az + b) ad bc f 0(z)= = =0. (cz + d)2 (cz + d)2 6 Suppose we are asked to find a conformal function that maps ⇧+ := z C { 2+ | Im(z) > 0 onto B1(0) = w C w < 1 .Thecrucialobservationisthat⇧ can be| written| as} the set of points{ whose2 || distance| } to i is smaller than their distance to i, i.e., as z i z C z i < z + i = z C < 1 { 2 || | | |} { 2 ||z + i| } z i Hence, since B1(0) = w C w < 1 ,itsucestotakef(z)= ,whichisin { 2 || | } z + i fact a Möbius transformation since i + i =0. 6 Similarly,

z + i f(z)= maps Im(z) < 0ontoB (0), • z i 1 z 1 f(z)= maps Re(z) > 0ontoB (0), and • z +1 1 z +1 f(z)= maps Re(z) < 0ontoB (0). • z 1 1 Also, it is not dicult to check that

z +1 f(z)= maps z :0< z < 1, 0 < arg(z) <⇡ onto z :0< arg(z) < • 1 z { | | } { ⇡/2 . }

61 As another example of conformal functions, given n N consider the function sending 2 z to zn.ThismapisconformalonC 0 since it is analytic and has non-zero derivative there. zn maps a sector z : ↵ 0 . \{ | } We may be interested in ﬁnding a conformal function which maps some given subset of the complex plane into some other given set. Sometimes we can solve this problem by considering compositions of simple known conformal maps. Let us see an example:

Example 11.7 Find a function conformal on C 0 and mapping the region X1 = z : 0 < z < 1, 0 < arg(z) <⇡/2 onto Y = B (0). \{ } { | | } 1 Well, we can do this by first mapping conformally X onto X = z : z < 1, 0 < 1 2 { | | arg(z) <⇡ , then mapping conformally X2 onto the sector X3 = z :0< arg(z) <⇡/2 , } { + } then mapping X3 conformally onto the half plane (and sector) X4 =⇧ = z :Im(z) > 0 , and finally mapping conformally X onto Y = B (0). { } 4 1 2 For the first step we can certainly take the function f1(z)=z (which is conformal z +1 on C 0 ). For the second step we may take f2(z)= , for the third step we may \{ } 1 z z i take again f (z)=z2, and for the forth step we may take f (z)= (we have already 1 3 z + i seen this one before). Hence the function h = f3 f1 f2 f1 will work, i.e., the function z4 +2iz2 +1 h(z)=i . z4 2iz2 +1 Note, incidentally, that the function f(z)=z4,whichisconformalonC 0 ,maps the above region X not onto B (0) but onto B (0) z Im(z)=0, Re(z) \{> 0 }. 1 1 1 \{ | } Example 11.8 Find a conformal function on C 0 mapping the region X = z : z < 2, 0 < arg(z) <⇡/4 onto Y = B (0). \{ } { | | } 1 1 2 Note that for this, it suces to compose the function sending z to 2 z with the function h from Example 11.7; in other words, it suces to take f = h g, where 1 2 g(z)= 2 z .

Example 11.9 We have seen that there are Möbiustransformations mapping the unit disk B1(0) onto any given half–plane. On the other hand, note that there is no Möbius transformation mapping B1(0) onto the whole plane C, and that there is no Möbius transformation mapping any given half–plane again onto C. The reason is simply that every Möbiustransformation is an injective map from C⇤ into C⇤. Suppose there was, for example, a Möbiustransformation f mapping B1(0) onto C. Let z be any point in C B1(0) and let w = f(z). Let then z0 B1(0) be such \ 2 that f(z0)=w. That is a contradiction since z = z0. A similar proof works of course to 6 rule out Möbiustransformations mapping a given half–plane onto C.

62 Finally, it is worth pointing out the following: We know that if (z1,z2,z3)and (w ,w ,w )aretriplesofcomplexnumberssuchthatz = z and w = w for all 1 2 3 i 6 j i 6 j i = j,thenthereisauniqueMöbiustransformationsf mapping zi to wi for i =1,2,3. We6 also know that every Möbius transformation is conformal, i.e., it preserves angles. However, it could be that, for example, z1, z2, z3 are collinear and w1, w2, w3 are not, i.e., the angle between the segments [z2,z1]and[z2,z3]is⇡,whereastheanglebetween the segments [w2,w1]and[w2,w3]isnot⇡.Soitwouldseemthatf does not preserve angles after all. How is this possible? Well, if we look closely at our definition of conformal function we will see that there is not contradiction here. The fact that f preserves angles implies only that, in our situation, the image of the segments [z2,z1]and[z2,z3]–letuscallthem and 0, respectively – meet at their origins at an angle ⇡.Inthatcaseitisofcoursenotpossible –inourcurrentexample–thatboth and 0 are line segments. It could happen, for example, that is a line segment (jointing w2 and w1)but0 is a circular arc (joining 25 w2 and w3).

12 Curves, paths, and connectedness

In this section we introduce paths in the complex plane and other related notions. This piece of theory will be needed for the development of the theory of integration along paths. The central notion here, namely the notion of (smooth) curve, has already come up in Section 11.

Deﬁnition 12.1 A curve (with parameter interval [↵,]) is a continuous function : [↵,] C, where [↵,] is a closed real interval. We call (↵) the initial point of and (!) the ﬁnal point of .

Fact 12.2 Every curve is compact.

Proof. Every curve is the continuous image of a compact set (namely, of a closed interval of reals), and the continuous image of a compact set is compact (Proposition 3.22). ⇤

is closed if its initial point and its ﬁnal point are the same.26 We say that :[↵,] C is simple if for all s

25See the next section on curves, paths, and connectedness for the formal deﬁnition of ‘line segment’ and ‘circular arc’. 26This notion of closed has of course nothing to do with the topological notion of closed set.

63 Given a curve ,wewillusuallydenotetherangeof by ⇤. Also, given a set S C, ✓ we will say that lies in S if ⇤ S.Asimplecurve is of course determined by ⇤,its initial point, and its orientation✓ (from its initial point to its ﬁnal point). We can revert a curve, by considering the curve with the same range with the opposite orientation:

Deﬁnition 12.3 Given a curve :[↵,] C, :[↵,] C is the curve given by (t)=( + ↵ t). is called the inversion! of . !

Note that ( )⇤ = ⇤ for every curve .Giventwocurves1 :[↵1,1] C, ! 2 :[↵2,2] C,iftheﬁnalpointof1 and the initial point of 2 coincide, then the ! natural concatenation of 1 and 2 will be continuous and so will be a curve by deﬁnition (it need not be smooth in general, though).

Deﬁnition 12.4 Suppose 1 :[↵1,1] C and 2 :[↵2,2] C are curves such ! ! that 1(1)=2(↵2). The join of 1 and 2 is the curve with parameter interval [↵1,1 + 2 ↵2] given by (t)=1(t) if t [↵1,1] and (t)=2(t + ↵2 1) if t [ , + ↵ ]. The join of and is2 denoted by .27 2 1 1 2 2 1 2 1 [ 2

Note: If is the join of 1 :[↵1,1] C and 2 :[↵2,2] C,then [↵1,1]= ! ! 1,butitisnotnecessarilytruethat [↵2,2]=2.

Given any ﬁnite sequence 1,...,n of curves, the join of 1,...,n,denotedby ... ,canbeformedprovidedtheﬁnalpointof coincides with the initial point 1 [ [ n i of i+1 for all i

Deﬁnition 12.5 A path is a curve which is the join of ﬁnitely many smooth curves.

The following is the simplest example of path (and of curve).

Deﬁnition 12.6 Given z1, z2 C, the line segment [z1,z2] is the curve with interval parameter [0, 1] given by [z ,z ](2t)=z (1 t)+z t. 1 2 1 2

In contexts where the orientation of a line segment [z1,z2]isirrelevant,wesometimes identify this line segment with its range and write [z1,z2]= z1(1 t)+z2t t [0, 1] . Note that this is an abuse of notation, though. { | 2 } Next we consider joins of line segments.

Deﬁnition 12.7 Let a = z0,z1,...zn 1,b = zn be points in C. Then [z0,z1] [z1,z2] [ [ ... [zn 1,zn] is the polygonal route (or polygonal path) from a to b. [ 27 This is just a piece of notation. The join of 1 and 2 of course is not the union of 1 and 2.

64 a = b and n = 0 are both allowed in the above deﬁnition. Next we are going to introduce some related notions of “connectedness” for subsets of the complex plane.

Deﬁnition 12.8 A set S C is polygonally connected if for all a, b S there is a polygonal path from a to b lying✓ inside S. S is path–connected if for all a,2b S there is a path from a to b lying inside S. S is convex if for all a, b S, [a, b] lies inside2 S. 2 Of course, every convex set is polygonally connected, and every polygonally connected set is path–connected. These implications cannot be reversed in general: An arc such as i✓ D1(0) = e ✓ R is certainly path–connected but not polygonally connected, and an annulus{ such| as2 z} 1 < z < 2 is polygonally connected but not convex. { | | | }

Deﬁnition 12.9 A set S C is disconnected if there are nonempty disjoint open sets ✓ G1, G2 C such that S G1 G2, S G1 = and S G2 = . S C is connected if and only✓ if it is not disconnected.✓ [ A region\ is6 a; nonempty\ connected6 ; ✓ open set.

The next theorem shows that all notions of connectedness we have seen coincide for open sets.28

Theorem 12.10 Let G C be a nonempty open set. Then the following are equivalent. ✓ 1. G is a region.

2. G is polygonally connected.

3. G is path–connected.

Proof. It will suce to prove that if G is connected, then it is polygonally connected and that if G is path–connected, then it is connected. Let us start with the ﬁrst implication. Suppose G is connected. Let a G,let 2 G = z G there is a polygonal path from a to z 1 { 2 | } and let G = G G . 2 \ 1 G = since a G . Hence, since G = G G ,itwillsucetoshowthatG and 1 6 ; 2 1 1 [ 2 1 G2 are both open (it will then follow that G2 = since G is connected, and therefore G = G ,whichmeansthatforeveryz G there is; a polygonal path from a to z,from 1 2 z 28This is not true in general. For example, 0 x + i sin(1/x) x R,x>0 is connected but not path–connected; in fact there is no path from{ 0} to[ any{ other point.| 2 }

65 which it follows that for every two points z, w in G there is a polygonal path ( z) w from z to w). [

To see that both G1 and G2 are open, let z G and let >0besuchthatB(z) G. It suces to show that if ✏ 0, 1 is such that2 z G ,thenB (z) G .Forthis,✓ 2{ } 2 ✏ ✓ ✏ note that for every w B(z)thereisapolygonalpathfromz to w inside B(z)(in fact, [z,w] B (z)). It2 follows that if z G ,thenw G (if is a polygonal route ✓ 2 1 2 1 from a to z,then [z,w]isapolygonalroutefroma to w), and that if z/G1,then w/G (if w G ,thenthereisapolygonalroute[ from a to w,butthen2 [w, z] 2 1 2 1 [ is a polygonal route from a to z and therefore z is in G1). This finishes the proof of the first implication. As to the proof of the second implication, suppose G is path–connected and let us assume, towards a contradiction, that G = G G ,whereG and G are nonempty 1 [ 2 1 2 open sets. Let a G1 and b G2.SinceG is path–connected, we may find a path ... from2 a to b lying2 inside G,wherethe ’s are smooth curves. There is 0 [ 1 [ [ n i then some i n such that ⇤ has nonempty intersection with both G and G .Ifi is  i 1 2 0 the first such i, then the initial point of i0 is in G1.Saythatthedomainofi0 is [↵,].

Claim 12.11 For every t [↵,] there is some real >0 such that if i0 (t) G✏ (for ✏ 0, 1 ), then every point2 s in [↵,] such that s t <is such that (s)2 G . 2{ } | | i0 2 ✏

Proof. Obvious, since i0 is continuous. ⇤

Let X = t [↵,]:range( [↵, t]) G .SinceX is nonempty (as a X), { 2 i0 ✓ 1} 2 = sup(X)existsbythecompletenessofR and obviously29 ↵ .   Suppose () G .Wemayassume<since otherwise ⇤ G by our i0 2 1 i0 ✓ 1 deﬁnition of ,whichcontradictsi⇤0 G2 = .Let be given by the above claim for and let s [↵,], s>,besuchthat\ 6 s; <.Thenx X by the choice of and the deﬁnition2 and therefore sup(X) >|,whichisacontradiction. | 2

Finally suppose i0 () G2. Again let be given by the above claim for and let s [↵,], s<,besuchthat2 s <.Theneveryt X is such that t

Note: We didn’t use the smoothness of the i’s in the proof of the second implication, but only the fact that they are continuous. Therefore the theorem can be actually improved: An open set is connected if and only if every two points in that set can be joined by a curve (which, in general, is a weaker notion than the notion of path– connectedness). The characterisation of connectedness for open sets in the above theorem is not true 1 >0 in general for non–open sets: For example, the set X = 0 x + i sin( x ):x R is connected but it is not path–connected. In fact, there{ is} no[ path{ lying inside X2 from} 0toanyotherpointinX. 29Try to prove this from what we have seen.

66 Deﬁnition 12.12 A circular arc traced anti-clockwise is a curve with parameter • it interval [✓1,✓2] or some reals ✓1 ✓2 sending t to a + re , for some ﬁxed a C  2 and r R>0. 2 A circular arc traced clockwise is a curve of the form for some circular arc • traced anti-clockwise .

We will often use the notation (a; r)toindicatethecurvea + reit,fort [0, 2⇡], 2 for some ﬁxed a C and r R>0.(thatis,(a; r)isthecirclecentredina with radius r traced anti-clockwise2 starting2 from the point a + r).

Deﬁnition 12.13 A circline path is a path which is a join of ﬁnitely many line segments and/or• circular arcs.

A contour is a simple closed circline path. • Lemma 12.14 Let be a path lying in an open set G. Then there is a real constant K>0 such that B (z) G for every z ⇤. K ✓ 2

Proof. For every z ⇤ we can ﬁnd a positive z R such that Bz (z) G. Bz/2(z): 2 2 ✓ { z ⇤ is an open cover of. Therefore, since ⇤ is compact by Fact 12.2, there are 2 } ﬁnitely many z1,...,zn ⇤ such that ⇤ B /2(z1) ... B /2(zn). Let now 2 ✓ z1 [ [ zn K =min /2,...,z /2 . Now it is easy to check that for every z ⇤, B (z) G: { z1 n } 2 K ✓

Let z ⇤.Leti be such that z zi <zi /2(whichexistsbecause B (z ),...,B2 (z ) covers ). Then| ifw is| such that w z

Theorem 12.15 (Covering theorem) Suppose G C is open and is a path lying in G with parameter interval [↵,]. There is then a real✓ constant K>0 and ﬁnitely many discs B1,...,BN such that

⇤ B ... B , • ✓ 1 [ [ N

Bi Bi+1 = for all i

there are a t

67 Proof. Let K be the constant given by Lemma 12.14. Let = B (z):z ⇤ . U { K/2 2 } Since is an open cover of ⇤,bycompactnessof⇤ (by Fact 12.2) we can ﬁnd ﬁnitely U many t0,...tn [↵,] such that BK/2((t0)) ... BK/2((tn)). Now it suces to set 2 30 [ [ Bi = BK ((ti)) for all i, noting that each BK ((ti)) is still contained in G by the choice of K. ⇤

The covering theorem (together with Theorem 12.10) can be used in order to extend certain results that are easy to prove for open discs to results that apply to arbitrary regions. Let us see an example. First we consider the following:

Proposition 12.16 Suppose B is an open ball and f : B C is an analytic function. Suppose that either !

1. f 0(z)=0for every z G, 2 2. f(z) is constant on G, or | | 3. f(z) R for every z G. 2 2 Then f is constant.

We saw the proof of this proposition in the lectures (actually we saw the proof in the case B = B1(0) but the proof in the general case is the same modulo small notational changes).

Corollary 12.17 Suppose G is a region and f : G C is an analytic function. Suppose that either !

1. f 0(z)=0for every z G, 2 2. f(z) is constant on G, or | | 3. f(z) R for every z G. 2 2 Then f is constant.

This corollary follows from Proposition 12.16 thanks to Theorems 12.10 and 12.15: Let a, b be two points in G.Wewanttoseef(a)=f(b). For this, let be a path from a to b lying inside G.LetB0,...,Bn be a sequence of open discs covering ⇤ and such that each one of them overlaps with the next one, as given by Theorem 12.15. Now apply Proposition N 1manytimeswiththerestrictionoff to B (for each i). i Our ﬁnal result is quite intuitive but not easy to prove. It says that a closed simple curve divides the complex plane into the inside and the outside.

30 Note that we are “fattening” each Bk/2((ti)) by doubling its radius to make sure that the resulting balls overlap. This last step is not really necessary thanks to Theorem 12.10 but it simpliﬁes the proof.

68 Theorem 12.18 (Jordan curve theorem) Let be a closed simple curve. Then C = ⇤ I() O(), where [ [ I() is a bounded open set (this is called the inside of ), and • O() is an unbounded open set (this is called the outside of ). •

References

[1] Lars V. Ahlfors, Complex analysis

[2] W W Chen, Introduction to Complex Analysis, web lecture notes.

[3] Ruel V. Churchill, James W. Brown and Roger F. Verhey, Complex Variables and Applications, 3rd edn., McGraw–Hill, 1976.

[4] H.A. Priestley, Introduction to complex analysis, 2nd edn., Oxford Univ. Press (1985).

[5] W. Rudin, Real and complex analysis, 3rd edn., McGraw–Hill (1987).