DOCSLIB.ORG

Lecture #31: the Cauchy Residue Theorem

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Lecture #31: the Cauchy Residue Theorem Mathematics312(Fall2013) November25,2013 Prof. Michael Kozdron Lecture #31: The Cauchy Residue Theorem Recall that last class we showed that a function f(z)hasapoleoforderm at z0 if and only if g(z) f(z)= (z z )m − 0 for some function g(z)thatisanalyticinaneighbourhoodofz and has g(z ) =0.Wealso 0 0 derived a formula for Res(f; z0). Theorem 31.1. If f(z) is analytic for 0 < z z0 <Rand has a pole of order m at z0, then | − | m 1 m 1 1 d − m 1 d − m Res(f; z0)= m 1 (z z0) f(z) = lim m 1 (z z0) f(z). (m 1)! dz − − (m 1)! z z0 dz − − z=z0 → − − In particular, if z0 is a simple pole, then Res(f; z0)=(z z0)f(z) =lim(z z0)f(z). − z z0 − z=z0 → Example 31.2. Suppose that sin z f(z)= . (z2 1)2 − Determine the order of the pole at z0 =1. Solution. Observe that z2 1=(z 1)(z +1)andso − − sin z sin z sin z/(z +1)2 f(z)= = = . (z2 1)2 (z 1)2(z +1)2 (z 1)2 − − − Since sin z g(z)= (z +1)2 2 is analytic at 1 and g(1) = 2− sin(1) =0,weconcludethatz =1isapoleoforder2. 0 Example 31.3. Determine the residue at z0 =1of sin z f(z)= (z2 1)2 − and compute f(z)dz C where C = z 1 =1/2 is the circle of radius 1/2centredat1orientedcounterclockwise. {| − | } 31–1 Solution. Since we can write (z 1)2f(z)=g(z)where − sin z g(z)= (z +1)2 is analytic at z =1withg(1) =0,theresidueoff(z)atz =1is 0 0 1 d2 1 d Res(f;1)= − (z 1)2f(z) = (z 1)2f(z) (2 1)! dz2 1 − dz − − − z=1 z=1 d sin z = dz (z +1)2 z=1 (z +1)2 cosz 2(z +1)sinz = − (z +1)4 z=1 4cos1 4sin1 = − 16 cos 1 sin 1 = − . 4 Observe that if C = z 1 =1/2 oriented counterclockwise, then the only singularity of {| − | } f(z)insideC is at z0 =1.Therefore, sin z (cos 1 sin 1)πi dz =2πi Res(f;1)= − . (z2 1)2 2 C − It is worth pointing out that we could have also obtained this solution using the Cauchy Integral Formula; that is, sin z g(z) cos 1 sin 1 (cos 1 sin 1)πi dz = dz =2πig(1) = 2πi − = − (z2 1)2 (z 1)2 · 4 2 C − C − as above. Remark. Suppose that C is a closed contour oriented counterclockwise. If f(z)isanalytic inside and on C except for a single point z0 where f(z)hasapoleoforderm,thenboth the Cauchy Integral Formula and the residue formula will require exactly the same work, namely the calculation of the m 1derivativeof(z z )mf(z). − − 0 Recall that there are two other types of isolated singular points to consider, namely removable singularities and essential singularities. If the singularity is removable, then the residue is obviously 0. Unfortunately, there is no direct way to determine the residue associated with an essential singularity. The coefficient a 1 of the Laurent series must be determined explicitly. − In summary, suppose that f(z)isanalyticfor0< z z <Rand has an isolated singularity | − 0| at z0.Bydirectinspectionofthefunction,onemaymakeaneducatedguessastowhether the isolated singularity is removable, a pole, or essential. If it believed to be either removable or essential, then compute the Laurent series to determine Res(f; z0). If it is believed to be a pole, then attempt to compute Res(f; z0)usingTheorem30.6. 31–2 Theorem 31.4 (Cauchy Residue Theorem). Suppose that C is a closed contour oriented counterclockwise. If f(z) is analytic inside and on C except at a finite number of isolated singularities z1,z2,...,zn, then n f(z)dz =2πi Res(f; zj). C j=1 Proof. Observe that if C is a closed contour oriented counterclockwise, then integration over C can be continuously deformed to a union of integrations over C1,C2,...,Cn where Cj is a circle oriented counterclockwise encircling exactly one isolated singularity, namely zj,and not passing through any of the other isolated singular points. This yields f(z)dz = f(z)dz + + f(z)dz. ··· C C1 Cn Since f(z)dz =2πi Res(f; zj), Cj the proof is complete. Remark. Note that if the isolated singularities of f(z)insideC are all either removable or poles, then the Cauchy Integral Formula can also be used to compute f(z)dz. C If any of the isolated singularities are essential, then the Cauchy Integral Formula does not apply. Moreover, even when f(z)hasonlyremovablesingularitiesorpoles,theResidue Theorem is often much easier to use than the Cauchy Integral Formula. Example 31.5. Compute 3z3 +4z2 5z +1 − dz (z 2i)(z3 z) C − − where C = z =3 is the circle of radius 3 centred at 0 oriented counterclockwise. {| | } Solution. Observe that 3z3 +4z2 5z +1 3z3 +4z2 5z +1 f(z)= − = − (z 2i)(z3 z) z(z 1)(z +1)(z 2i) − − − − has isolated singular points at z1 =0,z2 =1,z3 = 1, and z4 =2i.Moreover,eachisolated singularity is a simple pole, and so − 3z3 +4z2 5z +1 1 i Res(f;0)= − = = , (z 1)(z +1)(z 2i) 1 2i −2 z=0 − − − ·− 3z3 +4z2 5z +1 3+4 5+1 3 3(1 + 2i) Res(f;1)= − = − = = , z(z +1)(z 2i) 1 2 (1 2i) 2(1 2i) 10 z=1 − · · − − 31–3 3z3 +4z2 5z +1 3+4+5+1 7 7(2i 1) Res(f; 1) = − = − = = − , − z(z 1)(z 2i) 1 2 ( 1 2i) −2(1 + 2i) 10 z= 1 − − − − ·− · − − 3z3 +4z2 5z +1 3(2i)3 +4(2i)2 5(2i)+1 34 15i Res(f;2i)= − = − = − . z(z 1)(z +1) 2i(2i 1)(2i +1) 10 z=2i − − By the Cauchy Residue Theorem, 3z3 +4z2 5z +1 i 3(1 + 2i) 7(2i 1) 34 15i − dz =2πi + + − + − =6πi. (z 2i)(z3 z) −2 10 10 10 C − − 31–4.
Load more

Recommended publications
  • Global Subanalytic Cmc Surfaces
    GLOBALLY SUBANALYTIC CMC SURFACES IN R3 WITH SINGULARITIES JOSE´ EDSON SAMPAIO Abstract. In this paper we present a classification of a class of globally sub- 3 analytic CMC surfaces in R that generalizes the recent classification made by Barbosa and do Carmo in 2016. We show that a globally subanalytic CMC 3 surface in R with isolated singularities and a suitable condition of local con- nectedness is a plane or a finite union of round spheres and right circular cylinders touching at the singularities. As a consequence, we obtain that a 3 globally subanalytic CMC surface in R that is a topological manifold does not have isolated singularities. It is also proved that a connected closed glob- 3 ally subanalytic CMC surface in R with isolated singularities which is locally Lipschitz normally embedded needs to be a plane or a round sphere or a right circular cylinder. A result in the case of non-isolated singularities is also pre- sented. It is also presented some results on regularity of semialgebraic sets and, in particular, it is proved a real version of Mumford's Theorem on regu- larity of normal complex analytic surfaces and a result about C1 regularity of minimal varieties. 1. Introduction The question of describing minimal surfaces or, more generally, surfaces of con- stant mean curvature (CMC surfaces) is known in Analysis and Differential Geom- etry since the classical papers of Bernstein [4], Bombieri, De Giorgi and Giusti [9], Hopf [26] and Alexandrov [1]. Recently, in the paper [2], Barbosa and do Carmo showed that the connected algebraic smooth CMC surfaces in R3 are only the planes, round spheres and right circular cylinders.
    [Show full text]
  • MATH 305 Complex Analysis, Spring 2016 Using Residues to Evaluate Improper Integrals Worksheet for Sections 78 and 79
    MATH 305 Complex Analysis, Spring 2016 Using Residues to Evaluate Improper Integrals Worksheet for Sections 78 and 79 One of the interesting applications of Cauchy's Residue Theorem is to find exact values of real improper integrals. The idea is to integrate a complex rational function around a closed contour C that can be arbitrarily large. As the size of the contour becomes infinite, the piece in the complex plane (typically an arc of a circle) contributes 0 to the integral, while the part remaining covers the entire real axis (e.g., an improper integral from −∞ to 1). An Example Let us use residues to derive the formula p Z 1 x2 2 π 4 dx = : (1) 0 x + 1 4 Note the somewhat surprising appearance of π for the value of this integral. z2 First, let f(z) = and let C = L + C be the contour that consists of the line segment L z4 + 1 R R R on the real axis from −R to R, followed by the semi-circle CR of radius R traversed CCW (see figure below). Note that C is a positively oriented, simple, closed contour. We will assume that R > 1. Next, notice that f(z) has two singular points (simple poles) inside C. Call them z0 and z1, as shown in the figure. By Cauchy's Residue Theorem. we have I f(z) dz = 2πi Res f(z) + Res f(z) C z=z0 z=z1 On the other hand, we can parametrize the line segment LR by z = x; −R ≤ x ≤ R, so that I Z R x2 Z z2 f(z) dz = 4 dx + 4 dz; C −R x + 1 CR z + 1 since C = LR + CR.
    [Show full text]
  • Topic 7 Notes 7 Taylor and Laurent Series
    Topic 7 Notes Jeremy Orloff 7 Taylor and Laurent series 7.1 Introduction We originally defined an analytic function as one where the derivative, defined as a limit of ratios, existed. We went on to prove Cauchy's theorem and Cauchy's integral formula. These revealed some deep properties of analytic functions, e.g. the existence of derivatives of all orders. Our goal in this topic is to express analytic functions as infinite power series. This will lead us to Taylor series. When a complex function has an isolated singularity at a point we will replace Taylor series by Laurent series. Not surprisingly we will derive these series from Cauchy's integral formula. Although we come to power series representations after exploring other properties of analytic functions, they will be one of our main tools in understanding and computing with analytic functions. 7.2 Geometric series Having a detailed understanding of geometric series will enable us to use Cauchy's integral formula to understand power series representations of analytic functions. We start with the definition: Definition. A finite geometric series has one of the following (all equivalent) forms. 2 3 n Sn = a(1 + r + r + r + ::: + r ) = a + ar + ar2 + ar3 + ::: + arn n X = arj j=0 n X = a rj j=0 The number r is called the ratio of the geometric series because it is the ratio of consecutive terms of the series. Theorem. The sum of a finite geometric series is given by a(1 − rn+1) S = a(1 + r + r2 + r3 + ::: + rn) = : (1) n 1 − r Proof.
    [Show full text]
  • Math 336 Sample Problems
    Math 336 Sample Problems One notebook sized page of notes will be allowed on the test. The test will cover up to section 2.6 in the text. 1. Suppose that v is the harmonic conjugate of u and u is the harmonic conjugate of v. Show that u and v must be constant. ∞ X 2 P∞ n 2. Suppose |an| converges. Prove that f(z) = 0 anz is analytic 0 Z 2π for |z| < 1. Compute lim |f(reit)|2dt. r→1 0 z − a 3. Let a be a complex number and suppose |a| < 1. Let f(z) = . 1 − az Prove the following statments. (a) |f(z)| < 1, if |z| < 1. (b) |f(z)| = 1, if |z| = 1. 2πij 4. Let zj = e n denote the n roots of unity. Let cj = |1 − zj| be the n − 1 chord lengths from 1 to the points zj, j = 1, . .n − 1. Prove n that the product c1 · c2 ··· cn−1 = n. Hint: Consider z − 1. 5. Let f(z) = x + i(x2 − y2). Find the points at which f is complex differentiable. Find the points at which f is complex analytic. 1 6. Find the Laurent series of the function z in the annulus D = {z : 2 < |z − 1| < ∞}. 1 Sample Problems 2 7. Using the calculus of residues, compute Z +∞ dx 4 −∞ 1 + x n p0(z) Y 8. Let f(z) = , where p(z) = (z −z ) and the z are distinct and zp(z) j j j=1 different from 0. Find all the poles of f and compute the residues of f at these poles.
    [Show full text]
  • Complex Analysis Class 24: Wednesday April 2
    Complex Analysis Math 214 Spring 2014 Fowler 307 MWF 3:00pm - 3:55pm c 2014 Ron Buckmire http://faculty.oxy.edu/ron/math/312/14/ Class 24: Wednesday April 2 TITLE Classifying Singularities using Laurent Series CURRENT READING Zill & Shanahan, §6.2-6.3 HOMEWORK Zill & Shanahan, §6.2 3, 15, 20, 24 33*. §6.3 7, 8, 9, 10. SUMMARY We shall be introduced to Laurent Series and learn how to use them to classify different various kinds of singularities (locations where complex functions are no longer analytic). Classifying Singularities There are basically three types of singularities (points where f(z) is not analytic) in the complex plane. Isolated Singularity An isolated singularity of a function f(z) is a point z0 such that f(z) is analytic on the punctured disc 0 < |z − z0| <rbut is undefined at z = z0. We usually call isolated singularities poles. An example is z = i for the function z/(z − i). Removable Singularity A removable singularity is a point z0 where the function f(z0) appears to be undefined but if we assign f(z0) the value w0 with the knowledge that lim f(z)=w0 then we can say that we z→z0 have “removed” the singularity. An example would be the point z = 0 for f(z) = sin(z)/z. Branch Singularity A branch singularity is a point z0 through which all possible branch cuts of a multi-valued function can be drawn to produce a single-valued function. An example of such a point would be the point z = 0 for Log (z).
    [Show full text]
  • Residue Theorem
    Topic 8 Notes Jeremy Orloff 8 Residue Theorem 8.1 Poles and zeros f z z We remind you of the following terminology: Suppose . / is analytic at 0 and f z a z z n a z z n+1 ; . / = n. * 0/ + n+1. * 0/ + § a ≠ f n z n z with n 0. Then we say has a zero of order at 0. If = 1 we say 0 is a simple zero. f z Suppose has an isolated singularity at 0 and Laurent series b b b n n*1 1 f .z/ = + + § + + a + a .z * z / + § z z n z z n*1 z z 0 1 0 . * 0/ . * 0/ * 0 < z z < R b ≠ f n z which converges on 0 * 0 and with n 0. Then we say has a pole of order at 0. n z If = 1 we say 0 is a simple pole. There are several examples in the Topic 7 notes. Here is one more Example 8.1. z + 1 f .z/ = z3.z2 + 1/ has isolated singularities at z = 0; ,i and a zero at z = *1. We will show that z = 0 is a pole of order 3, z = ,i are poles of order 1 and z = *1 is a zero of order 1. The style of argument is the same in each case. At z = 0: 1 z + 1 f .z/ = ⋅ : z3 z2 + 1 Call the second factor g.z/. Since g.z/ is analytic at z = 0 and g.0/ = 1, it has a Taylor series z + 1 g.z/ = = 1 + a z + a z2 + § z2 + 1 1 2 Therefore 1 a a f .z/ = + 1 +2 + § : z3 z2 z This shows z = 0 is a pole of order 3.
    [Show full text]
  • A Formal Proof of Cauchy's Residue Theorem
    A Formal Proof of Cauchy's Residue Theorem Wenda Li and Lawrence C. Paulson Computer Laboratory, University of Cambridge fwl302,[email protected] Abstract. We present a formalization of Cauchy's residue theorem and two of its corollaries: the argument principle and Rouch´e'stheorem. These results have applications to verify algorithms in computer alge- bra and demonstrate Isabelle/HOL's complex analysis library. 1 Introduction Cauchy's residue theorem | along with its immediate consequences, the ar- gument principle and Rouch´e'stheorem | are important results for reasoning about isolated singularities and zeros of holomorphic functions in complex anal- ysis. They are described in almost every textbook in complex analysis [3, 15, 16]. Our main motivation of this formalization is to certify the standard quantifier elimination procedure for real arithmetic: cylindrical algebraic decomposition [4]. Rouch´e'stheorem can be used to verify a key step of this procedure: Collins' projection operation [8]. Moreover, Cauchy's residue theorem can be used to evaluate improper integrals like Z 1 itz e −|tj 2 dz = πe −∞ z + 1 Our main contribution1 is two-fold: { Our machine-assisted formalization of Cauchy's residue theorem and two of its corollaries is new, as far as we know. { This paper also illustrates the second author's achievement of porting major analytic results, such as Cauchy's integral theorem and Cauchy's integral formula, from HOL Light [12]. The paper begins with some background on complex analysis (Sect. 2), fol- lowed by a proof of the residue theorem, then the argument principle and Rouch´e'stheorem (3{5).
    [Show full text]
  • Chapter 2 Complex Analysis
    Chapter 2 Complex Analysis In this part of the course we will study some basic complex analysis. This is an extremely useful and beautiful part of mathematics and forms the basis of many techniques employed in many branches of mathematics and physics. We will extend the notions of derivatives and integrals, familiar from calculus, to the case of complex functions of a complex variable. In so doing we will come across analytic functions, which form the centerpiece of this part of the course. In fact, to a large extent complex analysis is the study of analytic functions. After a brief review of complex numbers as points in the complex plane, we will ¯rst discuss analyticity and give plenty of examples of analytic functions. We will then discuss complex integration, culminating with the generalised Cauchy Integral Formula, and some of its applications. We then go on to discuss the power series representations of analytic functions and the residue calculus, which will allow us to compute many real integrals and in¯nite sums very easily via complex integration. 2.1 Analytic functions In this section we will study complex functions of a complex variable. We will see that di®erentiability of such a function is a non-trivial property, giving rise to the concept of an analytic function. We will then study many examples of analytic functions. In fact, the construction of analytic functions will form a basic leitmotif for this part of the course. 2.1.1 The complex plane We already discussed complex numbers briefly in Section 1.3.5.
    [Show full text]
  • Lecture Note for Math 220A Complex Analysis of One Variable
    Lecture Note for Math 220A Complex Analysis of One Variable Song-Ying Li University of California, Irvine Contents 1 Complex numbers and geometry 2 1.1 Complex number field . 2 1.2 Geometry of the complex numbers . 3 1.2.1 Euler's Formula . 3 1.3 Holomorphic linear factional maps . 6 1.3.1 Self-maps of unit circle and the unit disc. 6 1.3.2 Maps from line to circle and upper half plane to disc. 7 2 Smooth functions on domains in C 8 2.1 Notation and definitions . 8 2.2 Polynomial of degree n ...................... 9 2.3 Rules of differentiations . 11 3 Holomorphic, harmonic functions 14 3.1 Holomorphic functions and C-R equations . 14 3.2 Harmonic functions . 15 3.3 Translation formula for Laplacian . 17 4 Line integral and cohomology group 18 4.1 Line integrals . 18 4.2 Cohomology group . 19 4.3 Harmonic conjugate . 21 1 5 Complex line integrals 23 5.1 Definition and examples . 23 5.2 Green's theorem for complex line integral . 25 6 Complex differentiation 26 6.1 Definition of complex differentiation . 26 6.2 Properties of complex derivatives . 26 6.3 Complex anti-derivative . 27 7 Cauchy's theorem and Morera's theorem 31 7.1 Cauchy's theorems . 31 7.2 Morera's theorem . 33 8 Cauchy integral formula 34 8.1 Integral formula for C1 and holomorphic functions . 34 8.2 Examples of evaluating line integrals . 35 8.3 Cauchy integral for kth derivative f (k)(z) . 36 9 Application of the Cauchy integral formula 36 9.1 Mean value properties .
    [Show full text]
  • Residue Thry
    17 Residue Theory “Residue theory” is basically a theory for computing integrals by looking at certain terms in the Laurent series of the integrated functions about appropriate points on the complex plane. We will develop the basic theorem by applying the Cauchy integral theorem and the Cauchy integral formulas along with Laurent series expansions of functions about the singular points. We will then apply it to compute many, many integrals that cannot be easily evaluated otherwise. Most of these integrals will be over subintervals of the real line. 17.1 Basic Residue Theory The Residue Theorem Suppose f is a function that, except for isolated singularities, is single-valued and analytic on some simply-connected region R . Our initial interest is in evaluating the integral f (z) dz . C I 0 where C0 is a circle centered at a point z0 at which f may have a pole or essential singularity. We will assume the radius of C0 is small enough that no other singularity of f is on or enclosed by this circle. As usual, we also assume C0 is oriented counterclockwise. In the region right around z0 , we can express f (z) as a Laurent series ∞ k f (z) ak (z z0) , = − k =−∞X and, as noted earlier somewhere, this series converges uniformly in a region containing C0 . So ∞ k ∞ k f (z) dz ak (z z0) dz ak (z z0) dz . C = C − = C − 0 0 k k 0 I I =−∞X =−∞X I But we’ve seen k (z z0) dz for k 0, 1, 2, 3,..
    [Show full text]
  • Laurent Series and Residue Calculus
    Laurent Series and Residue Calculus Nikhil Srivastava March 19, 2015 If f is analytic at z0, then it may be written as a power series: 2 f(z) = a0 + a1(z − z0) + a2(z − z0) + ::: which converges in an open disk around z0. In fact, this power series is simply the Taylor series of f at z0, and its coefficients are given by I 1 (n) 1 f(z) an = f (z0) = n+1 ; n! 2πi (z − z0) where the latter equality comes from Cauchy's integral formula, and the integral is over a positively oriented contour containing z0 contained in the disk where it f(z) is analytic. The existence of this power series is an extremely useful characterization of f near z0, and from it many other useful properties may be deduced (such as the existence of infinitely many derivatives, vanishing of simple closed contour integrals around z0 contained in the disk of convergence, and many more). The situation is not much worse when z0 is an isolated singularity of f, i.e., f(z) is analytic in a puncured disk 0 < jz − z0j < r for some r. In this case, we have: Laurent's Theorem. If z0 is an isolated singularity of f and f(z) is analytic in the annulus 0 < jz − z0j < r, then 2 b1 b2 bn f(z) = a0 + a1(z − z0) + a2(z − z0) + ::: + + 2 + ::: + n + :::; (∗) z − z0 (z − z0) (z − z0) where the series converges absolutely in the annulus. In class I described how this can be done for any annulus, but the most useful case is a punctured disk around an isolated singularity.
    [Show full text]
  • 21. the Residue Theorem Definition 21.1. Let F Be a Holomorphic Function with an Isolated Singularity at A. the Residue of F At
    21. The residue theorem Definition 21.1. Let f be a holomorphic function with an isolated singularity at a. The residue of f at a, denoted Resa f(z), is equal to a−1, the coef- ficient of 1=(z − a) in the Laurent series expansion of f in a punctured neighbourhood of a. Theorem 21.2 (Residue Theorem). Let U be a bounded domain with piecewise differentiable boundary and let f be a holomorphic function on U [@U except at finitely many isolated singular points a1; a2; : : : ; an belonging to U. We have n Z X f(z) dz = 2πi Resai f(z): @U j=1 Before we prove this theorem, we give some applications. Example 21.3. Compute: I 1 dz jzj=3 (z − 1)(z + 1) The function 1 (z − 1)(z + 1) has isolated singularities at z = 1 and z = −1 and is otherwise holo- morphic. Both of these points belong to the open unit disk of radius 3 centred at 0. The boundary of this disk is the circle of radius 3, centred at 0. So we have to compute the residue at these two points and then apply the residue theorem. We use the method of partial fractions 1 1 1 1 1 = − : (z − 1)(z + 1) 2 z − 1 2 z + 1 We have 1 1 Res = : 1 (z − 1)(z + 1) 2 and 1 1 Res = − : −1 (z − 1)(z + 1) 2 Thus I 1 dz = 0; jzj=3 (z − 1)(z + 1) by the residue theorem. 1 To apply the residue theorem, we need to be able to compute residues efficiently.
    [Show full text]