MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE Contents 1. Introduction 1 2. A fr-reching little integrl 4 3. Invrince of the complex integrl 5 4. The bsic complex integrl estimte 6 5. Comptibility 8 6. Comptibility of rc length 9 7. Existence of the integrl 11 1. Introduction The dt for complex pth integrl f(z) dz nd for relted integrl f(z) dz re s follows. Ω C is region, f : Ω C is continuous function, : [, b] Ω is continuous pth. However, the ssumption tht the pth is continuous is not strong enough to gurntee tht these integrls re sensible. This writeup discusses two wys to ddress this issue, nd how they relte. The first pproch to complex pth integrls is tht is ssumed to be piecewise C 1. This pproch is more thn dequte for every computtion tht we will do, becuse our pths of integrtion lwys will conctente finitely mny line segments nd circulr rcs. In this cse, ssuming without loss of generlity tht is C 1 by working with its pieces one t time, the complex integrl cn be treted s the integrl of differentil form, b f(z) dz f((t)) (t) dt. Here the integrnd f((t)) (t) is complex-vlued, but we simply work with it componentwise. Tht is, for continuous function ϕ(t) U(t) + iv (t) on [, b], the inevitble definition is b ϕ(t) dt b U(t) dt + i b V (t) dt. In our cse, the integrnd ϕ(t) f((t)) (t), with f u + iv nd x + iy, hs components 1

2 MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE U ux vy nd V vx uy, with u u(x(t), y(t)), v v(x(t), y(t)), x x (t), nd y y (t). Similrly, the pproprite definition here is b f(z) dz f((t)) (t) dt, this time integrting the function ϕ U + iv where U u x 2 + y 2 nd V v x 2 + y 2. As prticulr cse of the second integrl, the length of is defined s b length() dz (t) dt. t The second pproch to complex pth integrls is tht is ssumed to be rectifible. Here rectifible mens tht hs finite rc length, with rc length defined in nturl wy; this will be explined ust below. Now the integrl definitions re n f(z) dz f((c ))((t ) (t 1 )) nd f(z) dz 1 1 n f((c )) (t ) (t 1 ), with the its being tken over prtitions P {t } of [, b], nd corresponding smples S P {c }, t 0 c 1 t 1 c 2 t 2 t n 1 c n t n b with the t distinct, nd the mesh of prtition being mesh(p ) mx 1 n (t t 1 ). The fct tht these its exist needs to be estblished. In this setting, the rc length of is n length() dz (t ) (t 1 ). 1 The ssumption tht is rectifible is tht the sums in the previous disply re bounded bove. These sums constitute the set of lengths of polygons inscribed in. If is ssumed to be C 1, so tht b f(z) dz f((t)) (t) dt, f(z) dz b f((t)) (t) dt, nd we freely extend to the piecewise C 1 cse, some questions present themselves. Invrince: Do these integrls depend on the prmetriztion of? A short clcultion with the chin rule shows tht the integrl is invrint under order-preserving reprmetriztion. Tht is, ll tht mtters is the direction of pth-trversl. As for wht hppens when the direction is

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE 3 reversed, for ny pth, let denote the sme pth but trversed in the opposite direction. Then unsurprisingly, f(z) dz f(z) dz, but on the other hnd, f(z) dz f(z) dz. Size estimtes: How do the two nonnegtive rel numbers f(z) dz nd f(z) dz compre? A slightly clever rgument shows tht f(z) dz f(z) dz. Trdeoffs: Wht re the dvntges of defining complex pth integrls in wy tht depends on hving C 1 -pths? As lredy mentioned, defining the complex contour integrl for piecewise C 1 pths lets us compute every exmple tht we need. The sitution is different when is ssumed to rectifible. Invrince. The integrl s invrince under monotoniclly incresing reprmetriztions is essentilly wired into its definition. Prtitions pss through such reprmetriztions, preserving the property of their meshes going to zero. Size estimtes. Similrly, the estimte f(z) dz f(z) dz is essentilly utomtic, in consequence of the tringle inequlity. Trdeoffs. The topology theory tht underlies complex nlysis ddresses questions of deforming one pth to nother through succession of pths, nd those pths re known only to be continuous; it is convenient not to worry whether they re piecewise C 1, lthough still we hve to worry whether they re rectifible. We mke two more comments before moving on to specifics. First, the existence of the integrls being discussed here is substntive question regrdless of whether is C 1 or only rectifible. If is C 1 then the existence relies on the existence of the integrl of continuous rel-vlued function over compct intervl. This existence must be invoked until person is redy to pprecite tht it relies on the continuity being uniform, in consequence of the intervl being compct. On the other hnd, if is rectifible nd f is continuous then the existence of the integrls f(z) dz nd f(z) dz doesn t reduce to the rel cse. Insted vrint existence rgument is required, not using the notions of lower sum nd upper sum, becuse the complex number system is not ordered. This vrint rgument gin boils down to uniform continuity. It will be given below. Second, mintining two notions of complex pth integrls rises the question of their comptibility. Tht is, do the it of sums definitions of f(z) dz nd f(z) dz reduce to their differentil form definitions when the rectifible pth is further C 1? They do, nd confirming so is good prctice with beginning rel

4 MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE nlysis technique. In prticulr, the two notions of the length of C 1 curve gree, but one inequlity between them is esier to show thn the other. It is recommended conceptul exercise to speculte which inequlity should be the esy one, nd then to find the nice geometric proof tht it is. Here the relevnt technicl skill is the Cuchy Schwrz inequlity. By contrst, the proof of the hrder inequlity is more nlytic. We will estblish both inequlities below. 2. A fr-reching little integrl Before going into generlities, we work prticulr pth integrl tht lies t the hert of complex nlysis. Our dt re s follows. r is ny positive rel number, possibly very lrge nd possibly very smll, r is the circle of rdius r centered t the origin, trversed once counterclockwise, n is ny integer, nd f n (z) z n. This function is undefined t z 0 if n is negtive. The nturl prmetriztion of r is nd so the integrl of f n over r is Tht is, the integrl r : [0, 2π] C, r (t) re it, r f n (z) dz r z n dz 2π t0 2π t0 2π t0 f n ((t)) r(t) dt (re it ) n d(re it ) r n e int ire it dt 2π ir n+1 e i(n+1)t dt t0 { ir n+1 2π if n 1, 0 otherwise { 2πi if n 1, 0 otherwise. { 2πi if n 1, 0 otherwise is independent of r nd nerly independent of n. The preceding formul hs enormous consequences. For exmple, nïvely ssuming tht some function f hs representtion in integer powers of z, f(z) n z n, n nd nïvely ssuming tht the sum psses through integrtion over r, it follows tht integrting f over r (for ny suitble r > 0) picks off the coefficient 1

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE 5 of 1/z in f nd ignores everything else, 1 f(z) dz 1. 2πi Mking these ides precise requires some cre, nd it isn t quite this simple, but things pretty much work out s the clcultion here suggests. 3. Invrince of the complex integrl Let Ω be region in C, let f : Ω C be continuous function, nd consider two rectifible continuous curves in Ω, : [, b] Ω, : [c, d] Ω. Suppose tht is n orienttion-preserving reprmetriztion of, mening tht there exists continuous incresing biection r : [, b] [c, d] such tht r. Let P denote ny prtition of [, b] with subordinte smple S, P {t 0,..., t n }, S {c 1,..., c n }, nd similrly for [c, d], with the sme number of prtition points, P { t 0,..., t n }, S { c1,..., c n }, The integrls of f over the two curves re by definition f(z) dz f((c ))((t ) (t 1 )). nd f(z) dz mesh( P ) 0 f( ( c ))( ( t ) ( t 1 )). We show tht essentilly by definition, the two integrls re equl. The bsic ide is tht prtition-smple pirs for [c, d] nd the prtition-smple pirs for [c, d] re in biective correspondence vi r nd r 1, ( P, S) (r(p ), r(s)) nd (P, S) (r 1 ( P ), r 1 ( S)). Thus the sets of Riemnn sums for the two integrls re the sme. For exmple, ech term ( t ) where t P is ( t ) (r(t )) (t ) where t P, nd similrly ech (t ) where t P is ( t ) where t P. The nice little technicl point here is tht the inverse biection r 1 is lso continuous. To show this, it suffices to show tht r tkes closed sets to closed sets. But in this context, closed nd compct men the sme thing, nd indeed r tkes compct sets to compct sets becuse the continuous imge of compct set is compct. Now, becuse r nd r 1 re uniformly continuous, it follows tht if {(P m, S m )} nd {( P m, S m )} re prtition-smple sequences relted to ech other vi r nd r 1 then {mesh(p m )} 0 {mesh( P m )} 0.

6 MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE These considertions show tht the two integrls re equl, f(z) dz f((c ))((t ) (t 1 )) mesh( P ) 0 f( ( c ))( ( t ) ( t 1 )) f(z) dz. If the curves nd re C 1 then the invrince result is utomtic, using the chin rule for the third equlity to follow, nd then the chnge of vrible theorem from one vrible clculus for the fifth, the ltter theorem used twice becuse of the complex integrnd, f(z) dz b b b b d c (f ) (f r) ( r) (f r) ( r) r (((f ) ) r) r (f ) f(z) dz. But this immedite rgument is subsumed by the rgument for rectifible continuous pths. We show the following result. 4. The bsic complex integrl estimte Let Ω be region, let f : Ω C be continuous function, nd let : [, b] Ω be rectifible pth. Then f(z) dz f(z) dz. To see this, recll tht the integrl is the it of Riemnn sums, nd compute (using the fct tht the bsolute vlue function is continuous nd using the tringle

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE 7 inequlity) tht f(z) dz 1 Truly, this is ll there is to it. 1 f(z) dz. n f((c ))((t ) (t 1 )) n f((c ))((t ) (t 1 )) 1 n f((c )) ((t ) (t 1 )) Here lso is the stndrd rgument (in our text, for exmple) given the stronger hypothesis tht is C 1 -pth. This rgument is required if one hs defined the complex integrl only for (piecewise) C 1 -pths, by prmetriztion, rther thn for rectifible curves, by Riemnn sums. It is striking, t lest to the uthor of this note, tht the following proof, despite hving stronger hypotheses thn the previous one, is more complicted. In this instnce, voiding the Riemnn sum definition of the integrl mkes things hrder rther thn esier. The comprble result in the rel setting, b b ϕ(t) dt ϕ(t) dt, (ϕ : [, b] R integrble), t t is esy: simply integrte the reltion ϕ(t) ϕ(t) ϕ(t) over the intervl [, b]. And the complex cse should be essentilly no hrder. However, reducing the complex cse to the rel cse poses two obstcles: first, f(z) tkes complex vlues, nd second, dz is complex differentil. The following rgument ddresses these issues one t time, first reducing the problem to the cse of complex-vlued function to tht of rel-vlued one, ssuming tht the differentil is rel-vlued, nd then using the definitions dz (t) dt, dz (t) dt to reduce the cse of complex differentil to tht of rel one. So, begin by considering continuous complex-vlued function on rel intervl, The c is tht b t ϕ : [, b] C. ϕ(t) dt b t ϕ(t) dt. If b t ϕ(t) dt 0 then the c holds, so we my tke b t ϕ(t) dt reiθ, r > 0. Thus b b b ϕ(t) dt r e iθ re iθ e iθ ϕ(t) dt e iθ ϕ(t) dt t b t t t (Re(e iθ ϕ(t)) + iim(e iθ ϕ(t))) dt.

8 MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE But the integrl is rel, so its imginry prt is zero, leving us in position to quote the inequlity from the rel cse nd then quote the fct tht the size of the rel component is t most the size of the complex number, b b b ϕ(t) dt Re(e iθ ϕ(t)) dt Re(e iθ ϕ(t)) dt t t t t b t e iθ ϕ(t) dt Now the generl result follows. Let ϕ (f ). Then b b f(z) dz ϕ(t) dt t ϕ(t) dt t b f((t)) (t) dt 5. Comptibility b t ϕ(t) dt. f(z) dz. Let Ω C be region, let f : Ω C be continuous function, nd let : [, b] Ω be C 1 pth. Recll our two definitions of the integrl of f over, the first using the derivtive of nd then definition of the Riemnn integrl over rel intervl, f(z) dz b f((t)) (t) dt f((c )) (c )(t t 1 ) nd the second mking no reference to the derivtive, f(z) dz f((c ))((t ) (t 1 )). Note tht the Riemnn integrl definition here is not the version tht uses lower nd upper sums, but insted defines the integrl s common it over ll sequences of prtitions whose meshes go to zero. The next section will show tht this integrl exists, in consequence of showing tht complex integrl exists with no ssumption of differentibility. Here, tking the mesh definition of the rel Riemnn integrl for grnted, we sketch the rgument tht the two definitions ust given for the complex integrl re comptible. Prt of the summnd in the first definition is, letting (x, y), (c )(t t 1 ) ( x (c ) + iy (c ) ) (t t 1 ), while two pplictions of the Men Vlue Theorem show tht prt of the summnd in the second definition is (t ) (t 1 ) ( x (d ) + iy (e ) ) (t t 1 ), for some d, e (t 1, t ). Thus the difference of the summnds is f((c )) ( x (c ) x (d ) + i(y (c ) y (e )) ) (t t 1 ). Becuse f is continuous nd the trce of is compct (it is the continuous imge of the compct set [, b]), f is bounded on the trce of. Also, becuse is C 1, its component function derivtives x nd y re continuous on [, b], nd becuse

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE 9 [, b] is compct, they re uniformly continuous there. Therefore, given ε > 0, if the prtition P is fine enough then for ech, f((c )) ( x (c ) x (d ) + i(y (c ) y (e )) ) (t t 1 ) < ε(t t 1 ). b This mkes the two sums within ε of ech other. Thus the integrls re equl. 6. Comptibility of rc length Let Ω C be region, nd let : [, b] Ω be C 1 pth. Recll our two definitions of the length of, the first using the derivtive of, length() b (t) dt, nd the second being the supremum of inscribed polygonl pth-lengths, mking no reference to the derivtive, length() sup (t ) (t 1 ). P We show tht the definitions re comptible. It is not hrd to estblish tht the integrl of is t lest the length of ny inscribed polygonl pth, becuse these lengths grow under refinement nd the integrl is conceptully their it. Indeed, tke prtition of [, b], P {t 0, t 1,..., t n }, nd ssume tht no consecutive pir of division points t 1 nd t hve the sme imge under. (If (t 1 ) (t ) then the pir contributes nothing to the length of the polygonl pth, nd so we my drop its second point from the overll clcultion.) Fix ny {1,..., n}. Consider the unit vector in the direction between the th pir of consecutive polygon points, ˆv ((t ) (t 1 ))/ (t ) (t 1 ). The trivil estimte ˆv (t) ˆv (t), then the Cuchy Schwrz inequlity ˆv (t) ˆv (t) (t) give the inequlity in the clcultion (t ) (t 1 ) ˆv ((t ) (t 1 )) ˆv t t 1 (t) dt t t 1 ˆv (t) dt t t 1 (t) dt. Tht is, the th inscribed polygonl segment length is t most the th piece of the integrl. Sum over to get the inequlity b (t ) (t 1 ) (t) dt. This holds for ny prtition P, nd the right side is independent of P. It follows tht b (1) sup (t ) (t 1 ) (t) dt. P The opposite inequlity is more delicte. The ide is to get polygonl pthlengths s close to the integrl of s desired. The rgument to follow cn prove

10 MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE both inequlities, but the esier direction deserved its correspondingly smoother proof. The derivtives x (t) nd y (t) of the component functions of (t) re continuous, nd their domin [, b] is compct, so they re uniformly continuous on their domin. Thus, given ny ε > 0, there exists some δ > 0 so tht { } t, t [, b], mx{ x ( t) x (t), y ( t) y (t) } < t t < δ ε 4(b ). So if P prtitions [, b] more finely thn δ then for ny {1,..., n} nd for ny s, s (t 1, t ), the reverse tringle inequlity gives ( x (s ), y ( s ) ) ( x (t ), y (t ) ) ( x (t ) x (s ), y (t ) y ( s ) ) ( x (t ), y (t ) ) ( x (t ) x (s ), y (t ) y ( s ) ) > ( x (t ), y (t ) ) ε 2(b ) ε (t ) 2(b ). Now compute for ny, using the Men Vlue Theorem twice t the first step nd using the previous clcultion t the lst step, tht (t ) (t 1 ) ( x (s ), y ( s ) ) (t t 1 ) for some s, s (t 1, t ) ( x (s ), y ( s ) ) (t t 1 ) > ε (t ) (t t 1 ) 2(b ) (t t 1 ). Sum over to get (t ) (t 1 ) > (t ) (t t 1 ) ε 2. But if the prtition is fine enough, then by the definition of the Riemnn integrl we lso hve b (t ) (t t 1 ) > (t) dt ε 2. Combining the lst two displys gives nd it follows trivilly tht (t ) (t 1 ) > sup P Becuse this holds for ll ε > 0, (2) sup P b (t ) (t 1 ) > b (t ) (t 1 ) Equtions (1) nd (2) together give the result. (t) dt ε, b (t) dt ε. (t) dt.

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE 11 7. Existence of the integrl The outstnding issue is tht for region Ω C, continuous function f : Ω C, nd rectifible continuous pth : [, b] Ω, the integrl f(z) dz exists. Recll tht prtition of [, b] is set where P {t 0, t 1,..., t n } t 0 < t 1 < < t n b. The number n 0 cn vry from prtition to prtition. The mesh of P is the mximum of the lengths of the subintervls determined by P, mesh(p ) mx{t t 1 }. Refining the prtition P cnnot increse its mesh. A smple subordinte to P is set S P {c 1,..., c n } where t 0 c 1 t 1 t n 1 c n t n. As bove, the definition of the integrl of f over is very generl it, if it exists, n f(z) dz f((c ))((t ) (t 1 )). 1 The sum on the right is denoted Σ(P, S P ). Existence Theorem. Let Ω C be region, let f : Ω C be continuous, nd let : [, b] Ω be continuous nd rectifible. Then the integrl f(z) dz exists. Proof. Tke sequence of prtitions {P 1, P 2,..., P N,... } such tht N mesh(p N ) 0. We need to show tht the it N Σ(P N, S PN ) exists independently of the prticulr sequence {P N } of prtitions used nd independently of the smple S PN chosen for ech prtition P N. Becuse [, b] is compct nd is continuous, the trce of, ˆ {(t) : t [, b]} is compct. Consequently f is uniformly continuous on ˆ. Let ε > 0 be given, nd let L denote the length of. Then there exists some δ > 0 such tht for ll z, z ˆ, z z < δ f(z) f(z ) < ε 2L. Also becuse [, b] is compct, is uniformly continuous on [, b], so there exists some ρ > 0 such tht for ll t, t [, b], t t < ρ (t) (t ) < δ.

12 MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE Consequently, for ll t, t [, b], Tke n index Ñ lrge enough tht t t < ρ f((t)) f((t )) < ε 2L. mesh(p N ) < ρ for ny N > Ñ. C: For ny N > Ñ, for ny refinement Q of P N, nd for ny smple S Q, Σ(P N, S PN ) Σ(Q, S Q ) < ε 2. To prove this, note tht corresponding to ech th term of Σ(P N, S N ), f((c ))((t ) (t 1 )), we hve th sum of terms in Σ(Q, S Q ), m f((c k))((t k) (t k 1)), k1 with t 0 t 1 nd t m t. The th term f((c ))((t ) (t 1 )) of Σ(P N, S PN ) rewrites s telescoping sum, m f((c ))((t k) (t k 1)). Thus, k1 th term in Σ(P N, S PN ) th sum in Σ(Q, S Q ) m (f((c )) f((c k)))((t k) (t k 1)) k1 m f((c )) f((c k)) (t k) (t k 1 ) k1 < ε 2L length( [t 1,t ]), where we hve used the fct tht mesh(p N ) < ρ t the lst step. Now sum over to get Σ(P N, S PN ) Σ(Q, S Q ) < ε 2L length() ε 2. This proves the c. Now, the c proves tht for ny N nd M greter thn Ñ, Σ(P N, S PN ) Σ(P M, S PM ) < ε. To see this, let Q be the common refinement of P N nd P M, mening their union, nd let S Q be ny corresponding smple. Then by the c nd the tringle inequlity, Σ(P N, S PN ) Σ(P M, S PM ) Σ(P N, S PN ) Σ(Q, S Q ) + Σ(Q, S Q ) Σ(P M, S PM ) < ε 2 + ε 2 ε.

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE 13 This pretty much proves the theorem in turn. We now know tht the complex sequence {Σ(P N, S PN )} is Cuchy sequence. By the completeness of C, the sequence converges to some complex number N {Σ(P N, S PN )}. Wht remins to be shown is tht this it is independent of which sequence of prtitions P N we chose nd of which subordinte smple S PN we chose for ech P N. But if {P N } is nother such sequence of prtitions, or the sme sequence but with different smples, then the complex sequence lso converges, to some complex number The blended sequence {Σ(P N, S P N )} N {Σ(P N, S P N )}. {Σ(P 1, S P1 ), Σ(P 1, S P 1 ), Σ(P 2, S P2 ), Σ(P 2, S P 2 ),... } gin converges becuse the meshes of the blended sequence of prtitions go to 0, nd its it must be both of the previous its becuse ech is the it of subsequence. Thus the it hs the desired independence properties, mking it suitble definition of the integrl.