+Complex analysis math532notes

Complex Analysis Summer 2001 Kenneth Kuttler July 17, 2003 Contents General topology 1.1 Compactness in metric space 1.2 Connected sets 1.3 Exercises 11 14 17 Spaces of Continuous Functions 21 2.1 Compactness in spaces of continuous functions 21 2.2 Stone Weierstrass theorem 23 2.3 Exercises 27 The 3.1 3.2 3.3 complex numbers 31 Exercises 34 The extended complex plane 35 Exercises 36 Riemann Stieltjes integrals 37 4.1 Exercises 46 Analytic functions 47 5.1 Exercises 49 5.2 Examples of analytic functions 50 5.3 Exercises 51 Cauchy’s formula for a disk 53 6.1 Exercises 58 The 7.1 7.2 7.3 general Cauchy integral formula 63 The Cauchy Goursat theorem 63 The Cauchy integral formula 66 Exercises 72 The 8.1 8.2 8.3 8.4 8.5 8.6 open mapping theorem Zeros of an analytic function The open mapping theorem Applications of the open mapping theorem Counting zeros The estimation of eigenvalues Exercises 75 75 76 78 79 82 85 Singularities 9.1 The Concept Of An Annulus 9.2 The Laurent Series 9.3 Isolated Singularities 9.4 Partial Fraction Expansions 9.5 Exercises CONTENTS 87 87 88 91 93 95 10 Residues and evaluation of integrals 10.1 The argument principle and Rouche’s theorem 10.2 Exercises 10.3 The Poisson formulas and the Hilbert transform 10.4 Exercises 10.5 Infinite products 10.6 Exercises 97 107 108 110 113 114 120 11 Harmonic functions 123 11.1 The Dirichlet problem for a disk 123 11.2 Exercises 128 12 Complex mappings 12.1 Fractional linear transformations 12.2 Some other mappings 12.3 The Dirichlet problem for a half plane 12.4 Other problems involving Laplace’s equation 12.5 Exercises 12.6 Schwarz Christoffel transformation 12.7 Exercises 12.8 Riemann Mapping theorem 12.9 Exercises 131 131 133 134 136 137 138 141 141 147 13 Approximation of analytic functions 149 13.1 Runge’s theorem 152 13.2 Exercises 154 General topology This chapter is a brief introduction to general topology Topological spaces consist of a set and a subset of the set of all subsets of this set called the open sets or topology which satisfy certain axioms Like other areas in mathematics the abstraction inherent in this approach is an attempt to unify many different useful examples into one general theory For example, consider Rn with the usual norm given by 1/2 n |x| ≡ |xi | i=1 We say a set U in Rn is an open set if every point of U is an “interior” point which means that if x ∈U , there exists δ > such that if |y − x| < δ, then y ∈U It is easy to see that with this definition of open sets, the axioms (1.1) - (1.2) given below are satisfied if τ is the collection of open sets as just described There are many other sets of interest besides Rn however, and the appropriate definition of “open set” may be very different and yet the collection of open sets may still satisfy these axioms By abstracting the concept of open sets, we can unify many different examples Here is the definition of a general topological space Let X be a set and let τ be a collection of subsets of X satisfying ∅ ∈ τ, X ∈ τ, (1.1) If C ⊆ τ, then ∪ C ∈ τ If A, B ∈ τ, then A ∩ B ∈ τ (1.2) Definition 1.1 A set X together with such a collection of its subsets satisfying (1.1)-(1.2) is called a topological space τ is called the topology or set of open sets of X Note τ ⊆ P(X), the set of all subsets of X, also called the power set Definition 1.2 A subset B of τ is called a basis for τ if whenever p ∈ U ∈ τ , there exists a set B ∈ B such that p ∈ B ⊆ U The elements of B are called basic open sets The preceding definition implies that every open set (element of τ ) may be written as a union of basic open sets (elements of B) This brings up an interesting and important question If a collection of subsets B of a set X is specified, does there exist a topology τ for X satisfying (1.1)-(1.2) such that B is a basis for τ ? Theorem 1.3 Let X be a set and let B be a set of subsets of X Then B is a basis for a topology τ if and only if whenever p ∈ B ∩ C for B, C ∈ B, there exists D ∈ B such that p ∈ D ⊆ C ∩ B and ∪B = X In this case τ consists of all unions of subsets of B GENERAL TOPOLOGY Proof: The only if part is left to the reader Let τ consist of all unions of sets of B and suppose B satisfies the conditions of the proposition Then ∅ ∈ τ because ∅ ⊆ B X ∈ τ because ∪B = X by assumption If C ⊆ τ then clearly ∪C ∈ τ Now suppose A, B ∈ τ, A = ∪S, B = ∪R, S, R ⊆ B We need to show A ∩ B ∈ τ If A ∩ B = ∅, we are done Suppose p ∈ A ∩ B Then p ∈ S ∩ R where S ∈ S, R ∈ R Hence there exists U ∈ B such that p ∈ U ⊆ S ∩ R It follows, since p ∈ A ∩ B was arbitrary, that A ∩ B = union of sets of B Thus A ∩ B ∈ τ Hence τ satisfies (1.1)-(1.2) Definition 1.4 A topological space is said to be Hausdorff if whenever p and q are distinct points of X, there exist disjoint open sets U, V such that p ∈ U, q ∈ V U V · p · q Hausdorff Definition 1.5 A subset of a topological space is said to be closed if its complement is open Let p be a point of X and let E ⊆ X Then p is said to be a limit point of E if every open set containing p contains a point of E distinct from p Theorem 1.6 A subset, E, of X is closed if and only if it contains all its limit points Proof: Suppose first that E is closed and let x be a limit point of E We need to show x ∈ E If x ∈ / E, then E C is an open set containing x which contains no points of E, a contradiction Thus x ∈ E Now suppose E contains all its limit points We need to show the complement of E is open But if x ∈ E C , then x is not a limit point of E and so there exists an open set, U containing x such that U contains no point of E other than x Since x ∈ / E, it follows that x ∈ U ⊆ E C which implies E C is an open set Theorem 1.7 If (X, τ ) is a Hausdorff space and if p ∈ X, then {p} is a closed set C Proof: If x = p, there exist open sets U and V such that x ∈ U, p ∈ V and U ∩ V = ∅ Therefore, {p} is an open set so {p} is closed Note that the Hausdorff axiom was stronger than needed in order to draw the conclusion of the last theorem In fact it would have been enough to assume that if x = y, then there exists an open set containing x which does not intersect y Definition 1.8 A topological space (X, τ ) is said to be regular if whenever C is a closed set and p is a point not in C, then there exist disjoint open sets U and V such that p ∈ U, C ⊆ V The topological space, (X, τ ) is said to be normal if whenever C and K are disjoint closed sets, there exist disjoint open sets U and V such that C ⊆ U, K ⊆ V U V · p C Regular U V C K Normal ¯ is defined to be the smallest closed set containing E Note that Definition 1.9 Let E be a subset of X E this is well defined since X is closed and the intersection of any collection of closed sets is closed Theorem 1.10 E = E ∪ {limit points of E} Proof: Let x ∈ E and suppose that x ∈ / E If x is not a limit point either, then there exists an open set, U,containing x which does not intersect E But then U C is a closed set which contains E which does not contain x, contrary to the definition that E is the intersection of all closed sets containing E Therefore, x must be a limit point of E after all Now E ⊆ E so suppose x is a limit point of E We need to show x ∈ E If H is a closed set containing E, which does not contain x, then H C is an open set containing x which contains no points of E other than x negating the assumption that x is a limit point of E Definition 1.11 Let X be a set and let d : X × X → [0, ∞) satisfy d(x, y) = d(y, x), (1.3) d(x, y) + d(y, z) ≥ d(x, z), (triangle inequality) d(x, y) = if and only if x = y (1.4) Such a function is called a metric For r ∈ [0, ∞) and x ∈ X, define B(x, r) = {y ∈ X : d(x, y) < r} This may also be denoted by N (x, r) Definition 1.12 A topological space (X, τ ) is called a metric space if there exists a metric, d, such that the sets {B(x, r), x ∈ X, r > 0} form a basis for τ We write (X, d) for the metric space Theorem 1.13 Suppose X is a set and d satisfies (1.3)-(1.4) Then the sets {B(x, r) : r > 0, x ∈ X} form a basis for a topology on X Proof: We observe that the union of these balls includes the whole space, X We need to verify the condition concerning the intersection of two basic sets Let p ∈ B (x, r1 ) ∩ B (z, r2 ) Consider r ≡ (r1 − d (x, p) , r2 − d (z, p)) and suppose y ∈ B (p, r) Then d (y, x) ≤ d (y, p) + d (p, x) < r1 − d (x, p) + d (x, p) = r1 and so B (p, r) ⊆ B (x, r1 ) By similar reasoning, B (p, r) ⊆ B (z, r2 ) This verifies the conditions for this set of balls to be the basis for some topology Theorem 1.14 If (X, τ ) is a metric space, then (X, τ ) is Hausdorff, regular, and normal Proof: It is obvious that any metric space is Hausdorff Since each point is a closed set, it suffices to verify any metric space is normal Let H and K be two disjoint closed nonempty sets For each h ∈ H, there exists rh > such that B (h, rh ) ∩ K = ∅ because K is closed Similarly, for each k ∈ K there exists rk > such that B (k, rk ) ∩ H = ∅ Now let U ≡ ∪ {B (h, rh /2) : h ∈ H} , V ≡ ∪ {B (k, rk /2) : k ∈ K} then these open sets contain H and K respectively and have empty intersection for if x ∈ U ∩ V, then x ∈ B (h, rh /2) ∩ B (k, rk /2) for some h ∈ H and k ∈ K Suppose rh ≥ rk Then d (h, k) ≤ d (h, x) + d (x, k) < rh , a contradiction to B (h, rh ) ∩ K = ∅ If rk ≥ rh , the argument is similar This proves the theorem GENERAL TOPOLOGY Definition 1.15 A metric space is said to be separable if there is a countable dense subset of the space This means there exists D = {pi }∞ i=1 such that for all x and r > 0, B(x, r) ∩ D = ∅ Definition 1.16 A topological space is said to be completely separable if it has a countable basis for the topology Theorem 1.17 A metric space is separable if and only if it is completely separable Proof: If the metric space has a countable basis for the topology, pick a point from each of the basic open sets to get a countable dense subset of the metric space Now suppose the metric space, (X, d) , has a countable dense subset, D Let B denote all balls having centers in D which have positive rational radii We will show this is a basis for the topology It is clear it is a countable set Let U be any open set and let z ∈ U Then there exists r > such that B (z, r) ⊆ U In B (z, r/3) pick a point from D, x Now let r1 be a positive rational number in the interval (r/3, 2r/3) and consider the set from B, B (x, r1 ) If y ∈ B (x, r1 ) then d (y, z) ≤ d (y, x) + d (x, z) < r1 + r/3 < 2r/3 + r/3 = r Thus B (x, r1 ) contains z and is contained in U This shows, since z is an arbitrary point of U that U is the union of a subset of B We already discussed Cauchy sequences in the context of Rp but the concept makes perfectly good sense in any metric space Definition 1.18 A sequence {pn }∞ n=1 in a metric space is called a Cauchy sequence if for every ε > there exists N such that d(pn , pm ) < ε whenever n, m > N A metric space is called complete if every Cauchy sequence converges to some element of the metric space Example 1.19 Rn and Cn are complete metric spaces for the metric defined by d(x, y) ≡ |x − y| ≡ ( yi |2 )1/2 n i=1 |xi − Not all topological spaces are metric spaces and so the traditional − δ definition of continuity must be modified for more general settings The following definition does this for general topological spaces Definition 1.20 Let (X, τ ) and (Y, η) be two topological spaces and let f : X → Y We say f is continuous at x ∈ X if whenever V is an open set of Y containing f (x), there exists an open set U ∈ τ such that x ∈ U and f (U ) ⊆ V We say that f is continuous if f −1 (V ) ∈ τ whenever V ∈ η Definition 1.21 Let (X, τ ) and (Y, η) be two topological spaces X × Y is the Cartesian product (X × Y = {(x, y) : x ∈ X, y ∈ Y }) We can define a product topology as follows Let B = {(A × B) : A ∈ τ, B ∈ η} B is a basis for the product topology Theorem 1.22 B defined above is a basis satisfying the conditions of Theorem 1.3 More generally we have the following definition which considers any finite Cartesian product of topological spaces Definition 1.23 If (Xi , τi ) is a topological space, we make n be i=1 Ai where Ai ∈ τi n i=1 Xi into a topological space by letting a basis Theorem 1.24 Definition 1.23 yields a basis for a topology The proof of this theorem is almost immediate from the definition and is left for the reader The definition of compactness is also considered for a general topological space This is given next Definition 1.25 A subset, E, of a topological space (X, τ ) is said to be compact if whenever C ⊆ τ and E ⊆ ∪C, there exists a finite subset of C, {U1 · · · Un }, such that E ⊆ ∪ni=1 Ui (Every open covering admits a finite subcovering.) We say E is precompact if E is compact A topological space is called locally compact if it has a basis B, with the property that B is compact for each B ∈ B Thus the topological space is locally compact if it has a basis of precompact open sets In general topological spaces there may be no concept of “bounded” Even if there is, closed and bounded is not necessarily the same as compactness However, we can say that in any Hausdorff space every compact set must be a closed set Theorem 1.26 If (X, τ ) is a Hausdorff space, then every compact subset must also be a closed set Proof: Suppose p ∈ / K For each x ∈ X, there exist open sets, Ux and Vx such that x ∈ Ux , p ∈ Vx , and Ux ∩ Vx = ∅ Since K is assumed to be compact, there are finitely many of these sets, Ux1 , · · ·, Uxm which cover K Then let V ≡ ∩m i=1 Vxi It follows that V is an open set containing p which has empty intersection with each of the Uxi Consequently, V contains no points of K and is therefore not a limit point This proves the theorem Lemma 1.27 Let (X, τ ) be a topological space and let B be a basis for τ Then K is compact if and only if every open cover of basic open sets admits a finite subcover The proof follows directly from the definition and is left to the reader A very important property enjoyed by a collection of compact sets is the property that if it can be shown that any finite intersection of this collection has non empty intersection, then it can be concluded that the intersection of the whole collection has non empty intersection Definition 1.28 If every finite subset of a collection of sets has nonempty intersection, we say the collection has the finite intersection property Theorem 1.29 Let K be a set whose elements are compact subsets of a Hausdorff topological space, (X, τ ) Suppose K has the finite intersection property Then ∅ = ∩K Proof: Suppose to the contrary that ∅ = ∩K Then consider C ≡ KC : K ∈ K It follows C is an open cover of K0 where K0 is any particular element of K But then there are finitely many K ∈ K, K1 , · · ·, Kr such that K0 ⊆ ∪ri=1 KiC implying that ∩ri=0 Ki = ∅, contradicting the finite intersection property It is sometimes important to consider the Cartesian product of compact sets The following is a simple example of the sort of theorem which holds when this is done Theorem 1.30 Let X and Y be topological spaces, and K1 , K2 be compact sets in X and Y respectively Then K1 × K2 is compact in the topological space X × Y Proof: Let C be an open cover of K1 × K2 of sets A × B where A and B are open sets Thus C is a open cover of basic open sets For y ∈ Y , define Cy = {A × B ∈ C : y ∈ B}, Dy = {A : A × B ∈ Cy } 10 GENERAL TOPOLOGY Claim: Dy covers K1 Proof: Let x ∈ K1 Then (x, y) ∈ K1 × K2 so (x, y) ∈ A × B ∈ C Therefore A × B ∈ Cy and so x ∈ A ∈ Dy Since K1 is compact, {A1 , · · ·, An(y) } ⊆ Dy covers K1 Let n(y) By = ∩i=1 Bi Thus {A1 , · · ·, An(y) } covers K1 and Ai × By ⊆ Ai × Bi ∈ Cy Since K2 is compact, there is a finite list of elements of K2 , y1 , · · ·, yr such that {By1 , · · ·, Byr } covers K2 Consider n(y ) r {Ai × Byl }i=1l l=1 If (x, y) ∈ K1 × K2 , then y ∈ Byj for some j ∈ {1, · · ·, r} Then x ∈ Ai for some i ∈ {1, · · ·, n(yj )} Hence (x, y) ∈ Ai × Byj Each of the sets Ai × Byj is contained in some set of C and so this proves the theorem Another topic which is of considerable interest in general topology and turns out to be a very useful concept in analysis as well is the concept of a subbasis Definition 1.31 S ⊆ τ is called a subbasis for the topology τ if the set B of finite intersections of sets of S is a basis for the topology, τ Recall that the compact sets in Rn with the usual topology are exactly those that are closed and bounded We will have use of the following simple result in the following chapters Theorem 1.32 Let U be an open set in Rn Then there exists a sequence of open sets, {Ui } satisfying · · ·Ui ⊆ Ui ⊆ Ui+1 · · · and U = ∪∞ i=1 Ui Proof: The following lemma will be interesting for its own sake and in addition to this, is exactly what is needed for the proof of this theorem Lemma 1.33 Let S be any nonempty subset of a metric space, (X, d) and define dist (x,S) ≡ inf {d (x, s) : s ∈ S} Then the mapping, x → dist (x, S) satisfies |dist (y, S) − dist (x, S)| ≤ d (x, y) Proof of the lemma: One of dist (y, S) , dist (x, S) is larger than or equal to the other Assume without loss of generality that it is dist (y, S) Choose s1 ∈ S such that dist (x, S) + > d (x, s1 ) 15–2 W W L Chen : Introduction to Complex Analysis We summarize the above discussion in the pictures below: z w i -1 -1 -i It is easy to check that the function ψ(w) = π 5π − arg w is harmonic in this region and satisfies ψ(w) = when arg w = 3π/4 and ψ(w) = when arg w = 5π/4 It follows that our required harmonic function is given by φ(z) = 15.2 π 5π − arg z+i z−i Use of Schwarz-Christoffel Transformations We now discuss three examples which use Schwarz-Christoffel transformations Example 15.2.1 We wish to find a non-constant harmonic function in the region above the polygonal path given in Example 14.4.2, with boundary condition φ = on the polygonal path Here φ = const can be interpreted as lines of flow on a river over a step on the river bed Recall that the Schwarz-Christoffel transformation f (z) = (z − 1)1/2 + log(z + (z − 1)1/2 ) π maps the upper half plane onto the region in question We now need to find a non-constant harmonic function ψ on the upper half plane with boundary condition ψ = on the real line For example, the function ψ(z) = Imz satisfies the requirements We now need to invert the function f (z) to obtain a harmonic function φ(w) = Im(f −1 (w)) in the original region Example 15.2.2 We wish to find a harmonic function in the slit plane given in Example 14.4.4, with boundary conditions φ = on the upper slit and φ = −1 on the lower slit Here φ = const can be interpreted as equipotential lines in a region around two semi-infinite conducting plates with opposite charges Recall that the Schwarz-Christoffel transformation f (z) = − π z2 − log z + −i π Chapter 15 : Laplace’s Equation Revisited 15–3 maps the upper half plane onto the region in question Furthermore, it maps the negative and positive real axis onto the upper and lower slits respectively We now need to find a harmonic function ψ on the upper half plane with boundary conditions ψ = on the negative real axis and ψ = −1 on the positive real axis For example, the function ψ(z) = arg z − π satisfies the requirements (here we take the principal value of the argument) We now need to invert the function f (z) to obtain a harmonic function φ(w) = arg(f −1 (w)) − π in the original region Example 15.2.3 We wish to find a non-constant harmonic function in the slit upper half plane given in Example 14.4.5, with boundary condition φ = on the slit and the real axis Here φ = const can be interpreted as lines of flow past an obstacle Recall that the Schwarz-Christoffel transformation f (z) = (z − 1)1/2 maps the upper half plane onto the region in question We now need to find a harmonic function ψ on the upper half plane with boundary conditions ψ = on the real line For example, the function ψ(z) = Imz satisfies the requirements Note now that f −1 (w) = (w2 + 1)1/2 We therefore obtain the harmonic function φ(w) = Im((w2 + 1)1/2 ) in the original region Here we choose a branch of the square root that is positive for large positive w INTRODUCTION TO COMPLEX ANALYSIS W W L CHEN c W W L Chen, 1996, 2003 This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990 It is available free to all individuals, on the understanding that it is not to be used for financial gains, and may be downloaded and/or photocopied, with or without permission from the author However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners Chapter 16 UNIFORM CONVERGENCE 16.1 Uniform Convergence of Sequences Recall that if a sequence an of complex numbers converges to a, then, given any N ∈ R such that |an − a| < whenever n > N > 0, there exists We can extend this to pointwise convergence in a region D ⊆ C in a natural way A sequence of complex valued functions an (z) defined on D converges pointwise to a function a(z) defined on D if, given any > and any z ∈ D, there exists N ∈ R such that |an (z) − a(z)| < whenever n > N Here the value of N may depend on the choice of z ∈ D Indeed, for any fixed z ∈ D, we simply consider the convergence of the sequence an (z) of complex numbers to the complex number a(z) The region D does not play any essential part in the argument apart from providing the complex numbers z in question In this chapter, we introduce the idea of uniformity to the question of convergence Put simply, uniformity transfers the dependence of N on z to dependence of N only on the region D containing the complex numbers z in question More precisely, we have the following definition Definition Suppose that D ⊆ C is a region We say that a sequence of complex valued functions an (z) converges uniformly in D to a function a(z), denoted by an (z) → a(z) as n → ∞ uniformly in D, if, given any > 0, there exists N ∈ R such that for every z ∈ D, |an (z) − a(z)| < whenever n > N Remark Note that N no longer depends on the choice of z ∈ D Note also that a precise definition can be given by requiring N ∈ R to satisfy sup |an (z) − a(z)| < z∈D whenever n > N 16–2 W W L Chen : Introduction to Complex Analysis Example 16.1.1 Consider the sequence an (z) = z n in the region D = {z : |z| < 1} Note first of all that for every fixed z ∈ D, we have an (z) → as n → ∞ On the other hand, given any > 0, we have, for every z ∈ D, that |an (z) − 0| = |z| < < n n whenever n > 1/ Hence an (z) → as n → ∞ uniformly in D Now consider the same sequence in the region D = C Note that |an (z) − 0| < if and only if n> |z| It is therefore impossible to find a suitable N independent of the choice of z ∈ C Hence an (z) converges to 0, but not uniformly, in C 16.2 Consequences of Uniform Convergence In this section, we show that uniform convergence carries a number of properties of the sequence over to the limit function The following three results concern respectively continuity, integrability and differentiability THEOREM 16A Suppose that for every n ∈ N, the function an (z) is continuous in a region D ⊆ C Suppose further that an (z) → a(z) as n → ∞ uniformly in D Then a(z) is continuous in D Proof Suppose that z0 ∈ D is fixed For every z ∈ D, we have a(z) − a(z0 ) = a(z) − an (z) + an (z) − an (z0 ) + an (z0 ) − a(z0 ), so that (1) Given any (2) |a(z) − a(z0 )| ≤ |an (z) − a(z)| + |an (z) − an (z0 )| + |an (z0 ) − a(z0 )| > 0, there exists N (independent of the choice of z ∈ D) such that |an (z) − a(z)| < and |an (z0 ) − a(z0 )| < whenever n > N We now choose any n > N and consider the function an (z) Clearly this function is continuous at z0 Hence given any > 0, there exists δ > such that (3) |an (z) − an (z0 )| < whenever |z − z0 | < δ Combining (1)–(3), we conclude that |a(z) − a(z0 )| < whenever |z − z0 | < δ, so that a(z) is continuous at z0 Since z0 ∈ D is arbitrary, the result follows Example 16.2.1 Consider the sequence an (z) = z n on the real interval [0, 1] Each function an (z) is clearly continuous in [0, 1] Also an (z) → as n → ∞ if z ∈ [0, 1) and an (1) → as n → ∞, so that the limit function is not continuous in [0, 1] In view of Theorem 16A, it is clear that this discontinuity is caused by the lack of uniform convergence of an (z) in [0, 1] Chapter 16 : Uniform Convergence 16–3 THEOREM 16B Suppose that for every n ∈ N, the function an (z) is continuous in a region D ⊆ C Suppose further that an (z) → a(z) as n → ∞ uniformly in D Then for any contour C lying in D, we have lim n→∞ an (z) dz = C a(z) dz C Proof Note first of all that the integrals exist, since integrability over C is a consequence of continuity in D Suppose now that the contour C has length L Given any > 0, there exists N ∈ R such that for every z ∈ D, |an (z) − a(z)| < /L whenever n > N Then an (z) dz − C a(z) dz ≤ L sup |an (z) − a(z)| ≤ z∈C C whenever n > N THEOREM 16C Suppose that for every n ∈ N, the function an (z) is analytic in a disc D = {z : |z − z0 | < R} Suppose further that an (z) → a(z) as n → ∞ uniformly in Dr = {z : |z − z0 | ≤ r} for every r ∈ [0, R) Then a(z) is analytic in D, and an (z) → a (z) as n → ∞ uniformly in Dr for every r ∈ [0, R) Proof Suppose that T is any triangular path in D We now choose r ∈ [0, R) so that T ⊆ Dr Then a(z) dz = lim an (z) dz = n→∞ T T Here the second equality follows from Cauchy’s integral theorem, while the first equality follows from Theorem 16B, in view of uniform convergence in Dr The assertion that a(z) is analytic in D now follows from Morera’s theorem (Theorem 6G) Suppose next that r ∈ [0, R) is fixed We now choose ρ = (r + R)/2, so that r < ρ < R, and let Cρ denote the circle {ζ : |ζ − z0 | = ρ}, followed in the positive (anticlockwise) direction (the reader is advised to draw a picture) For every z ∈ Dr , we have, by Cauchy’s integral formula, that an (z) − a (z) = 2πi Cρ an (ζ) − a(ζ) dζ (ζ − z)2 Note that for every ζ ∈ Cρ , we have |ζ − z| ≥ ρ − r Also, in view of the uniform convergence of the sequence an (z) in Dρ , we have, given any > 0, there exists N such that for every z ∈ Dρ , |an (z) − a(z)| < (ρ − r)2 ρ whenever n > N It follows that for every z ∈ Dr , we have |an (z) − a (z)| < ρ sup ζ∈Cρ an (ζ) − a(ζ) ≤ (ζ − z)2 whenever n > N Hence an (z) → a (z) as n → ∞ uniformly in Dr Note that Theorem 16C is restricted to discs However, as far as application is concerned, this is not a serious restriction For any point z in an arbitrary domain D ⊆ C, we can always find an open disc D such that z ∈ D ⊆ D, and so we can apply Theorem 16C to the disc D We immediately have the following result THEOREM 16D Suppose that for every n ∈ N, the function an (z) is analytic in a domain D ⊆ C Suppose further that an (z) → a(z) as n → ∞ uniformly in D Then a(z) is analytic in D Furthermore, (k) for every z ∈ D and every k ∈ N, we have an (z) → a(k) (z) as n → ∞ 16–4 W W L Chen : Introduction to Complex Analysis 16.3 Cauchy Sequences Suppose that a sequence of complex numbers an converges to a Then given any N ∈ R such that |an − a| < /2 whenever n > N It follows that > 0, there exists |an − am | ≤ |an − a| + |am − a| < whenever m, n > N Definition We say that a sequence of complex numbers an is a Cauchy sequence if, given any > 0, there exists N ∈ R such that |an − am | < whenever m, n > N In the last section of this chapter, we shall prove the following result THEOREM 16E (GENERAL PRINCIPLE OF CONVERGENCE) A sequence of complex numbers an is convergent if and only if it is a Cauchy sequence In other words, a sequence an of complex numbers is convergent if and only if, given any > 0, there exists N ∈ R such that |an − am | < whenever m, n > N Definition Suppose that D ⊆ C is a region We say that a sequence of complex valued functions an (z) is a uniform Cauchy sequence in D, if, given any > 0, there exists N ∈ R such that for every z ∈ D, |an (z) − am (z)| < whenever m, n > N We have the following important result THEOREM 16F (GENERAL PRINCIPLE OF UNIFORM CONVERGENCE) Suppose that D ⊆ C is a region A sequence of complex valued functions an (z) converges uniformly in D if and only if it is a uniform Cauchy sequence in D Proof It is simple to show that uniform convergence implies uniform Cauchy To prove the converse, note that for every fixed z ∈ D, the sequence of complex numbers an (z) is a Cauchy sequence It follows from Theorem 16E that an (z) converges to a(z), say Since an (z) is a uniform Cauchy sequence in D, it follows that, given any > 0, there exists N ∈ R such that for every z ∈ D, |an (z) − am (z)| < whenever m, n > N Letting m → ∞, we conclude that |an (z) − a(z)| ≤ whenever n > N 16.4 Uniform Convergence of Series Recall that the convergence of a series depends on the convergence of the sequence of partial sums Definition Suppose that D ⊆ C is a region We say that a series of complex valued functions ∞ an (z) n=1 converges uniformly in D if the sequence of partial sums N sN (z) = an (z) n=1 converges uniformly in D We immediately have the following analogues of Theorems 16A, 16B, 16D, 16E and 16F They can be established by applying the earlier results to the sequence of partial sums Chapter 16 : Uniform Convergence 16–5 THEOREM 16G Suppose that for every n ∈ N, the function an (z) is continuous in a region D ⊆ C Suppose further that the series ∞ an (z) n=1 converges uniformly to a function s(z) in D Then s(z) is continuous in D THEOREM 16H Suppose that for every n ∈ N, the function an (z) is continuous in a region D ⊆ C Suppose further that the series ∞ an (z) n=1 converges uniformly to a function s(z) in D Then for any contour C lying in D, we have ∞ an (z) dz = n=1 C s(z) dz C In other words, we can interchange the order of summation and integration THEOREM 16J Suppose that for every n ∈ N, the function an (z) is analytic in a domain D ⊆ C Suppose further that the series ∞ an (z) n=1 converges uniformly to a function s(z) in D Then s(z) is analytic in D Furthermore, for every z ∈ D and every k ∈ N, we have ∞ an(k) (z) = s(k) (z) n=1 In other words, we can interchange the order of summation and differentiation THEOREM 16K (GENERAL PRINCIPLE OF CONVERGENCE) A series ∞ an n=1 of complex numbers converges if and only if, given any > 0, there exists N0 ∈ R such that N2 an < n=N1 +1 whenever N2 > N1 > N0 16–6 W W L Chen : Introduction to Complex Analysis THEOREM 16L (GENERAL PRINCIPLE OF UNIFORM CONVERGENCE) Suppose that D ⊆ C is a region A series ∞ an (z) n=1 of complex valued functions converges uniformly in D if and only if, given any > 0, there exists N0 ∈ R such that for every z ∈ D, N2 an (z) < n=N1 +1 whenever N2 > N1 > N0 We can also establish the following uniform versions of the Comparison test and the Ratio test THEOREM 16M (WEIERSTRASS M -TEST) Suppose that D ⊆ C is a region Suppose further that an (z) is a sequence of complex valued functions such that |an (z)| ≤ Mn for every z ∈ D, where the real series ∞ Mn n=1 of non-negative terms is convergent Then the series ∞ an (z) n=1 converges uniformly (and absolutely) in D Proof Using the Triangle inequality, we have N2 N2 N2 an (z) ≤ n=N1 +1 Given any |an (z)| ≤ n=N1 +1 Mn n=N1 +1 > 0, it follows from Theorem 16K that there exists N0 such that N2 Mn < n=N1 +1 whenever N2 > N1 > N0 It follows that for every z ∈ D, N2 an (z) < n=N1 +1 whenever N2 > N1 > N0 The result now follows from Theorem 16L Chapter 16 : Uniform Convergence 16–7 THEOREM 16N (RATIO TEST) Suppose that D ⊆ C is a region Suppose further that an (z) is a sequence of complex valued functions such that a1 (z) is bounded in D, and an+1 (z) ≤R To see this, note that writing z = x + iy, where x, y ∈ R, we have 1 1 = x+iy = x n−iy = x e−iy log n = x (cos(y log n) − i sin(y log n)), nz n n n n so that 1 = x z n n Since x > 1, the series ∞ x n n=1 of non-negative terms is convergent It follows from the Comparison test that the series (5) converges absolutely Suppose now that δ > is fixed Consider the region D = {z : Rez > + δ} Then for every z ∈ D, we have 1 = x < 1+δ nz n n The series ∞ 1+δ n n=1 16–8 W W L Chen : Introduction to Complex Analysis of non-negative terms is convergent It follows from the Weierstrass M -test that the series (5) converges uniformly in D We comment here that the series (5) is called the Riemann zeta function, and is crucial in the study of the distribution of prime numbers Indeed, the study of this function has led to much of the development in complex analysis Example 16.4.2 In Chapter 10, we discussed the function π cot πz, and showed that it has simple poles at the (real) integers with residue Here we shall make a more detailed study Consider the function ∞ 2z f (z) = + z n=1 z − n2 Let us first of all study this function in the region DR = {z : |z| < R}, where R > is fixed Let N ∈ N satisfy N > 2R, and write f (z) = f1 (z) + f2 (z), where ∞ N f1 (z) = 2z + z n=1 z − n2 and f2 (z) = n=N +1 2z z − n2 Clearly the function f1 (z) is analytic in DR , with the exception of simple poles at the (real) integers in DR Consider next the function f2 (z) in DR For every z ∈ DR and every n > N > 2R, we have 2z 2R 2R 8R ≤ = < z − n2 n − R2 n − (R/n)2 3n It follows from the Weierstrass M -test that the series for f2 (z) converges uniformly in DR , and is analytic in DR in view of Theorem 16J Hence f (z) is analytic in DR , with the exception of simple poles at the (real) integers in DR It follows that f (z) is meromorphic in C, with simple poles at the (real) integers It is easy to check that all these simple poles have residue Note also that we can write f (z) = z n∈Z z2 − n2 We shall show that f (z) = π cot πz For convenience, we shall change notation, and show that (6) n∈Z π cot πa = a2 − n2 a whenever a ∈ Z Consider the function g(z) = π cot πz a2 − z Since the function π cot πz has simple poles at every n ∈ Z with residue 1, and since a ∈ Z, it follows that g(z) has simple poles at every n ∈ Z and at z = ±a, with residues res(g, n) = a2 − n2 and res(g, ±a) = − π cot πa 2a For every N ∈ N, let CN denote the boundary of the rectangular domain z = x + iy : |x| < N + and |y| < N , followed in the positive (anticlockwise) direction If N > |a|, then we have 2πi CN π cot πz dz = a2 − z −N ≤n≤N a2 π cot πa − −n a Chapter 16 : Uniform Convergence 16–9 Clearly (6) will follow if we show that the integral on the left hand side converges to as N → ∞ It can be shown that | cot πz| ≤ coth π for every z ∈ CN Hence for every N > |a|, we have CN 16.5 π cot πz (8N + 2)π coth π π cot πz ≤ dz ≤ (8N + 2) sup →0 − z2 a2 − z a N − |a|2 z∈CN as N → ∞ Application to Power Series Let z, α ∈ C In this section, we shall study series of the type ∞ an (z − α)n (7) (a0 , a1 , a2 , ∈ C), n=0 known commonly as power series THEOREM 16P Suppose that the series given by (7) converges for a particular value z = z0 Then, for every r < |z0 − α|, the series converges uniformly (and absolutely) in the disc Dr = {z : |z − α| ≤ r} Proof Suppose that ∞ an (z0 − α)n n=0 converges Then an (z0 − α)n → as n → ∞, and so there exists M ∈ R such that |an (z0 − α)n | ≤ M for every n ∈ N ∪ {0} For every z ∈ Dr , we have |an (z − α)n | ≤ M z−α z0 − α n ≤M r z0 − α n for every n ∈ N ∪ {0} The result now follows from the Weierstrass M -test, noting that the geometric series ∞ M n=0 r z0 − α n converges THEOREM 16Q (CONVERGENCE THEOREM FOR POWER SERIES) For the power series given by (7), exactly one of the following holds: (a) The series converges absolutely for every z ∈ C (b) There exists a positive real number R such that the series converges absolutely for every z ∈ C satisfying |z − α| < R and diverges for every z ∈ C satisfying |z − α| > R (c) The series diverges for every z = α Sketch of Proof In the notation of Theorem 16P, consider S = {r ≥ : (7) converges absolutely in Dr } Then S contains the number In view of Theorem 16P, S must be an interval with lower end-point 0, so that S = [0, ∞), S = {0} or there exists some positive number R such that S = [0, R) or S = [0, R] The first two possibilities correspond to (a) and (c) respectively, while the last possibility corresponds to (b) 16–10 W W L Chen : Introduction to Complex Analysis Definition The number R in Theorem 16Q is called the radius of convergence of the series (7) We also say that R = if case (c) occurs, and that R = ∞ if case (a) occurs We now show that differentiation of a power series can be carried out term by term, and that the series so obtained converges to the derivative THEOREM 16R Suppose that the power series given by (7) has radius of convergence R > Then it represents an analytic function f (z) in the open disc D = {z : |z − α| < R} Furthermore, the derivatives of f (z) can be obtained by differentiating the series term by term Proof For every r < R, it follows from Theorem 16P that the series converges uniformly in the disc Dr = {z : |z −α| < r} It now follows from Theorem 16J that the series converges to an analytic function f (z) in Dr , and the derivatives of f (z) can be obtained by differentiating the series term by term Since the above holds for any r < R, the result follows Example 16.5.1 Suppose that f (t) is a complex valued function continuous (and so bounded) on the closed real interval [0, 1] Consider the function F (z) = e−zt f (t) dt For any fixed z ∈ C, we have the power series (here t is the variable) e−zt = (8) ∞ (−zt)n , n! n=0 with infinite radius of convergence It follows from Theorem 16P that the series (8) converges uniformly in [0, 1], and so can be multiplied by the bounded function f (t) and integrated term by term, in view of Theorem 16H Hence ∞ (9) F (z) = n=0 ∞ (−zt)n (−z)n f (t) dt = n! n! n=0 tn f (t) dt Furthermore, if |f (t)| ≤ M , where M is a fixed positive number, then 1 tn f (t) dt ≤ M tn dt = M n+1 Suppose now that R > is fixed If |z| < R, then (−z)n n! tn f (t) dt ≤ M Rn (n + 1)! Note that the series ∞ M Rn (n + 1)! n=0 converges, so it follows from the Weierstrass M -test that the series in (9) converges uniformly in the disc {z : |z| < R} By Theorem 16J, the function F (z) is analytic in {z : |z| < R} Since R > is arbitrary, it follows that F (z) is entire Chapter 16 : Uniform Convergence 16.6 16–11 Cauchy Sequences In this section, we shall prove Theorem 16E Clearly a convergent sequence of complex numbers is Cauchy It remains to show that a Cauchy sequence of complex numbers is convergent The proof of this result usually involves the Bolzano-Weierstrass theorem which states that every bounded sequence of complex numbers has a convergent subsequence Here, we shall give a proof without using the Bolzano-Weierstrass theorem Assume, first of all, that the sequence an is real Since an is a Cauchy sequence, it follows that there exists an increasing sequence of natural numbers N1 < N2 < < Np < such that |an − am | < whenever n, m ≥ Np (we simply take 2p = 2−p for every p ∈ N) In particular, we have |aNp+1 − aNp | < 2p for every p ∈ N For every p ∈ N, let bp = aNp − 2p−1 Then bp+1 − bp = aNp+1 − aNp + 1 ≥ p − |aNp+1 − aNp | > 0, 2p so that the sequence bp is increasing Note next that |bp | = aNp − 1 1 ≤ |aNp − aN1 | + |aN1 | + p−1 ≤ + |aN1 | + p−1 , 2p−1 2 so that the sequence bp is bounded Hence the sequence bp converges to L, say, as p → ∞ We now show that an → L as n → ∞ Given any < 2p and > 0, we now choose p ∈ N so large that |bp − L| < Suppose that n ≥ Np Then |an − L| ≤ |an − aNp | + |aNp − bp | + |bp − L| < 1 + p−1 + < p 2 as required Suppose now that the sequence an is complex valued Then we can write an = xn + iyn , where xn , yn ∈ R If an is a Cauchy sequence, then it is easy to see that the real sequences xn and yn are real Cauchy sequences It follows that both xn and yn converge, and so an converges 16–12 W W L Chen : Introduction to Complex Analysis Problems for Chapter 16 Suppose that an (z) → a(z) and bn (z) → b(z) as n → ∞ uniformly in a region D a) Show that an (z) + bn (z) → a(z) + b(z) as n → ∞ uniformly in D b) Suppose that f (z) is bounded in D Show that an (z)f (z) → a(z)f (z) as n → ∞ uniformly in D c) Write f (z) = 1/z and an (z) = 1/n Find a region D such that an (z) converges uniformly in D but an (z)f (z) does not converge uniformly in D For each of the following power series, find a number R such that the series converges for |z| < R and diverges for |z| > R: ∞ n=0 ∞ c) ∞ 2n z n a) n2 z n b) n=1 ∞ n 2n z 2+n n n=1 d) n=0 3n z n + 5n 4n Show that each of the following represents an entire function: ∞ a) ∞ zn (n!)1/2 n=1 b) zn 2n2 n=1 ∞ c) n nz n=1 Show that each of the following functions is meromorphic in C, and find the residues at the poles: ∞ a) ∞ (−1)n n!(n + z) n=0 b) ∞ Show that for every z ∈ Z, we have π = (n + z) sin πz n=−∞ ∞ (z + n)2 n=1 a) Show that except at the poles, we have n=−∞ n2 z π = +z πz ∞ b) By writing the series as 1/z plus a sum over all natural numbers, evaluate ∞ c) By letting z → 0, show that + n2 z n=1 π2 = n n=1 Consider the exponential series ∞ zn n! n=0 which converges for every z ∈ C Suppose further that e(z) is the sum of the series a) Show that the series converges uniformly in the disc DR = {z : |z| < R} for every real number R > b) Suppose that D is a bounded region in C Explain why the series converges uniformly in D c) Show that for every z ∈ C satisfying |z| = R, we have M n=N +1 RM zn ≥ − RM n! M! 1 + + + M −N −1 R R R ≥ RM 1 − M! R − Chapter 16 : Uniform Convergence 16–13 d) Use (c) to show that the series does not converge uniformly in C e) Explain carefully why e(z) is an entire function in C [Remark: In view of the unfavourable conclusion of (d), you should take extra care here.] f) Show that e (z) = e(z) for every z ∈ C and e(0) = g) Let g(z) = e(−z)e(z) Show that g (z) = for every z ∈ C, and deduce that e(−z)e(z) = for every z ∈ C h) Suppose that a ∈ C is fixed By studying the function ga (z) = e(−z)e(z + a), show that e(z + a) = e(z)e(a) for every z ∈ C This question makes use of the function e(z) discussed in Problem Suppose that for every z ∈ C, we write c(z) = e(iz) + e(−iz) and s(z) = e(iz) − e(−iz) 2i a) By using the Taylor series for e(iz) and e(−iz), find the Taylor series for c(z) and s(z) b) Show that c (z) = −s(z) and s (z) = c(z) for every z ∈ C c) By studying the function h(z) = c2 (z) + s2 (z), show that c2 (z) + s2 (z) = for every z ∈ C ... which is of considerable interest in general topology and turns out to be a very useful concept in analysis as well is the concept of a subbasis Definition 1.31 S ⊆ τ is called a subbasis for the... and also, · · ·Ui ⊆ Ui ⊆ Ui+1 · · · 1.1 Compactness in metric space Many existence theorems in analysis depend on some set being compact Therefore, it is important to be able to identify compact