Chapter 3:Complex Integralssha Uni. of Sci & Tech)FCV&ITNovember 5.20193/53angWan
Chapter 3: Complex Integrals Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 3 / 53
g3.1 The concept of complex integralshaUni.ofSci&TechFCV&ITNovember5,20194/53angWan
§3.1 The concept of complex integral Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 4 / 53
Thedefinitionofcomplexintegraly252xoangWatshaUni.ofSci&Tech)FCV&ITNovember5.20195/53
The definition of complex integral O x y z0 z1 z2 z3 z4 z5 z6 Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 5 / 53
The definLet C be a piecewise smoth curve,zo is its beginning point and Z is itsend point. Let f(z) be defined in C.Take the partition pointsZo,Z1,*+,Zn=ZalongthecurveC,wehaveSn==1f(k)zkwhere △zk = Zk-Zk-1, Vk E zk-1zk, let8 = max[|zk - zk-1l,k = 1,2, .-. ,n] ,if the limit of this sumlimSn=limf(x)Azk5→080k=1exists,f if called integrable along C.The limit is called the integral offalong C, denoted fcf(z)dz.FCV&ITUni.of Sci &Tech)November5,20196/53
The definition of complex integral Let C be a piecewise smoth curve, z0 is its beginning point and Z is its end point. Let f(z) be defined in C. Take the partition points z0, z1, · · · , zn = Z along the curve C, we have Sn = ∑ n k=1 f(ζk)∆zk , where ∆zk = zk − zk−1 , ∀ζk ∈ _ zk−1zk , let δ = max{|zk − zk−1 |, k = 1, 2, · · · , n} ,if the limit of this sum lim δ→0 Sn = lim δ→0 n ∑ k=1 f(ζk)∆zk exists, f if called integrable along C. The limit is called the integral of f along C, denoted R C f(z)dz. Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 6 / 53
The definition of conplexintegralThepropertiesoftheintegralsLetf,gbeintegrablealongC,(1) Jcαf +βg=α Jcf +βJc8(2) Jef= Je,f + Je,f, c=CiuC2(3) Jcf = - Jc(4)For continuous function f,if|f(z)/<M,the length of C isL,then/, f(2)dz/≤ /,1f(2)]ds ≤ MLSincef(R)Az≤EIf(3K)/Azk/≤If(3K)[Askand-|f()Ask≤M"-Ask=ML.Take→0bothsidesoftheaboveinequality,wehaveJ f(2)dz ≤ J/,1(2)ds ≤ MLNovember 5, 2019of Sci &Tech)FCV&IT7/53
The definition of complex integral The properties of the integrals Let f , g be integrable along C, (1) R C α f + βg = α R C f + β R C g (2) R C f = R C1 f + R C2 f , C = C1 ∪ C2 (3) R C f = − R C− f (4) For continuous function f , if | f(z)| ≤ M, the length of C is L, then Z C f(z)dz ≤ Z C | f(z)|ds ≤ ML Since n ∑ k=1 f(ζk)∆zk ≤ n ∑ k=1 | f(ζk)||∆zk | ≤ n ∑ k=1 | f(ζk)|∆sk and ∑ n k=1 | f(ζk)|∆sk ≤ M ∑ n k=1 |∆sk = ML. Take δ → 0 both sides of the above inequality, we have Z C f(z)dz ≤ Z C | f(z)|ds ≤ ML Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 7 / 53
ThedeLet C bea piecewisesmooth curvethat itsparametric equation isz=z(t)=x(t)+y(t),(αa<t<β)If f(z) =u(x,y) +iv(x,y) is continuous in C, u(x,y) and v(x,y) arealsocontinuous inC.Denotek=Sk+ink,Azk = zk- zk-1 = (x -xk-1) +i(yk-yk-1) = Axk +iAykE f(k)Ax=E(u(Sk, n) + io(sk, n)(Axk + iAyk)k=1nE(u(Sk, nk)Axk - o(sk, k)Ayk)k=1+iE(u(Sk,k)Ayk+v(sk,nk)Axk)k=1As →0,takete limit f(z)dz = /, udx-ody +i /,odx+ udyFCV&ITci&Tech)November5.20198/53
The definition of complex integral Let C be a piecewise smooth curve that its parametric equation is z = z(t) = x(t) + y(t), (α ≤ t ≤ β) If f(z) = u(x, y) + iv(x, y) is continuous in C, u(x, y) and v(x, y) are also continuous in C. Denote ζk = ξk + iηk , ∆zk = zk − zk−1 = (x − xk−1) + i(yk − yk−1) = ∆xk + i∆yk n ∑ k=1 f(ζk)∆k = n ∑ k=1 (u(ξk , ηk) + iv(ξk , ηk))(∆xk + i∆yk) = n ∑ k=1 (u(ξk , ηk)∆xk − v(ξk , ηk)∆yk) + i n ∑ k=1 (u(ξk , ηk)∆yk + v(ξk , ηk)∆xk) As δ → 0, take te limit Z C f(z)dz = Z C udx − vdy + i Z C vdx + udy Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 8 / 53
ThedefinWe see that if f is continuous in a piecewise smooth curve:z = z(t) = x(t) +iy(t,t e (α,β)), then f is integrable[ (z)dz = / (u(x(t),y(t)x(t) - 0(x(t),y(t)y(t)datti[P(o(x(t),y(t)x'(t) + u(x(t),y(t)y(t)dt(u(x(t),y(t) + io(x(t),y(t)(x(t) +iy'(t)dtf(z(t)z(t)dtFCV&ITof Sci & Tech)November5,20199/53
The definition of complex integral We see that if f is continuous in a piecewise smooth curve: z = z(t) = x(t) + iy(t, t ∈ (α, β)), then f is integrable. Z C f(z)dz = Z β α (u(x(t), y(t))x 0 (t) − v(x(t), y(t))y 0 (t))dt + i Z β α (v(x(t), y(t))x 0 (t) + u(x(t), y(t))y 0 (t))dt = Z β α (u(x(t), y(t)) + iv(x(t), y(t)))(x 0 (t) + iy0 (t))dt = Z β α f(z(t))z 0 (t)dt Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 9 / 53
The definitixintearalExample(3.1.1)Evaluatefzdz,asshowninthefigure,(1) C is the straight line joining (0,0) to (1,1)(2) C is the broken line joining (0,0) to (1,0) and then to (1,1)Solution.o Let C:x =t,y =t,o<t≤1, thenyzdz=(t-it)(1+i)dt=2tdt=1(1,1)O Let Ci : x = t,y = 0, C2 : x = 1,y = t,C2zdz+zdz=zdzIC2xoC1(1,0)tdt-(1-it)idt011+=+i=1+iFCV&IT10/53of Sci &Tech)November5,2019angW
The definition of complex integral Example (3.1.1) Evaluate R C zdz ¯ , as shown in the figure, (1) C is the straight line joining (0, 0) to (1, 1). (2) C is the broken line joining (0, 0) to (1, 0) and then to (1, 1). C1 C C2 O x y (1, 1) (1, 0) Solution. 1 Let C : x = t, y = t, 0 ≤ t ≤ 1, then Z C zdz ¯ = Z 1 0 (t − it)(1 + i)dt = Z 1 0 2tdt = 1 2 Let C1 : x = t, y = 0, C2 : x = 1, y = t, Z C zdz ¯ = Z C1 zdz ¯ + Z C2 zdz ¯ = Z 1 0 tdt + Z 1 0 (1 − it)idt = 1 2 + 1 2 + i = 1 + i Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 10 / 53
Thedefinition of coxintegralExampleEvaluate fxdz, where C is the circumference of the unit squate.Solution.DefineC:[o,4]→CasfollowsyC=Ci+C2+C3+C4,wherethefoursidesof theC3 (1,1)unit square are:Ci(t)=t+0i0≤t≤1C4C2C2(t) = 1+ (t - 1)i1<t<2C3(t)= (3-t)+i2≤t≤3xoGi (1,0)C4(t) = 0 + (4 - t)t,3<t<4.Wecomputeasfollows11xdzxdztdt =dt = 12ICC21xdz =xdz:3-t)(-1)dt0(-1)dt = 03二2JCa1JC3FCV&ITNovember.5, 201911/53FangWara Uni.of Sci & Tech)
The definition of complex integral Example Evaluate H C xdz, where C is the circumference of the unit squate. C1 C2 C3 C4 O x y (1, 1) (1, 0) Solution. Define C : [0, 4] → C as follows C = C1 + C2 + C3 + C4, where the four sides of the unit square are: C1(t) = t + 0i 0 ≤ t ≤ 1 C2(t) = 1 + (t − 1)i 1 ≤ t ≤ 2 C3(t) = (3 − t) + i, 2 ≤ t ≤ 3 C4(t) = 0 + (4 − t)t, 3 ≤ t ≤ 4. We compute as follows Z C1 xdz = Z 1 0 tdt = 1 2 Z C2 xdz = Z 2 1 dt = 1 Z C3 xdz = Z 3 2 (3 − t)(−1)dt = − 1 2 Z C4 xdz = Z 4 3 0(−1)dt = 0 Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 11 / 53
The definiteRrExample(3.1.3)Let C be the circle of radius r around a E C. Evaluate fc(z -a)"dz for allintegersn=0,±1,±2,.Solution.C:z=z(0)=a+rei?,0<<2元27neineireiede-a)"dz=d0=2元n=-1i(n+1)edo=0,n1FCV&IThaUni.ofSci&TechNovember5.201912/53
The definition of complex integral Example (3.1.3) Let C be the circle of radius r around a ∈ C. Evaluate H C (z − a) ndz for all integers n = 0, ±1, ±2, · · · . Solution. C : z = z(θ) = a + reiθ , 0 ≤ θ ≤ 2π. I C (z − a) n dz = Z 2π 0 r n e inθ ireiθ dθ = i Z 2π 0 dθ = 2πi, n = −1 r n+1 Z 2π 0 e i(n+1)θ dθ = 0, n 6= −1 Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 12 / 53