4. Complex integration: Cauchy integral theorem and …

1 4. Complex integration : Cauchy integral theorem and Cauchyintegral formulasDefinite integral of a Complex -valued function of a real variableConsider a Complex valued functionf(t) of a real variablet:f(t) =u(t) +iv(t),which is assumed to be a piecewise continuous function defined inthe closed intervala t b. The integral off(t) fromt=atot=b, is defined as baf(t)dt= bau(t)dt+i bav(t) of a Complex integral with real variable of baf(t)dt= baRef(t)dt= bau(t) baf(t)dt= baImf(t)dt= bav(t) ba[ 1f1(t) + 2f2(t)]dt= 1 baf1(t)dt+ 2 baf2(t)dt,where 1and 2are any Complex baf(t)dt ba|f(t)| prove (4), we consider baf(t)dt =e i baf(t)dt= bae i f(t)dt,where = Arg( baf(t)dt). Since baf(t)dt is real, we deduce that baf(t)dt = Re bae i f(t)dt= baRe [e i f(t)]dt ba|e i f(t)|dt= ba|f(t)| is real, show that|e2 i 1| 2 | |.SolutionLetf(t) =ei t, andtare real. We obtain 2 0ei tdt 2 0|ei t|dt= 2.

2 The left-hand side of the above inequality is equal to 2 0ei tdt = ei ti 2 0 =|e2 i 1|| |.Combining the results, we obtain|e2 i 1| 2 | |, is of a contour integralConsider a curveCwhich is a set of pointsz= (x, y) in the complexplane defined byx=x(t), y=y(t), a t b,wherex(t) andy(t) are continuous functions of the real parametert. One may writez(t) =x(t) +iy(t), a t b. The curve is said to besmoothifz(t) has continuous derivativez (t)6= 0 for all points along the curve. Acontouris defined as a curve consisting of a finite numberof smooth curves joined end to end. A contour is said to be asimple closed contourif the initial and final values ofz(t) arethe same and the contour does not cross Letf(z) be any Complex function defined in a domainDin thecomplex plane and letCbe any contour contained inDwithinitial pointz0and terminal pointz. We divide the contourCintonsubarcs by discrete pointsz0, z1, z2.

3 , zn 1, zn=zarranged consecutively along the direction of in-creasingt. Let kbe an arbitrary point in the subarczkzk+1and form thesumn 1 k=0f( k)(zk+1 zk).6 Subdivision of the contour intonsubarcs by discrete pointsz0, z1, ,zn 1, zn= write zk=zk+1 zk. Let = maxk| zk|and take the limitlim 0n n 1 k=0f( k) above limit is defined to be thecontour integraloff(z) alongthe the above limit exists, then the functionf(z) is said to beinte-grable along the we writedz(t)dt=dx(t)dt+idy(t)dt, a t b,then Cf(z)dz= baf(z(t))dz(t) (z) =u(x, y) +iv(x, y) anddz=dx+idy, we have Cf(z)dz= Cu dx v dy+i Cu dy+v dx= ba[u(x(t), y(t))dx(t)dt v(x(t), y(t))dy(t)dt]dt+i ba[u(x(t), y(t))dy(t)dt+v(x(t), y(t))dx(t)dt] usual properties of real line integrals are carried overto theircomplex counterparts. Some of these properties are:(i) Cf(z)dzis independent of the parameterization ofC;(ii) Cf(z)dz= Cf(z)dz, where Cis the opposite curve ofC;(iii) The integrals off(z) along a string of contours is equal to thesum of the integrals off(z) along each of these the integral C1z z0dz,whereCis a circle centered atz0and of any radius.

4 The path istraced out once in the anticlockwise circle can be parameterized byz(t) =z0+reit,0 t 2 ,whereris any positive real number. The contour integral becomes C1z z0dz= 2 01z(t) z0dz(t)dtdt= 2 0ireitreitdt= 2 value of the integral is independent of the the integral (i) C|z|2dzand (ii) C1z2dz,where the contourCis(a) the line segment with initial point 1 and final pointi;(b) the arc of the unit circle Imz 0 with initial point 1 and the two results agree?11 Solution(i) Consider C|z|2dz,(a) Parameterize the line segment byz= 1 + (1 +i)t,0 t 1,so that|z|2= ( 1 +t)2+t2anddz= (1 +i) value of the integral becomes C|z|2dz= 10(2t2 2t+ 1)(1 +i)dt=23(1 +i).12(b) Along the unit circle,|z|= 1 andz=ei ,dz=iei d . The initialpoint and the final point of the path correspond to = and = 2, respectively. The contour integral can be evaluated as C|z|2dz= 2 iei d =ei 2 = 1 + results in (a) and (b) do not agree. Hence, the value of thiscontour integral does depend on the path of (ii) Consider C1z2dz.

5 (a) line segment from 1 toi C1z2dz= 101 +i[ 1 + (1 +i)t]2dt= 1 1 + (1 +i)t 10= 1 1i= 1+i.(b) subarc from 1 toi C1z2dz= 2 1e2i iei d = e i ] 2 = 1 + of the absolute value of a Complex integralThe upper bound for the absolute value of a Complex integral canbe related to the length of the contourCand the absolute value off(z) alongC. In fact, Cf(z)dz M L,whereMis the upper bound of|f(z)|alongCandLis the arc lengthof the consider Cf(z)dz = baf(z(t))dz(t)dtdt ba|f(z(t))| dz(t)dt dt baM dz(t)dt dt=M ba (dx(t)dt)2+(dy(t)dt)2dt=M that C1z2dz 2, whereCis the line segment joining 1 +iand 1 + the contourC, we havez=x+i, 1 x 1, so that 1 |z| 2. Correspondingly,12 1|z|2 1. Here,M= maxz C1|z|2= 1and the arc lengthL= 2. We have C1z2dz M L= an upper bound of the modulus of the integralI= CLogzz 4idzwhereCis the circle|z|= , Logzz 4i ln|z| +|Argz|||z| |4i||so thatmaxz C Logzz 4i ln3 + |3 4|= ln3 + ;L= (2 )(3) = 6.

6 Hence, CLogzz 4idz 6 ( + ln3).18 ExampleFind an upper bound for ez/(z2+ 1)dz , where is the circle|z|= 2 traversed once in the counterclockwise path of integration has lengthL= 4 . Next we seek an upperboundMfor the functionez/(z2+ 1) when|z|= 2. Writingz=x+iy, we have|ez|=|ex+iy|=ex e2,for|z|= x2+y2= 2,and by the triangle inequality|z2+ 1| |z|2 1 = 4 1 = 3 for|z|= ,|ez/(z2+ 1)| e2/3 for|z|= 2, and so ezz2+ 1dz e23 4 .19 Path independenceUnder what conditions that C1f(z)dz= C2f(z)dz,whereC1andC2are two contours in a domainDwith the sameinitial and final points andf(z) is piecewise continuous property of path independence is valid forf(z) =1z2but it failswhenf(z) =|z|2. The above query is equivalent to the question:When does Cf(z)dz= 0hold, whereCis any closed contour lying completely insideD? Theequivalence is revealed if we treatCasC1 observe thatf(z) =1z2is analytic everywhere except atz= 0butf(z) =|z|2is nowhere integral theoremLetf(z) =u(x, y)+iv(x, y)be analytic on and inside a simple closedcontourCand letf (z)be also continuous on and inside C, then Cf(z)dz= proof of the Cauchy integral theorem requires the Green theo-rem for a positively oriented closed contourC: If the two real func-tionsP(x, y) andQ(x, y) have continuous first order partial deriva-tives on and insideC, then CP dx+Q dy= D(Qx Py)dxdy,whereDis the simply connected domain bounded we writef(z) =u(x, y) +iv(x, y), z=x+iy; we have Cf(z)dz= Cu dx v dy+i Cv dx+u can infer from the continuity off (z) thatu(x, y) andv(x, y)have continuous derivatives on and insideC.

7 Using the Green the-orem, the two real line integrals can be transformed into doubleintegrals. Cf(z)dz= D( vx uy)dxdy+i D(ux vy) integrands in the double integrals are equal to zero dueto theCauchy-Riemann relations, hence the 1903, Goursat was able to obtain the same resultwithout assum-ing the continuity off (z).22 Goursat TheoremIf a functionf(z) is analytic throughout a simply connected domainD, then for any simple closed contourClying completely insideD,we have Cf(z)dz= 1 The integral of a functionf(z) which is analytic throughout a simplyconnected domainDdepends on the end points and not on theparticular contour taken. Suppose and are insideD,C1andC2are any contours insideDjoining to , then C1f(z)dz= C2f(z) the curvey=x3 3x2+4x 1 joining points (1,1) and (2,3),find the value of C(12z2 4iz) 1. The integral is independent of the path joining (1,1)and (2,3). Hence any path can be chosen. In particular, let uschoose the straight line paths from (1,1) to (2,1) and then from(2,1) to (2,3).

8 Case 1 Along the path from (1,1) to (2,1), y= 1, dy= 0 so thatz=x+iy=x+i, dz=dx. Then the integral equals 21{12(x+i)2 4i(x+i)}dx={4(x+i)3 2i(x+i)2} 21= 20 + 2 Along the path from (2,1) to (2,3), x= 2, dx= 0 so thatz=x+iy= 2 +iy, dz=idy. Then the integral equals 31{12(2+iy)2 4i(2+iy)}i dy={4(2+iy)3 2i(2+iy)2} 31= 176+ adding, the required value = (20 + 30i) + ( 176 + 8i) = 156 + 2. The given integral equals 2+3i1+i(12z2 4iz)dz= (4z3 2iz2) 2+3i1+i= 156 + is clear that Method 2 is 2 Letf(z) be analytic throughout a simply connected domainD. Con-sider a fixed pointz0 D; by virtue of Corollary 1,F(z) = zz0f( )d ,for anyz D,is a well-defined function inD. ConsideringF(z+ z) F(z) z f(z) =1 z z+ zz[f( ) f(z)]d .By the Cauchy theorem , the last integral is independent of the pathjoiningzandz+ zso long as the path is completely insideD. Wechoose the path as the straight line segment joiningzandz+ zand choose| z|small enough so that it is completely continuity off(z), we have for all pointsuon this straight linepath|f(u) f(z)|< whenever|u z|<.

9 Note that| z|< is observed have z+ zz[f(u) f(z)]du < | z|so that F(z+ z) F(z) z f(z) =1| z| z+ zz[f(u) f(z)]du < for| z|< . This amounts to saylim z 0F(z+ z) F(z) z=f(z),that is,F (z) =f(z) for allzinD. Hence,F(z) is analytic inDsinceF (z) exists at all points inD(which is an open set).28 This corollary may be considered as a Complex counterpart ofthe fundamental theorem of real calculus. If we integratef(z) along any contour joining and insideD,then the value of the integral is given by f(z)dz= z0f( )d z0f( )d =F( ) F( ), and 3 LetC, C1, C2, .., Cnbe positively oriented closed contours, whereC1, C2, .., Cnare all insideC. ForC1, C2, .., Cn, each of thesecontours lies outside of the other contours. Let intCidenote thecollection of all points bounded insideCi. Letf(z) be analytic onthe setS:C intC\intC1\intC2\ \intCn(see the shadedarea in Figure), then Cf(z)dz=n k=1 Ckf(z) proof for the case whenn= 2 is presented The constructed boundary curve is composed ofC C1 C2and the cut lines, each cut line travels twice in opposite direc-tions.

10 To explain the negative signs in front ofC1andC2, we note thatthe interior contours traverse in the clockwise sense as parts ofthe positively oriented boundary curve. With the introduction of these cuts, the shaded region boundedwithin this constructed boundary curve becomes a simply con-nected have Cf(z)dz+ C1f(z)dz+ C2f(z)dz= 0,so that Cf(z)dz= C1f(z)dz+ C2f(z) the domain that contains the whole Complex plane exceptthe origin and the negative real axis. Let be an arbitrary contourlying completely insideD, and starts from 1 and ends at a point . Show that dzz= Log .SolutionLet 1be the line segment from 1 to| |along the real axis, and 2be a circular arc centered at the origin and of radius| |whichextends from| |to . The union 1 2 forms a closedcontour. Since the integrand1zis analytic everywhere insideD, bythe Cauchy integral theorem , we have dzz= 1dzz+ contour starts fromz= 1 and ends atz= . The arc 2ispart of the circle|z|=| |.