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CONTOUR INTEGRALS 1. Theory

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CONTOUR INTEGRALS 1. Theory CONTOUR INTEGRALS 1. Theory Denition 1. Let γ :[a; b] ! C be a C1 curve and let f : C ! C. Dene: b f(z)dz = f(γ(t))γ0(t)dt ˆγ ˆa Remark 2. This is just like a line integral in R, so we can use tools from there e.g. the fundemental theorem of calculus. One can check using the change of variables formula that if we re-parametrize the curve, then the value of the integral does not change. If you reverse the direction of the curve, you get a multiplicative factor of −1. Theorem 3. [Cauchy-Goursat Theorem] If f(z) is holomorphic, and γ is the closed curve over a triangle, then dz=0. γ f(z) ´ Proof. (super rough sketch) Cut triangle into four pieces (midpoints), and integrate along each piece. There is one piece so that j piecej > 1=4j whole trianglej. Doing this recursivly, gives a sequence of triangles ∆k withj ∆kj > 1 original triangle But since triangle pieces are closed sets, there is a single point in the intersection k . 4k j ´ j ´ ∆ ´! fz0g k Use´ the fact that f is holomorphic to write a Taylor series approx near z0, with error to 0 as ∆ as k ! 1. Then k plug approx into ∆ to get an upper bound estimate and conclude original = 0. ´ ´ Corollary 4. Holomorphic function have antiderivatives in convex (or starlike) domains. Proof. Dene F (z) = f(ζ)dζ. Show that F is continuous by the Cauchy Goursat theorem. The check F is [z0;z] dierentiable by the Cauchy´ Riemann Equantions (again, by C-G) and the derivative is f. Corollary 5. If holomorphic in a convex (or starlike) domain then d for any closed curve . f γ f(z) z = 0 γ ´ Proof. By last corolarry it has an antiderivative, so integral is zero by Fundemental Theorem of Calculus. Corollary 6. If γ1 and γ2 are homotopic (i.e. they can be continuously deforemed into each other) in a region Ω where f is holomorphic, then f(z)dz = f(z)dz γ1 γ2 ´ ´ Proof. Requires a little bit of work and set up....but not too hard from where we are. Theorem 7. [Cauchy Integral Formula] Suppose that f is holomorphic in a disk of radius R centered at some point z0. Then for any a in the disk and r < R: 1 f(z) f(a) = dz 2πi ˆ z − a jz−z0j=r Proof. By the previous theorem, the integral on the RHS does not depend on r. By taking r ! 0 we can see that this value has to be f(a). Corollary 8. Suppose f is holomorphic in Ar;R = fz : r < z < Rg :The f has Laurent Series expansion, f(a) = P1 n for . Moreover, the coecients are given explicitly in terms of an integrals of n=−∞ a cn a 2 Ar;R cn f(z) Proof. Suppose WOLOG we are working in Draw a picture then use the homotopic curves theorem and the Cauchy Integral Formula to see that for any a 2 Ar;R: 1 f(z) f(a) = dz 2πi ˆ z − a jz−aj= 1 f(z) 1 f(z) = dz − dz 2πi ˆ z − a 2πi ˆ z − a jzj=R jzj=r 1 CONTOUR INTEGRALS 2 Then do some manipulations to force a geometric sum out: f(z) d 1 f(z)d z = a z ˆ z − a ˆ 1 − z z jzj=R jzj=R 1 X an f(z) = dz ˆ z z n=0 jzj=R 0 1 1 X f(z) = an B dzC @ ˆ zn+1 A n=0 jzj=R Exchanging the sum with the integral here is ok because the sum is uniformly convergent as has a a 2 Ar;R z < 1 for . We do the same thing on the inner circle of radius , where this time z for : jzj = R r a < 1 jzj = r f(z) d −1 f(z)d − z = z z ˆ z − a ˆ a − 1 a jzj=r jzj=r 1 X z n f(z) = dz ˆ a a n=0 jzj=r 0 1 1 X = a−(n+1) B f(z)zndzC @ ˆ A n=0 jzj=r So nally then: 1 X n f(a) = a cn n=−∞ 1 f(z) c = dz for n ≥ 0 n 2πi ˆ zn+1 jzj=R 1 c = f(z)zn−1dz for n ≥ 1 −n 2πi ˆ jzj=r Corollary 9. In the above setting, d is the coecient of the exponent in the Laurent series γ f(z) z = 2πic−1 −1 expansion of f. ´ Denition 10. The coecient c−1 above is called the Residue of f at z = 0 This is because the annulus Ar;R was centered at . In general where is the Laurent series expansion for in an annulus z = 0 Resz=z0 f := c−1 c−1 f centered at z0. There are lots of dierent notations for the residue of f. Theorem 11. [Residue Theorem] f(z)dz = 2πi P W ind (z ) · Res f γ zk γ k z=zk ´ Proof. This is not very far from the corollary we just saw! 2. Computing Residues Method_1 Do some algebraic manipulations to get the Laurent series expansion of f and look at c−1. Usually, you do this by combining some power series you already know e.g. α or 1 3 (1 + z) = 1 + αz + ::: sin(z) = z − 3! z + 1 5 5! z + ::: CONTOUR INTEGRALS 3 Example Show 1 = 3! Resz=0 z−sin(z) 5·4 1 1 sin z −1 = 1 − z − sin(z) z z 1 1 1 −1 = 1 − 1 + z2 − z4 ::: z 3! 5! 1 3! 1 −1 = 1 − z2 + ::: z z2 5 · 4 3! 1 = 1 + z2 + ::: z3 5 · 4 3! 3! 1 = + + ::: z3 5 · 4 z −k Method_2 If you know the order of the singularity and it is nite, then you know that f(z) = c−k (z − z0) + k k−1 :::, so (z − z0) f(z) = c−k + ::: is an ordinary power series. The coecient c−1 is the coecient of (z − z0) in f(z), so its not hard to see that: k−1 1 d h k i Resz=z0 f = lim k−1 (z − z0) f(z) z!z0 (k − 1)! dz For a simple pole, this is: Resz=0f = lim zf(z) z!0 If g(z) and 0 then: f(z) = h(z) h(0) = 0; h (0) 6= 0 g(z) Resz=0f = lim z z!0 h(z) g(z) = lim z!0 h(z)=z g(0) = h0(0) 3. Making Estimates It often comes up that you want estimates on f(z)dz to show that certain contour integrals are small. Here γ are two that are good. ´ Theorem 12. [Estimation Lemma] Let and let 0 d be the length of the curve . M = supz2γ jf(z)j L = jγ (t)j t γ Then f(z)dz ≤ ML ´ γ ´ Proof. From the denition, f(z)dz = f(γ(t))γ0(t)dt ≤ jf(γ(t))j jγ0(t)j dt ≤ M jγ0(t)j dt = ML γ ´ ´ ´ ´ iaz iθ Theorem 13. [Jordan's Lemma] Suppose f(z) can be written f(z) = e g(z)Let CR = fz : z = Re ; θ 2 [0; π]g be a curve in the upper half plane. Then: π iθ f(z) dz ≤ max g Re θ2[0,π] ˆCR a iθ Proof. (From Wikipedia) Let IR = f(z) dz and MR = max g Re for convenience. Have: CR θ2[0,π] ´ π f(z) dz = g(Reiθ) eiaR(cos θ+i sin θ) iReiθ dθ ˆCR ˆ0 π = R g(Reiθ) eaR(i cos θ−sin θ) ieiθ dθ : ˆ0 CONTOUR INTEGRALS 4 So then: π iθ aR(i cos θ−sin θ) iθ IR = f(z) dz ≤ R g(Re ) e ie dθ ˆCR ˆ0 π iθ −aR sin θ = R g(Re ) e dθ ˆ0 π −aR sin θ ≤ MRR e dθ ˆ0 π=2 −aR sin θ = 2MRR e dθ ˆ0 Now use the inequality 2θ (This holds since is concave down for ; draw a picture to see sin θ ≥ π sin θ 2 [0; π=2] what the inequality is saying), to get: π=2 −aR2θ/π IR ≤ 2MRR e dθ ˆ0 π = 2M R 1 − e−aR R 2aR π ≤ M a R Remark 14. Jordan's lemma will work nicely for many applications. If you need a bit more power, it is possible to θR π=2 improve it. One way is to split the integral above into and then from . If you choose θR to be small (i.e. 0 θR θR ´ ´ π=2 θR ! 0 as R ! 1) then the integral is controlled since θR is small, while is controlled since y > θR here. 0 θR This works well for functions for which´ jf(x + iy)j is small for large y. ´ Remark 15. Wikipedia has a really really great section on methods of Contour Integration with a few nice worked examples. http://en.wikipedia.org/wiki/Methods_of_contour_integration Example 16. 1 log x d 0 (1+x2)2 x =? ´ Proof. (From Wikipedia) To do this integral, it turns out it is useful to look at the function log(z)2 . (We use f(z) = (1+z2)2 the branch of log where Arg(z) 2 (−π; π)) Draw a keyhole contour around this branch of logarithm. The outside and inside circular contours ! 0 by the estimation lemma.
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