# II. Integration and Cauchy s Theorem

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1 MTH6111 Complex Anlysis Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve. We follow Priestley s terminology. Definitions let U be n open set in C. A curve in C is continuous function : [, β] U, where [, β] is closed intervl in R. The imge of is = {(t) : t [, β]} U. Two curves 1 : [ 1, β 1 ] U nd 2 : [ 2, β 2 ] U cn be joined if 1 (β 1 ) = 2 ( 1 ). If this condition is stisfied we cn reprmetrize 2 s 2 : [β 1, β 1 + β 2 2 ] U by setting 2 (t) = 2 (t + 2 β 1 ) nd define the join of 1 nd 2 to be the pth : [ 1, β 1 + β 2 2 ] U which sends t 1 (t) if t [ 1, β 1 ] nd t 2 (t) if t [β 1, β 1 + β 2 2 ]. is smooth if it is C 1 (tht is, differentible, with its derivtive continuous). A pth is the join of finitely mny smooth curves (tht is, it is piecewise C 1 ). A complex-vlued function on rel intervl [, β] cn be written g = Re(g) + iim(g). We sy tht g is integrble if ech of the rel functions Re(g) nd Im(g) re integrble, nd we define: g = Reg(g) + i Im(g) We ssume the definition of rel integrl nd stndrd properties. In prticulr, if f : [, β] R is continuous, then f is integrble on [, β], nd if f 1 nd f 2 re continuous functions [, β] R then (f 1 + f 2 ) = f 1 + f 2. If : [, β] C is pth, so there exist points = t 0 < t 1 <... < t n = β such tht is smooth on ech [t j, t j+1 ], nd if f : C is continuous, we define n 1 f(z)dz = j=0 tj+1 t j f((t)) (t)dt In prticulr, if is C 1 we set f(z)dz = f((t)) (t)dt 1

2 1.1 Lemm If : [, β] C is pth nd f : C is continuous, then: (i) f(z)dz = f(z)dz, where ( )(t) = ( + β t) for t [, β]. (ii) If is the join of 1 nd 2 then f = 1 f + 2 f. (iii) Let = ψ, where ψ is C 1 function [, β] [, β] with ψ (t) > 0 for ll t [, β] (tht is, is reprmetriztion of ), then f(z)dz = f(z)dz. Proof (i) ( ) (t) = ( + β t), so f(z)dz = f(( + β t)) ( + β t)dt = (where s = + β t). s= s=β f((s)) (s)( ds) = f((s) (s)ds (ii) is immedite from the property of rel integrls tht c = b + c if < b < c. b (iii) f(z)dz = β f( (t)) (t)dt = β f((ψ(t))) (ψ(t))ψ (t)dt = f((s)) (s)ds = f(z)dz (where s = ψ(t)). Note tht (iii) tells us the integrl depends only on, not on the prticulr prmetriztion of the pth. In fct we could define f(z)dz to be the limit of the sum n f(z j )(z j z j 1 ) 1 over finer nd finer prtitions () = z 0 < z 1 <... < z n = (β) of. Now setting z = (t), so tht dz = (t)dt, we see tht this new definition grees with f((t)) (t)dt. We choose the ltter s our definition of f(z)dz for this course becuse we lwys compute pth integrls using this formul, nd it lso llows us to pply the theory of integrls of rel functions without hving to reprove everything from the beginning for complex functions. Exmple Let : [0, 2π] C be defined by (t) = + re it. This is pth which goes once nticlockwise round circle with centre nd rdius r. We denote this pth by C(, r). Then (z ) n dz = 2π r n e int rie it dt = 2π C(,r) 0 0 which is esily computed to be 0 if n 1 nd to be 2πi if n = 1. ir n+1 e (n+1)it dt 2

3 2. The fundmentl theorem of clculus In rel nlysis, this is the theorem tht integrtion is the opposite of differentition, tht is (i) If F is differentible then b F (x)dx = F (b) F (). (ii) If f is continuous then F (x) = x f(t)dt is differentioble, nd F (x) = f(x). In the complex cse we hve n nlogue of (i): 2.1 Fundmentl theorem of clculus Let be pth [, β] C, nd let F : U C be holomorphic, where U is n open set contining. Then F (z)dz = F ((b)) F (()) In prticulr, if is closed pth (this mens () = (β)) then F (z)dz = 0. Proof If we cn prove the result when is smooth, we cn deduce it for the cse tht is piecewise smooth, for then we just prtition [, β] up into = t 0 < t 1 <... < t n = β with smooth on ech [t j, t j+1 ], observe tht t j+1 t j F (z)dz = F ((t j+1 )) F ((t j )) by the result for smooth pths, nd dd up. So suppose is smooth. F (z)dz = F ((t)) (t)dt = (F ) (t)dt = Re(F ) (t)dt + i Im(F ) (t)dt = [Re(F )(t) + iim(f )(t)] β (by the fundmentl theorem of clculus for rel functions) = F ((β)) F (()) It is not so esy to formulte complex nlogue of prt(ii) of the fundmentl theorem of clculus for rel functions. The problem is tht z f(z)dz will in generl depend on the pth chosen in C between the points nd z. However Cuchy s Theorem will tell us tht if f is holomorphic then ny two pths from to z which re homotopic to one nother (in other words ech cn be deformed to the other) give the sme vlue for this integrl. 3. Lengths nd the estimtion lemm Definition The length of the pth [, β] C is defined to be Length() = (t) dt 3

4 Exmples (see exercise sheet) The length of the stright line pth from z 0 to z 1 is z 1 z 0. The length of C (,r) is 2πr. The length of pth is unchnged by reprmetriztion of the pth. 3.1 Estimtion Lemm Let f : U C be continuous (where U is some subset of C), let be pth in U, nd suppose f(z) < M for ll z. Then f(z)dz ML. Proof We first prove tht for rel-vlued functions u nd v of t R, ( ) b (u(t) + iv(t))dt To prove this, write b (u(t) + iv(t))dt = Reiφ. Then b b (u(t) + iv(t))dt = e iφ (u(t) + iv(t))dt = b b u(t) + iv(t) dt e iφ (u(t) + iv(t))dt = b Re(e iφ (u(t) + iv(t))dt (since the left-hnd side is rel number nd by definition g = Re(g) + i Im(g) for ny g : R C) b e iφ (u(t) + iv(t)) dt (since Re(z) z for ny z C) = b u(t) + iv(t) dt Now, pplying (*) to f = f((t)) (t)dt, we hve f b f((t)) (t) dt = f((t)). (t) dt M (t) dt = ML. We shll use the Estimtion Lemm mny times in this course. As first exmple we investigte when we cn integrte power series term by term (recll we hve lredy proved we cn differnetite term by term if the series is bsolutely convergent). A sequence (F n ) n 0 of continuous complex-vlued functions on some subset S C is sid to converge uniformly on S to function F : S C if given ny ɛ > 0 there exists n N such tht n > N, z S, F (z) F n (z) < ɛ. Another wy to sy this is tht sup z S F (z) F n (z) tends to zero s n tends to. 4

5 3.2 Proposition Let be pth in U C nd (f n ) n 0 be sequence of continuous functions U C such tht n=0 f n converges uniformly on to function F. Then ( f n ) = f n n=0 n=0 Proof Let F n = n j=0 f j. Then sying tht the series n=0 f n converges uniformly on to F mens tht the sequence (F n ) n 0 converges uniformly on to F, in other words tht So F n j=0 f j = F sup z F (z) F n (z) 0 s n. n j=0 which tends to zero s n tends to infinity. Thus F = lim f j = (F F n ) L( ).sup z F (z) F n (z) n n j=0 f j In prctice, we test for uniform convergence using the following criterion: 3.3 Weierstrss M-test Let (f n ) n 0 be complex-vlued functions defined on S C, nd suppose for ech n there exists positive rel number M n such tht sup z S f n (z) M n. If n=0 M n is convergent, then n=0 f n is uniformly convergent on S. Proof We omit the proof, but it s not hrd, nd cn be found in mny textbooks. 3.4 Proposition Let n=0 nz n hve rdius of convergence R > 0. Then for every r with 0 < r < R the series is uniformly convergent on the closed disc D(0, r). Proof For ny z D(0, r), we hve n z n n r n. Since n=0 nr n is convergent (s r < R) the result now follows by the M-test. 3.5 Corollry n=0 nz n is uniformly convergent on ny compct subset S of D(0, R). Proof z z is continuous function, so, since S is compct, sup z S z is ttined by some z S. Therefore sup z S z < R. Thus S D(0, r) for some r < R. The result now follows from 3, Corollry 0 nz n = 0 nz n for ny pth in D(0, R), where R is the rdius of convergence of the power series. Proof is compct, since [, β] is compct nd is continuous. The result now follows from 3.5 nd

6 4. Cuchy s Theorem Cuchy s Theorem for tringle If f is holomorphic inside nd on tringle then f(z)dz = 0. Proof Let be the (nticlockwise) tringle with vertice u, v, w nd let f(z)dz = I Subdivide the tringle into four smller tringles 0, 1, 2, 3 by tking new vertices u, v, w the midpoints of the edges of the tringle uvw. Orient ech j in n nticlockwise sense. Then f + f + f + f = f = I since the integrls long internl edges cncel. So t lest one of the smll tringles, cll it 1, hs f I Subdivide 1 by tking the midpoints of its edges, nd repet the procedure we just pplied to. Inductively we obtin sequence of tringles = 0, 1, 2,... such tht (i) for ech n, n+1 n, where n is the closed tringulr region hving n s its boundry. (ii) length( n )= 2 n L, where L =length(). (iii) n f 4 n I. The intersection n=0 n contins point z = (to prove this tke ny point z n in ech n : then (z n ) n 0 is Cuchy sequence nd so it hs limit, but the n s re closed so this limit is in ll of them). The function f is holomorphic, so it is differentible t z = nd hence given ny ɛ > 0 there exists δ > 0 such tht tht is, f(z) f() f () < ɛ z D(, δ) z (1) f(z) f() (z )f () < ɛ z z D(, δ) Now choose N such tht N Note tht D(, δ) (certinly possible s the n s get smller s n tends to ). (2) z 2 N L z N 6

7 We lso know tht (3) N (f() + (z )f ())dz = 0 by the fundmentl theorem of clculus (since the integrnd is polynomil of degree one in z nd therefore hs n ntiderivtive). Putting (1) nd (2) together we see tht (4) sup f(z) f() (z )f () < ɛ2 N z N nd so from (3) we get f(z)dz = N (f(z) f() (z )f ())dz ɛ2 N L length( N ) = ɛ(2 N L) 2 = ɛ4 N L 2 N But now from (iii) we deduce tht I < ɛl 2. Finlly, since ɛ is rbitrry nd I does not depend on ɛ, we deduce tht I = 0. A subset S C is clled convex if given ny two points nd b in S the line segment [, b] is contined in S, where 4.2 Theorem (Existence of Antiderivtives) [, b] = {tb + (1 t) : t R, 0 t 1} Let U be convex open subset of C nd let f be holomorphic on U. Then there exists holomorphic function F on U such tht F = f. Proof Choose point U, nd for ech z U define F (z) = f(w)dw We must prove F is differentible, with derivtive f. Given ny z U, z + h is lso in U for h sufficiently smll (s U is open). By 4.1, F (z + h) F (z) = f(w)dw So F (z + h) F (z) h [,z] = 1 h f(w)dw By continuity of f, given ɛ > 0 there exists δ > 0 such tht f(w) f(z) < ɛ whenever w z < δ. So for h < δ we hve (f(w) f(z))dw < ɛ. h 7

8 by the Estimtion Lemm. Since f(z) is constnt when we integrte with respect to w, by the fundmentl theorem of clculus we know tht f(z)dw = f(z) 1dw = f(z).[w] z+h z so the inequlity t the bottom of the previous pge gives us f(w)dw f(z).h < ɛ. h tht is, F (z + h) F (z) f(z) < ɛ h = f(z).h whenever h < δ. In other words F is differentible t z, with derivtive f(z).. Comment The theorem bove is the nerest nlogue tht we hve for complex functions of prt (ii) of the fundmentl theorem of clculus for rel functions. 4.3 Corollry (Cuchy s Theorem for convex region) Let U be n open convex subset of C nd f be holomorphic on U. Then f(z)dz = 0 for every closed pth in U (where : [, β] U is sid to be closed pth if () = (β)). Proof By 4.2 f hs n ntiderivtive F. f(z)dz = F ((β)) F (()) = 0. Now by the fundmentl theorem of clculus 2.1 we hve Finlly in this section, we consider (without proof) some more generl forms of Cuchy s Theorem. We shll use vrious of these sttements in the next section nd lter in the course. Definition A simple closed curve is continuous : [, β] C with () = (β) but with no other pirs of distinct t, t such tht (t) = (t ). The Jordn Curve Theorem sttes tht if is simple closed curve then C \ is the disjoint union of two open sets, I(), the inside (which is bounded), nd O(), the outside (which is unbounded). Comment In fct the Theorem sys more bout O() nd I(), for exmple tht they both re connected sets, nd tht there is continuous bijection between O() nd the open unit disc. The Jordn Curve Theorem is fmously hrd to prove correctly: most proofs use complex nlysis or lgebric topology. 4.4 Cuchy s theorem for simple closed pth If f is holomorphic everywhere on nd inside simple closed pth, then f(z)dz = 0. Definition Two closed curves 0 : [0, 1] U C nd 1 : [0, 1] U C re sid to be homotopic if 0 cn be continuously deformed to 1 through fmily of closed curves s in U. More precisely if there 8

9 exists continuous function H : [0, 1] [0, 1] U such tht H(0, t) = 0 (t) t [0, 1] H(1, t) = 1 (t) t [0, 1] H(s, 0) = H(s, 1) s [0, 1] (Then for ech s [0, 1], s : [0, 1] U is the closed curve defined by s (t) = H(s, t).) Two pths 0 : [0, 1] U C nd 1 : [0, 1] U C between the sme pir of points nd b re sid to be homotopic reltive to their end points if one cn be deformed to the other keeping their endpoints fixed, tht is, if there exists continuous function H : [0, 1] [0, 1] U such tht H(0, t) = 0 (t) t [0, 1] H(1, t) = 1 (t) t [0, 1] H(s, 0) = nd H(s, 1) = b s [0, 1] 4.5 Deformtion Principle (i) If 0 nd 1 re homotopic closed pths in U nd f is holomorphic on U then 1 f(z)dz = 2 f(z)dz. (ii) If 1 nd 2 re pths in U between the sme points nd b which re homotopic in U reltive to their end points, nd f is holomorphic on U, then 1 f(z)dz = 2 f(z)dz. An open set U C is clled pth-connected if every pir of points in U cn be joined by some curve in U. It is clled simply-connected if it is pth-connected nd every closed curve in U is homotopic to constnt curve [, β] {point}. 4.6 Cuchy s Theorem for simply-connected regions If f is holomorphic on simply-connected region U then f(z)dz = 0 for every closed pth in U. Comment Theorem 4.6 generlises 4.3 from convex region to ny simply-connected region. 4.7 Cuchy s Theorem for multiply-connected region If is simple closed pth, nd 1,..., n re disjoint simple closed pths in the interior of, such tht the interiors of 1,..., n re disjoint, then if f is holomorphic on, nd on 1,..., n, nd on the region between nd 1... n, we hve f(z)dz = f(z)dz f(z)dz 1 n (provided nd 1,..., n re ll prmetrized in the the sme direction e.g. positively, tht is, nticlockwise.) 9

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