p(t) dt + i 1 re it ireit dt =


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1 Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.) A curve is most conveniently defined by prmetristion. So curve is function : [, b] C (from finite closed rel intervle [, b] to the plne). We cn imgine the point (t) being trced out by pen which is t position (t) t time t. We cn write (t) = x(t) + iy(t) in terms of its rel n imginry prts. Then we define (t) = x (t) + iy (t) (cn be viewed s the tngent vector or velocity vector to the curve) nd we will only be deling with curves where (t) is defined nd continuous. A curve is clled closed if () = (b) (strt nd end point coincide). A curve is clled simple if it never goes though the sme point twice (with the possible exception tht () = (b) is llowed prt from this ll (t) hve to be different points). Exmple. A simple exmple to keep in mind is circle, sy the circle of rdius r > 0 bout the origin where we trvel once round it nticlockwise strting nd ending t the point r on the positive xis. Then r : [0, 2π] C, r (t) = re it = r cos t + ir sin t is one obvious prmetristion. Definition If : [, b] C is curve in C nd f(z) is complexvlued function defined t lest for ll z = (t), then we define b f((t)) (t) dt. (This lst is firly ordinry integrl, except tht f((t)) (t) will hve complex vlues. Sy f((t)) (t) = p(t) + iq(t) (in terms of the rel prt p(t) nd imginry prt q(t)). Then the complex integrl mens simply b p(t) + iq(t) dt = b p(t) dt + i b q(t) dt Techniclly we will require tht these ordinry integrls of p nd q should exist, but tht will be ok in ll our exmples. Continuity of f nd of (t) is enough to mke the integrl ok. Exmple. Tke r s in the exmple bove nd f(z) = 1/z. Then we cn explicitly compute r 1 z dz = 2π 0 1 re it ireit dt = 2π 0 idt = 2πi (In prctice, we cn rrely do the clcultion so directly. Typiclly we will use theorems to simplify the curve first.) Elementry properties (of complex integrls). The bsic properties re reminiscent of those for line integrls in R 2 (except tht we now hve complex vlues).
2 Complex integrls 2 1. The exct prmetristion of the curve is not importnt, lthough the direction is. So for exmple, if we tke the circle z = r but prmetrise it in different wy, while still going once round nticlockwise sy by σ r : [0, 1] C with σ r (t) = e 2πit, then the integrl will not chnge. So r f(z) dz σ r Chnging the direction of the curve chnges the the integrl by fctor 1. For exmple in the cse of the circle, µ r : [0, 2π] C with µ r (t) = e 2πit hs µ r r f(z) dz. These fct follow by ordinry substitution (or chnge of vribles). 2. If f(z) = F (z) for some nlytic F (z) nd : [, b] C is curve with ll points (t) in the set where F (z) is nlytic, then b F ((t)) (t) dt = b is the difference of the vlues F (end) F (strt). d dt F ((t)) dt = [F ((t))]b t= = F ((b)) F (()) 3. In prticulr, if the integrnd f(z) hs n nlytic ntiderivtive F (z) tht works ll long, then the exct pth does not enter in to the vlue of f(z) dz (s long s stys in the set where F is nlytic) nd the integrl will be 0 if the pth is closed (strt = end). If you look bck t the lst exmple you will see tht the integrl of f(z) = 1/z round the closed curve r ws not zero. Thus there is no ntiderivtive of 1/z tht works ll the wy round r. [Recll tht log z is n ntiderivtive of 1/z except on the negtive xis. The jump we mke in the rgument rg(z) t the negtive xis ctully corresponds to the vlue 2πi of the integrl.] 4. We cn use Green s theorem for complex vlued P (x, y) nd Q(x, y). Tht is ( Q P dx + Q dy = x P ) dx dy y R is true for complexvlued P, Q if is simple closed curve in R 2, R is the interior of nd both P nd Q re wellbehved inside nd on. Here we interpret the integrls of complex things s the integrl of the rel prt +i times the integrl of the imginry prt. Theorem 1 (Cuchy s theorem) If is simple closed nticlockwise curve in the complex plne nd f(z) is nlytic on some open set tht includes ll of the curve nd ll points inside, then 0
3 Complex integrls 3 Proof. We write dz = dx + idy nd use Green s theorem on ( (if(z)) f(z) dx + (if(z)) dy = x R f(z) ) dx dy y (with R denoting the interior of ). If you recll the proof of the CR equtions you will remember tht (becuse the limit defining f f(z+h) f(z) (z) = lim h 0 cn be tken in ny direction or in ll h directions t once f (z) = f(z) x = 1 f(z) i y It follows tht the integrnd of the double integrl we got from Green s theorem (if(z)) f(z) ( f(z) = i x y x 1 ) f(z) = 0 i y nd so we get 0. [Another wy to do this is to write f(z) = u + iv nd use the CR equtions to get the integrnd of the double integrl to be zero.] Remrk. It is vitl tht there re no bd points of f(z) inside or on. Look gin t the lst exmple nd see tht f(z) = 1/z is fine everywhere except t z = 0. This cn (nd does in tht cse) mke the integrl nonzero. Corollry 2 Suppose we hve two nticlockwise simple closed curves 1 nd 2 with one entirely contined in the interior of the other. Suppose f(z) is nlytic on some open set tht includes both 1 nd 2 nd the region between the two curves. Then f(z) dz 1 2 (The proof involves mking nrrow bridge between the two curves nd simple closed curve Γ tht goes lmost once round the outer curve, in cross one side of the bridge, the wrong wy round the inner curve nd bck cross the bridge. f(z) will be nlytic on nd inside Γ nd then 0 by Γ Cuchy s theorem. Let the width of the bridge tend to zero nd we find tht we get the result we wnt becuse the integrl long the bridge in different directions cncel.) Exmple. Looking bck t the exmple r 1/z dz we sw they ll turned out to be 2πi no mtter wht the rdius r > 0 ws. We cn now see tht this independence of r follows becuse of (the Corollry to) Cuchy s theorem. Also we cn see tht we will lso get the sme 2πi
4 Complex integrls 4 for integrls round more complicted curves tht go once nticlockwise round the origin. For exmple the ellipse σ : [0, 2π] C with σ(t) = 4 cos t + 3i sin t contins r if r < 3. So σ 1/z dz = 2 1/z dz = 2πi. Theorem 3 (Cuchy s integrl formul) Suppose tht is simple closed nticlockwise curve in the complex plne nd f(z) is nlytic on some open set tht includes ll of the curve nd ll points inside. Then for ny point z 0 inside we hve f(z) dz = 2πif(z 0 ) z z 0 (We will not proved this but the ide is tht the integrl will remin the sme if we replce by smll circle round z 0. If we let the rdius of the smll circle 0 then we cn show tht the integrl must be very close to 2πif(z 0 ). As the integrl is idependent of the rdius, it must ctully be 2πif(z 0 ).) Exmple. Consider z 2 + z z 2 + 2z 3 dz The integrnd is bd t two points becuse the denomintor is zero t two plces. z 2 + z + 1 z 2 + 2z 3 = z2 + z + 1 (z + 3)(z 1) (The bd points re z = 1 nd z = 3.) Only one of these bd points z = 1 is inside 2. We cn in fct write the integrl s z 2 + z z 2 + 2z 3 dz = f(z) z 1 dz where f(z) = z2 + z + 1 z nd then the Cuchy integrl formul gives the nswer 2πif(1) = 2πi 3 = 3πi/2 for the integrl 4 we strted with. It did not mtter tht the curve ws exctly the circle of rdius 2, only tht it went round 1 but not round 3. Power series. One cn mke use of Cuchy s integrl formul to prove tht every nlytic function f(z) cn be represented by power series in ny disc where it is nlytic. If f(z) is nlytic in n open set tht includes the disc {z C : z z 0 < r} of rdius r > 0 bout z 0, then f(z) = n (z z 0 ) n for ll z with z z 0 < r. n=0 The coefficients n cn be represented s integrls n = 1 f(z) 2πi (z z 0 ) dz for ny 0 < s < r, or s n+1 n = f (n) (z 0 ) n! z z 0 =s
5 Complex integrls 5 Exmple. Tke f(z) = 1/z nd z 0 = 2. Then f(z) is nlytic for z z 0 < 2 nd so there is power series for f(z) there. e z A more complicted exmple is f(z) = (z 1)(z 2). For ny z 0 different from 1 nd 2, there is power series for f(z) in the lrgest disc z z 0 < r tht misses 1 nd 2. Specificlly r is the shorter of the two distnces z 1 nd z 2.
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