5. COMPLEX INTEGRATION 5.1. Paths and Curves. A path is a continuous map of a real interval to the complex plane, γ : [a, b] C, (a < b)

Transcription

1 27 5. OMPLEX NTEGRATON 5.1. Pths nd urves. A pth is continuous mp of rel intervl to the complex plne, : [, b], ( < b) The set of points given by rnge of the pth (t) is the curve, nd the pth (t) is its prmetriztion. The incresing vlue of the prmeter ssigns n orienttion to the curve. The sme curve cn be given different prmetriztions: for exmple the complex segment [, ] cn be prmetrized by 1 (t) = t or 2 (t) = t 2, t [, 1]. Both pths give the sme segment; the difference lies in the time lw by which the mobile point z(t) covers it (in the exmple z(t) moves with uniform speed or uniform ccelertion). is geometric object, (t) is one of the possible lws tht produces the set of points. Since intervls re compct sets in R, nd pth is continuous function, the curve is compct set in 17. We list some definitions tht pply to pths, nd imply some geometric property of the curves (tht would be difficult to phrse otherwise). A pth is closed if () = (b) (we then sy tht curve is closed if it is the rnge of closed pth). A pth is simple if it does not self-intersect (except possibly for the endpoints): (t 1 ) = (t 2 ) t 1 = t 2. A Jordn curve J is simple closed curve. A theorem of topology (Jordn, Veblen) sttes tht /J is the union of two disjoint sets: one is bounded nd the other is unbounded. They re nmed the interior of J (nt J) nd the exterior of J (Ext J). A pth is smooth if it is differentible nd with continuous derivtive (t) for ll t [, b] (i.e. x = Re nd y = m re of clss 1 [, b]). The complex number (t) = ẋ(t) + iẏ(t) gives the components of the vector tngent to the curve in (t). A pth is piecewise smooth if it is continuous (it is pth) nd is smooth on ech subintervl of some prtition of [, b]. A curve cn be given different prmetriztions. Let (t) be smooth pth with t [, b] nd τ(s) rel mp of [, b ] to [, b] with dτ/ds >, τ( ) = nd τ(b ) = b. The smooth pth 2 (s) = (τ(s)) with support in [, b ] is reprmetriztion of the pth (t). The geometric object, the curve, is invrint under reprmetriztion. n geometric point z of the curve, with coordintes s nd t (τ(s) = t) the two pths hve tngent vectors 2 (s) nd (t) with the sme direction (dτ/ds is rel nd positive): 2 (ds) = (t) dτ, τ(s) = t. ds f smooth pth is not simple, the geometric point where the curve self-intersects is the imge of two or more vlues t, t,... (the point hs different coordintes) nd there re two or more tngent vectors (t), (t ), By the Heine Borel theorem, set in is compct iff it is closed nd bounded.

2 28 L. G. MOLNAR 5.2. omplex integrtion. Given continuous complex function on smooth pth, f :, the integrl of f on is defined s: b (44) f(ζ)dζ = dt (t)f((t)) By expnding the complex product b dt(ẋ + iẏ)[u(x, y) + i v(x, y)] one obtins combintion of four rel Riemnn integrls tht re well defined, since ll functions re continuous on closed intervls. The geometric picture of the integrl is s follows: choose n points of the curve, nd consider the sum (z k+1 z k )f( z k+1 z k ) 2 k For n nd z k+1 z k one my expnd z k+1 z k (t k+1 ) (t k ) (t k )(t k+1 t k ). The sum converges to the integrl. Note tht the integrl differs from the following one b (45) f(ζ) dz = dt ẋ 2 + ẏ 2 [u(x(t), y(t)) + i v(x(t), y(t))] which hs the mening of sum (with imginry fctor) of the res of two foils with bsis the curve nd heights u nd v. Exmple. Let s evlute ζ2 dζ on the following curves: i) the (oriented) segment from to b. A prmetriztion is (t) = +(b )t, with t 1. Then (t) = (b ) nd 1 ζ 2 dζ = (b ) dt [ + (b )t] 2 =... = b3 3 3 [,b] ii) the semicircle S with dimeter [, b] (from to b counterclockwise). A prmetriztion is (θ) = 1 2 ( + b) ( b)eiθ, θ π. t is () =, (π) = b, nd = i 2 ( b)eiθ S ζ 2 dζ = i π [ 1 2 ( b) dθe iθ 2 ( + b) + 1 ] 2 2 ( b)eiθ π π = i [ ( + b) 2 2 ( b) dθe iθ + (2 b 2 ) dθe 2iθ = i [ ] ( + b) 2 2 ( b) ( b)2 2i 2i + = b ( b)2 4 Note tht the two integrls on different pths gve the sme results! π dθe 3iθ ] f curve is prmetrized by (t), t [, b], the curve with opposite orienttion is prmetrized by the pth ( + b t) (on [, b]). Then: dζf(ζ) = b dt (+b t)f((+b t) = b dt (t)f((t)) = dζf(ζ)

3 Two useful inequlities. Proposition 5.1. f the complex function f is integrble on rel intervl, then: (46) dxf(x) dx f(x) Proof. Let f = eiθ f, then e iθ f is rel. Moreover, f(x)dx = e iθ fdx = Re[e iθ f] = Re[e iθ+iargf ] f f dx. We used Re fdx = Re (u + iv)dx = u dx = Refdx. f the curve is prmetrized s (t), the rel function f(ζ(t)) is continuous on [, b] nd hs mximum. The previous inequlity gives: Proposition 5.2 (Drboux s inequlity). f f is complex continuous function on smooth pth, then: (47) dζf(ζ) L() sup f(z) z where L() = b dt (t) is the length of the pth Primitive. As the exmple my suggest, the evlution of n integrl is gretly simplified if the function hs primitive. f f(z) is continuous on domin D, primitive of f is function F(z) tht is nlytic on D nd (48) F (z) = f(z), z D Then, if is contined in D the integrl of f does not depend on but only on the endpoints of the pth: b b (49) f(z)dz = dt (t)f ((t)) = dt df = F((b)) F(()). dt f the pth is closed, the integrl is zero. The existence of primitive s n nlytic function is delicte issue, s it is relted to globl properties: Theorem 5.3. Let f(z) be continuous function on n open connected set D. The following sttements re equivlent: 1) for ny two points nd b in D the integrl of f long (piecewise) smooth pth with endpoints nd b does not depend on the pth; 2) the integrl of f long ny closed (piecewise) smooth pth in D is zero; 3) there is function F(z) nlytic in ll points of D such tht F (z) = f(z) t ll points of D. Proof. The proof tht 1 implies 2, 2 implies 1 nd 3 implies 1 re trivil. The nontrivil sttement is 1 implies 3. Define F(z) s the integrl of f from point to point z long pth. By hypothesis the function only depends on z nd not on the pth. onsider point z nd disk centered on z contined in D (recll tht D is open). To prove nlyticity choose z + h in the disk (we shll tke the limit h ) nd consider two pths from in D tht end in z + h nd z, nd the segment [z, z + h]. Prmetrize the segment s ζ(t) = z + ht (dζ = hdt). The

4 3 L. G. MOLNAR integrl long the closed pth z z + h vnishes becuse 1 implies 2. Then: F(z + h) F(z) = dζf(ζ) = dtf(z+ht) = f(z)+ dt[f(z+ht) f(z)] h h [z,z+h] Since f is continuous, f(z + ht) f(z) < ǫ if t h < δ. Then, when h the lst term vnishes ( 1 dt[f(z + ht) f(z)] 1 dt f(z + ht) f(z) < ǫ) nd F (z) exists nd equls f(z). The following exmple is instructive. onsider the open disk D(, R), nd the circulr nticlockwise pths z = r (r < R). The integrl long ny such circle is computed exctly for ny integer n: 2π (5) dzz n = ir n+1 dθe i(n+1)θ = 2πri δ n, 1 Why does 1/z give non zero result? Also the functions with n = 2, 3,.. ren t nlytic in the origin. One cn remove the origin nd consider the punctured disk D = D(, R)/{} where they re ll nlytic. The point is tht primitives exist nd re nlytic in D for n 1. The primitive of 1/z is log z, but it is not nlytic in D becuse of the brnch cut tht termintes in the origin. A lot of effort will be mde to estblish when the integrl of holomorphic function on closed pth vnishes, i.e. the function dmits primitive. This will be pursued in hierrchy of uchy theorems uchy trnsform. Let be smooth pth (open or closed). f g is complex function, continuous on, the following uchy trnsform of g exists: (51) G(z) = dζ g(ζ) ζ z, z / Lemm 5.4. The functions (52) G n (z) = re nlytic, nd G n(z) = n G n+1 (z). g(ζ) dζ (ζ z) n Proof. Let z / nd choose δ such tht the disk centered in z, z z < δ, does not meet. For z in the smller disk z z < δ/2 we hve z ζ > δ/2 for ll points ζ. From (53) G 1 (z) G 1 (z ) = (z z ) g(ζ)dζ (ζ z)(ζ z ) we obtin G 1 (z) G 1 (z ) < z z 2 δ g dζ nd this proves continuity of G 2 1 t z, nd differentibility: the rtio (54) G 1 (z) G 1 (z ) z z = g(ζ)dζ (ζ z)(ζ z ) tends to the limit G 1 (z ) = G 2 (z ) s z z. n the sme mnner one proves tht G 2 is continuous nd G 2 = 2G 3, nd so on.

5 ndex of closed pths. Definition 5.5. f is (piecewise) smooth closed pth nd z /, the ndex (or winding number) of with respect to z is dζ 1 (55) nd(, z) = 2πi ζ z. The index is useful concept becuse it is topologicl property of the pth, nd does not depend on the prmetriztion. t enumertes the number of net windings of round the point z. This is clerly seen in the exmple. Exmple. k is circle of rdius R centered in the origin, tht winds k times nticlockwise on itself. t cn be prmetrized by (t) = Re i2πkt, t [, 1]. The ndex is nd( k, z) = k dζ 1 2πi ζ z = k 1 dt R e i2πkt 1 R e i2πkt re iθ f z is in the circle (r < R) one my expnd the denomintor in geometric series nd note tht only the l = term survives: nd( k, z) = k 1 1 dt 1 r = k r l R e i(2πkt θ) R leilθ l= 1 dte i2πlkt = k f r > R the nlogous expnsion does not hve the l = term nd lwys gives zero: R l 1 nd( k, z) = k r l e ilθ dte i2πlkt = l=1 Note tht until the point z does not cross the circle, the rdius R cn be modified without chnging the ndex. Actully, the ndex is invrint under deformtions of the circles tht do not cross z (the deformed pth is homotopiclly equivlent to the initil circles). (56) Note tht for smooth closed pth, it is nd(, z) = = 1 1 dt (t) 2πi (t) z = 1 dt d log((t) z) 2πi dt dt d 2πi dt [log (t) z + irg((t) z)] = n + n, where n + nd n re the numbers of windings (nti/clockwise) of the pth round z. The first term (log of modulus) cncels becuse the pth is closed. The term with rg requires cre. We need to show tht it is possible to choose determintion of θ(t) = rg((t) z) tht is continuous on [, 1] (for piecewise smooth pths the integrl splits on subintervls of smoothness. The problem of ssessing the existence of continuous θ(t) remins unvried). A rigorous tretment of the problem is now given. Proposition 5.6. Given (piecewise) smooth pth nd point z / the index nd (, z) is n integer.

6 32 L. G. MOLNAR Proof. onsider the function (57) λ(t) = + t (s) ds (s) z where is constnt. Since is continuous nd is piecewise continuous, λ is continuous nd piecewise differentible, nd λ(t)((t) z) (t) =. Then d/dt[e λ ((t) z)] =, i.e. e λ ( z) is constnt on ech intervl of smoothness. Since the function is continuous, the sme constnt holds in ll subintervls, i.e. on [, 1], nd cn be put equl to one by choice of : (58) e λ(t) ((t) z) = e (() z) = 1 Note tht 2πi nd(, z) = λ(1). Since (1) = () in eqution (58) for t = 1, it is e λ(1) = 1. Then λ(1) = 2πni. Proposition 5.7. The index function of pth is constnt in ech open set D tht does not contin the pth. Proof. Let us consider point z in D nd disk centered in it tht is contined in D. For h is smller then the rdius, let us evlute 1 dζ 1 [nd(, z + h) nd(, z)] = h 2πi (ζ z h)(ζ z) n the limit h, it is d nd(, z) = dz dζ 1 2πi (ζ z) 2 = becuse there is primitive. Then nd(, z) is constnt in every point in D. n prticulr nd(, z) = for z, mening tht the index is zero in Ext ().

7 33 6. ENTRE FUNTONS Entire functions re functions tht re nlytic on the whole complex plne. Polynomils, the exponentil nd its liner combintions re entire. The uchy theorem nd uchy s integrl formul will be proven first for entire functions, nd then extended to just holomorphic functions on domin. We begin by proving uchy s theorem on rectngles; this will enble us to construct the primitive of entire functions nd prove uchy s theorem for ll smooth closed pths. Lemm 6.1. f f is entire nd R is rectngle with boundry R: (59) dζf(ζ) = R Proof. Let L be the digonl length of R. Let us cut the sides of R in hlves, nd obtin four equl rectngles R 1k (the index 1 stnds for genertion). f we denote (R) the integrl on the boundry of rectngle R with some orienttion, it is (R) = k (R k) becuse of cncelltions of integrls on shred sides (hving opposite orienttions). Let 1 be the term mong (R 1k ) with lrgest modulus nd R 1 the corresponding rectngle. Then (R) k (R 1k) 4 1. The prtition is repeted gin on R 1 nd produces four rectngles R 2k, nd 1 = k (R 2k). Select the rectngle R 2, with lrgest vlue (R 2k ) nd cll the integrl 2. t is After n itertions we hve rectngle R n such tht 4 n n. Let be point of the boundry of R n ; since f(z) is nlytic it is true tht ǫ δ such tht f(z) f() = f () + r(z, ) z with reminder r(z, ) < ǫ if z < δ. Since in R n two points hve seprtion t most L/2 n, rephrsing is: ǫ N such tht f(z) = f()+f ()(z )+r(z, )(z ) where r(z, ) < ǫ if n > N. This gives the estimte: n = dzf() + f ()(z ) + r(z, )(z ) = dzr(z, )(z ) R n R n dz r(z, )(z ) ǫ sup z 4 L R n z R n 2 n < 4ǫL2 4 n (Note tht R dz(cz + d) = on ny rectngle, by use of the primitive). Then, 4ǫ L 2 for rbitrry ǫ, so it equls zero. With this result, primitive F(z) of f cn be evluted explicitly s n integrl long two sides of rectngle with opposite corners nd z = x + iy (6) F(z) = x Let us prove tht F (z) = f(z). dx f(x + i) + i Proof. Note tht, becuse of the Lemm, it is F(z + h) = F(z) + x+hx x y dx f(x + iy) + i dy f(x + iy ) y+hy y dy f(x + h x + iy )

8 34 L. G. MOLNAR Then we evlute F(z + h) F(z) f(z)h = x+h x dx [f(x + iy) f(x + iy)]+ +i y+hy y dy [f(x + h x + iy ) f(x + h x + iy) + f(x + h x + iy) f(x + iy)] Since f is continuous, for h smll enough it is f(z + h) f(z) < ǫ. Therefore: F(z + h) F(z) f(z)h h x ǫ + 2 h y ǫ 3ǫ h With primitive in our hnds, we conclude tht the integrl of f on ny closed smooth pth gives zero: Theorem 6.2 (uchy theorem for entire functions). (61) dζf(ζ) = n the proof of the Lemm 6.1 we only used the property of nlyticity of f on the rectngle. The primitive ws built s n integrl from reference point (the origin, but it cn be different). uchy s theorem for generl closed curves followed. At this stge we conlude sfely tht uchy s theorem holds lso for functions holomorphic on rectngle nd for ll closed smooth pths in it. The next step is to prove uchy s integrl formul for entire functions. We need generliztion of the proof of the Lemm 6.1 to llow for functions tht re nlytic everywhere but t point. Lemm 6.3. Let f(z) be function continuous on nd nlytic on /{}. Then dζf(ζ) = for ny rectngle R. R Proof. f / R nothing chnges in the proof of 6.1. f R the key ingredients for proof re: f(z) < M becuse f is continuous on compct set R nd Lemm 6.1. The trick is to decompose R = R r 1... r k where R is rectngle with liner sizes of order ǫ tht contins (s interior or boundry point if R), nd r i re rectngles tht fit the prtition. The contour integrl is (R) = (R ) + i (r i). By the Lemm 6.1 (r i ) = ; by the Drboux inequlity: (R ) cmǫ, where c is some constnt. The vlidity of the Lemm llows to construct primitive nd obtin uchy s theorem for functions continuous everywhere nd entire up to point. This is needed to prove the next Theorem 6.4 (uchy s integrl formul). f is n entire function, is smooth closed curve, /. uchy s integrl formul is: (62) dζ f(ζ) = f()nd(, ) 2πi ζ where nd(, ) is the index of the curve with respect to. Proof. onsider the function g(z) = f(z) f() z if z nd g() = f (). g is continuous on nd nlytic in /. Then, uchy s theorem gives dζ f(ζ) (63) = dζ g(ζ) = f()nd(, ) 2πi ζ x

9 35 orollry 6.5 (The fundmentl theorem of lgebr). (my proof) A polynomil of degree greter thn zero lwys hs zero. Proof. Suppose tht the polynomil P(z) hs no zeros. Then 1/P(z) is entire nd vnishes for z. For ny z nd we evlute by uchy s formul with cirle lrge enough to contin z nd : 1 P(z) 1 P() = ( dζ 1 1 2πi P(ζ) ζ z 1 ) ζ dζ 1 1 = (z ) 2πi P(ζ) (ζ z)(ζ ) The vlue of the integrl does not depend on the rdius of the circle round z nd. By tking R to we obtin zero: P(z) is constnt (degree zero). Theorem 6.6 (Liouville s theorem). (my proof) f f(z) is n entire function nd f(z) M for ll z, then f(z) is constnt. Proof. Given ny two points z nd z + h, nd circle with center z nd rdius R > h, by uchy integrl formul: [ 1 h [f(z+h) f(z)] = dζ 2πi f(ζ)1 1 h ζ z h 1 ] dζ f(ζ) = ζ z 2πi (ζ z h)(ζ z) Tke the limit h nd obtin f (z) (f is entire). Prmetrize integrtion vrible s ζ = z + Re iθ : f dζ f(ζ) (z) = 2πi (ζ z) 2 = 1 2π dθ R 2π f(z + Reiθ ) Finlly obtin f (z) M/R. Since R is rbitrrily lrge it is f (z) =, i.e. f(z) is constnt. Theorem 6.7 (Picrd s theorem 18 ). Let f be n entire function nd nd b two distinct complex vlues such tht, for ll z, f(z) nd f(z) b. Then f(z) is constnt. Proof Evlution of integrls. Difficult integrls of rel functions on the rel line my become simple in the complex plne. uchy s theorem yields null result for integrls of entire functions on closed pths; it cn be used to evlute n integrl on n open pth if the pth cn be closed, nd the dded integrls re known. (64) (65) (66) dx e ix2 = dx sin x x π 2 ei π 4 = π dx e x2 cos(2x) = π 2 2 e 18 hrles E. Picrd (1856, 1941).