u(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.


 Erick Hampton
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1 Lecture 4 Complex Integrtion MATHGA Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex plne is to link it to line integrls in R 2 s lredy seen in vector clculus. We my understnd this in two steps: A) Consider complex function f(t) = u(t) + iv(t), for t [, ] R, nd u nd v rel vlued functions. If f is continuous function, we my define := u(t) + i v(t) (1) This definition, comined with the elementry properties of ddition nd multipliction in C we sw in Lecture 1, mens tht the integrl hs mny intuitive properties tht re reminiscent of the properties of integrls of rel functions. Let us mention few without proof, s these proofs re elementry:  Let c [, ] nd f continuous on [, ] c + λ C, λ = λ ( ) R = c = ( ) R(f(t)), I = I(f(t))  Although the following property is lso intuitive, let us prove tht: f(t) (2) If = 0, the inequlity is trivil. For ( ) 0, let θ = rg ( ) ( ) = R e iθ = R e iθ = R(e iθ f(t)) f(t) With this preliminry step in plce, we re redy to define integrtion on generl curve in C. B) Let e piecewise differentile rc in the complex plne, with prmetric eqution : z = z(t), < t < If the function f is continuous on, then f(z(t)) is continuous on (, ), nd we define the integrl of f on s the line integrl f(z) := f(z(t)) (3) where the integrl my hve to e split to mtch the intervls in which z is differentile. The definition ove only mkes sense if the integrl is independent of the wy the rc is prmeterized. This is simple to check, using the rules for the chnge of vriles for integrls of rel vlued functions. Imgine tht nother prmeteriztion for is given y : τ (α, β) z(t(τ)) 1
2 with t : τ (α, β) t(τ) (, ) piecewise differentile. Then, f(z) = f(z(t)) β = = α β α f(z(t(τ))) dτ dτ f(z(t(τ))) (t(τ)) dτ dτ 1.2 Elementry properties Let : z = z(t), t (, ). We define the opposite rc, written, y : z = z( t), t (, ) Then, f(z) = f(z( t)) d [z( t)] = f(z( t)) ( t ) = f(z(t)) (t) where the lst equlity is otined with simple chnge of vrile. Hence f(z) = f(z) (4) Linerity s n opertor on functions Let f nd g e two continuous functions on the piecewise differentile rc, nd (α, β) C 2 (αf + βg) = α f + β g (5) Linerity s n opertor on curves Consider n rc which cn e sudivided into two piecewisedifferentile rcs 1 nd 2, nd f continuous function on. Then f = f + f = f + f (6) We cn use this property to show tht n integrl over closed curve does not depend on the strting point on the curve. Indeed, consider two such points P nd Q, corresponding to different prmeteriztions, s shown in Figure 1. If we cll 1 the prt of from P to Q, nd 2 the prt of from Q to P, Figure 1: Closed curve sudivided into the rcs 1 nd 2 f = f + 1 f = 2 f + 2 f 1 The expression in the middle corresponds to the evlution of the integrl strting from the point P, while the expression on the right corresponds to the evlution of the integrl strting from the point Q. 2
3 1.3 An importnt exmple We conclude this section with very simple exmple which will ply fundmentl role in the rest of this course. Let C, nd consider the integrl z where is the closed circle with rdius R nd centered in. A simple prmeteriztion for is : θ [0, 2π) z(θ) = Re iθ +. Thus z = 2π 0 idθ = 2πi 2 The fundmentl theorem of clculus for integrls in C 2.1 Line integrls with respect to x nd y The line integrl with respect to z is defined s f(z) := f(z) (7) Line integrls with respect to x = R(z) nd y = I(z) long the rc re then nturlly constructed s f(z)dx = 1 ( ) f(z) + f(z), f(z)dy = 1 ( ) f(z) f(z) 2 2i (8) If we then write f(z) = u(x, y) + iv(x, y), with z = x + iy, we hve f(z) = f(z)dx + i f(z)dy = (udx vdy) + i (udy + vdx) (9) which cn e viewed s nother definition for f(z), involving only line integrls of sclr functions, s lredy introduced in vector clculus. 2.2 Independence of pth We hve just reduced the complex integrl f(z) to line integrls of the form P (x, y)dx+q(x, y)dy. We will now recll wellknown result of vector clculus on independence of pth to determine when f(z) only depends on the endpoints of nd not the ctul pth descries. Theorem: Let Ω e n open connected set of R 2, nd P nd Q two functions tht re continuous on Ω, nd potentilly complex vlued. The integrl P dx + Qdy depends only on the end points of iff there exists function U(x, y) on Ω with the prtil derivtives P (x, y) = U, Q(x, y) = U. Proof : The sufficient condition is strightforwrd: if such U exists, then P dx + Qdy = ( U dx + U ) dy = for ny rc etween the points (x(), y()) nd (x(), y()). [ ] d U(x(t), y(t)) = U(x(), y()) U(x(), y())  Conversely, if P (x, y)dx + Q(x, y)dy only depends on the end points, we cn construct single vlued function U y fixing point (x 0, y 0 ) Ω, nd defining U(x, y) = P (x, y)dx + Q(x, y)dy 3
4 where is ny rc etween (x 0, y 0 ) nd (x, y). We now show tht U stisfies the conditions of the theorem. Consider the point (x + x, y), nd ny rc etween (x 0, y 0 ) nd (x + x, y). For x sufficiently smll, there exists n rc in Ω etween (x, y) nd (x + x, y), nd prllel to the xxis. By the independence of pth of the integrl, we cn write U(x + x, y) = P (x, y)dx + Q(x, y)dy = P (x, y)dx + Q(x, y)dy + P (x, y)dx = U(x, y) + P (x, y)dx Constructing rcs in this mnner for ll smll x, we my write U(x + x, y) U(x, y) 1 lim = lim x 0 x x 0 x x+ x x P (x, y)dx = P (x, y) where the lst eqution follows from the continuity of P. We thus hve U = P (x, y). With very similr proof, we would show tht U = Q(x, y), which concludes our proof. 2.3 The fundmentl theorem of clculus for integrls in C Consider f(z) = u(x, y) + iv(x, y), P (x, y) = u(x, y) + iv(x, y), nd Q(x, y) = i(u(x, y) + iv(x, y)). Then f(z) = P (x, y)dx + Q(x, y)dy The integrl on the righthnd side depends on the end points if nd only if there exists F (x, y) such tht P (x, y) = F F, nd Q(x, y) =. If such n F exists, then F = i F Writing F (z) = U(x, y) + iv (x, y), the equlity ove ecomes CuchyRiemnn equtions for U nd V. So F is nlytic, with derivtive f. We hve proven the following theorem. Theorem (Fundmentl theorem of clculus for integrls in C): The integrl f(z), with f continuous on n open connected set Ω contining, depends only on the end points of iff f is the derivtive of n nlytic function F in Ω. We sy tht F is primitive of f. Corollry: if f(z) = df where F is nlytic on n open connected set Ω nd if is closed curve in Ω, then f(z) = 0 (10) Conversely, if f is continuous function on n open connected set Ω nd is such tht f(z) = 0 for ny closed contour in Ω, then f hs primitive. The proof of the ltter is left to the reder s n enlightening exercise, close to wht we hve done previously in this lecture. Exmple: Let n N nd C [ ] (z ) n+1 (z ) n = d n + 1 nd (z ) n+1 /(n + 1) is entire, so (z )n = 0 for ll closed curves in C. For n = 1, we hve lredy seen tht the result does not hold For n = k, k N \ {1}, the result holds for ny curve in C tht does not go through 4
5 3 Integrtion with respect to rc length We will often encounter integrls with respect to rc length, defined y f(z)ds := f(z) = f(z(t)) z (t) (11) As efore, this only mkes sense if the integrl is independent of the prmeteriztion. This cn e verified esily, s well s the fct tht f(z)ds = f(z)ds The length of curve in the complex plne is given y L() = ds = Finlly, using Eq.(2) we hve the tringle inequlity: f(z) f(z) L() sup f(z) (12) z 5
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