DOCSLIB.ORG

Topic 9 Notes 9 Definite Integrals Using the Residue Theorem

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Topic 9 Notes 9 Definite Integrals Using the Residue Theorem Topic 9 Notes Jeremy Orloff 9 Definite integrals using the residue theorem 9.1 Introduction In this topic we'll use the residue theorem to compute some real definite integrals. Z b f(x) dx a The general approach is always the same 1. Find a complex analytic function g(z) which either equals f on the real axis or which is closely connected to f, e.g. f(x) = cos(x), g(z) = eiz. 2. Pick a closed contour C that includes the part of the real axis in the integral. 3. The contour will be made up of pieces. It should be such that we can compute Z g(z) dz over each of the pieces except the part on the real axis. Z 4. Use the residue theorem to compute g(z) dz. C 5. Combine the previous steps to deduce the value of the integral we want. 9.2 Integrals of functions that decay The theorems in this section will guide us in choosing the closed contour C described in the introduction. The first theorem is for functions that decay faster than 1=z. Theorem 9.1. (a) Suppose f(z) is defined in the upper half-plane. If there is an a > 1 and M > 0 such that M f(z) < j j z a j j for z large then j j Z lim f(z) dz = 0; R!1 CR where CR is the semicircle shown below on the left. Im(z) Im(z) R R − Re(z) CR CR Re(z) R R − 1 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 2 Semicircles: left: Reiθ, 0 < θ < π right: Reiθ, π < θ < 2π. (b) If f(z) is defined in the lower half-plane and M f(z) < ; j j z a j j where a > 1 then Z lim f(z) dz = 0; R!1 CR where CR is the semicircle shown above on the right. Proof. We prove (a), (b) is essentially the same. We use the triangle inequality for integrals and the estimate given in the hypothesis. For R large Z Z Z Z π M M Mπ f(z) dz f(z) dz dz = R dθ = : ≤ j j j j ≤ z a j j Ra Ra−1 CR CR CR j j 0 Since a > 1 this clearly goes to 0 as R . QED ! 1 The next theorem is for functions that decay like 1=z. It requires some more care to state and prove. Theorem 9.2. (a) Suppose f(z) is defined in the upper half-plane. If there is an M > 0 such that M f(z) < j j z j j for z large then for a > 0 j j Z lim f(z)eiaz dz = 0; x1!1; x2!1 C1+C2+C3 where C1 + C2 + C3 is the rectangular path shown below on the left. Im(z) Im(z) C x2 x1 2 − Re(z) i(x1 + x2) C3 C1 C3 C1 Re(z) x x1 i(x1 + x2) C2 − 2 − Rectangular paths of height and width x1 + x2. (b) Similarly, if a < 0 then Z lim f(z)eiaz dz = 0; x1!1; x2!1 C1+C2+C3 where C1 + C2 + C3 is the rectangular path shown above on the right. Note. In contrast to Theorem 9.1 this theorem needs to include the factor eiaz. Proof. (a) We start by parametrizing C1, C2, C3. C1: γ1(t) = x1 + it, t from 0 to x1 + x2 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 3 C2: γ2(t) = t + i(x1 + x2), t from x1 to x2 − C3: γ3(t) = x2 + it, t from x1 + x2 to 0. − Next we look at each integral in turn. We assume x1 and x2 are large enough that M f(z) < j j z j j on each of the curves Cj. Z Z Z iaz iaz M iaz f(z)e dz f(z)e dz e dz ≤ j j j j ≤ z j j j j C1 C1 C1 j j Z x1+x2 M = eiax1−at dt p 2 2 0 x1 + t j j M Z x1+x2 e−at dt ≤ x1 0 M = (1 e−a(x1+x2))=a: x1 − Since a > 0, it is clear that this last expression goes to 0 as x1 and x2 go to . 1 Z Z Z iaz iaz M iaz f(z)e dz f(z)e dz e dz ≤ j j j j ≤ z j j j j C2 C2 C2 j j Z x1 M = eiat−a(x1+x2) dt p 2 2 −x2 t + (x1 + x2) j j Me−a(x1+x2) Z x1+x2 dt ≤ x1 + x2 0 Me−a(x1+x2) ≤ Again, clearly this last expression goes to 0 as x1 and x2 go to . 1 The argument for C3 is essentially the same as for C1, so we leave it to the reader. The proof for part (b) is the same. You need to keep track of the sign in the exponentials and make sure it is negative. Example. See Example 9.16 below for an example using Theorem 9.2. Z 1 Z 1 9.3 Integrals and −∞ 0 Example 9.3. Compute Z 1 1 I = 2 2 dx: −∞ (1 + x ) Solution: Let f(z) = 1=(1 + z2)2: 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 4 It is clear that for z large f(z) 1=z4: ≈ In particular, the hypothesis of Theorem 9.1 is satisfied. Using the contour shown below we have, by the residue theorem, Z X f(z) dz = 2πi residues of f inside the contour. (1) C1+CR Im(z) CR i Re(z) R C R − 1 We examine each of the pieces in the above equation. Z f(z) dz: By Theorem 9.1(a), CR Z lim f(z) dz = 0: R!1 CR Z f(z) dz: Directly, we see that C1 Z Z R Z 1 lim f(z) dz = lim f(x) dx = f(x) dx = I: R!1 R!1 C1 −R −∞ So letting R , Equation 1 becomes ! 1 Z 1 X I = f(x) dx = 2πi residues of f inside the contour. −∞ Finally, we compute the needed residues: f(z) has poles of order 2 at i. Only z = i is ± inside the contour, so we compute the residue there. Let 1 g(z) = (z i)2f(z) = : − (z + i)2 Then 2 1 Res(f; i) = g0(i) = = −(2i)3 4i So, π I = 2πi Res(f; i) = : 2 Example 9.4. Compute Z 1 1 I = 4 dx: −∞ x + 1 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 5 Solution: Let f(z) = 1=(1 + z4). We use the same contour as in the previous example Im(z) CR ei3π/4 eiπ/4 Re(z) R C R − 1 As in the previous example, Z lim f(z) dz = 0 R!1 CR and Z Z 1 lim f(z) dz = f(x) dx = I: R!1 C1 −∞ So, by the residue theorem Z X I = lim f(z) dz = 2πi residues of f inside the contour. R!1 C1+CR The poles of f are all simple and at eiπ=4; ei3π=4; ei5π=4; ei7π=4: Only eiπ=4 and ei3π=4 are inside the contour. We compute their residues as limits using iπ=4 L'Hospital's rule. For z1 = e : −i3π=4 z z1 1 1 e Res(f; z1) = lim (z z1)f(z) = lim − 4 = lim 3 = i3π=4 = z!z1 − z!z1 1 + z z!z1 4z 4e 4 i3π=4 and for z2 = e : −iπ=4 z z2 1 1 e Res(f; z2) = lim (z z2)f(z) = lim − 4 = lim 3 = i9π=4 = z!z2 − z!z2 1 + z z!z2 4z 4e 4 So, 1 i 1 i 2i p2 I = 2πi(Res(f; z1) + Res(f; z2)) = 2πi − − + − = 2πi = π 4p2 4p2 −4p2 2 Example 9.5. Suppose b > 0. Show Z 1 cos(x) πe−b 2 2 dx = : 0 x + b 2b Solution: The first thing to note is that the integrand is even, so 1 Z 1 cos(x) I = 2 2 : 2 −∞ x + b 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 6 Also note that the square in the denominator tells us the integral is absolutely convergent. We have to be careful because cos(z) goes to infinity in either half-plane, so the hypotheses of Theorem 9.1 are not satisfied. The trick is to replace cos(x) by eix, so Z 1 ix ~ e 1 ~ I = 2 2 dx; with I = Re(I): −∞ x + b 2 Now let eiz f(z) = : z2 + b2 For z = x + iy with y > 0 we have ei(x+iy) e−y f(z) = j j = : j j z2 + b2 z2 + b2 j j j j Since e−y < 1, f(z) satisfies the hypotheses of Theorem 9.1 in the upper half-plane. Now we can use the same contour as in the previous examples Im(z) CR ib Re(z) R C R − 1 We have Z lim f(z) dz = 0 R!1 CR and Z Z 1 lim f(z) dz = f(x) dx = I:~ R!1 C1 −∞ So, by the residue theorem Z X I~ = lim f(z) dz = 2πi residues of f inside the contour. R!1 C1+CR The poles of f are at bi and both are simple. Only bi is inside the contour. We compute ± the residue as a limit using L'Hospital's rule eiz e−b Res(f; bi) = lim (z bi) = : z!bi − z2 + b2 2bi So, πe−b I~ = 2πi Res(f; bi) = : b Finally, 1 πe−b I = Re(I~) = ; 2 2b as claimed.
Load more

Recommended publications
  • 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]
  • 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]
  • 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]
  • 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]
  • 12 Lecture 12: Holomorphic Functions
    12 Lecture 12: Holomorphic functions For the remainder of this course we will be thinking hard about how the following theorem allows one to explicitly evaluate a large class of Fourier transforms. This will enable us to write down explicit solutions to a large class of ODEs and PDEs. The Cauchy Residue Theorem: Let C ⊂ C be a simple closed contour. Let f : C ! C be a complex function which is holomorphic along C and inside C except possibly at a finite num- ber of points a1; : : : ; am at which f has a pole. Then, with C oriented in an anti-clockwise sense, m Z X f(z) dz = 2πi res(f; ak); (12.1) C k=1 where res(f; ak) is the residue of the function f at the pole ak 2 C. You are probably not yet familiar with the meaning of the various components in the statement of this theorem, in particular the underlined terms and what R is meant by the contour integral C f(z) dz, and so our first task will be to explain the terminology. The Cauchy Residue theorem has wide application in many areas of pure and applied mathematics, it is a basic tool both in engineering mathematics and also in the purest parts of geometric analysis. The idea is that the right-side of (12.1), which is just a finite sum of complex numbers, gives a simple method for evaluating the contour integral; on the other hand, sometimes one can play the reverse game and use an `easy' contour integral and (12.1) to evaluate a difficult infinite sum (allowing m ! 1).
    [Show 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 = − .
    [Show full text]
  • Calculus of Complex Functions. Laurent Series and Residue Theorem
    Math Methods I Lia Vas Calculus of Complex functions. Laurent Series and Residue Theorem Review of complex numbers.A complex number is any expression of the form x+iy wherep x and y are real numbers, called the real part and the imaginary part of x + iy; and i is −1: Thus, i2 = −1. Other powers of i can be determined using the relation i2 = −1: For example, i3 = i2i = −i and i10 = (i2)5 = (−1)5 = −1: Two complex numbers can be added by adding similar terms and they can be multiplied by foiling. The complex number x − iy is said to be complex conjugate of the number x + iy: To find the quotient of two complex numbers, one multiplies both the numerator and the denominator by the complex conjugate of the denominator. In this way, the answer is again in x + iy form. Example 1. Find the sum, product and quotient of 2 + i and 3 − 4i: Solutions. The sum is 2 + i + 3 − 4i = 2 + 3 + i − 4i = 5 − 3i: The product is (2 + i)(3 − 4i) = 6 − 8i + 3i − 4i2 = 6 − 8i + 3i + 4 = 10 − 5i: The complex number 3 + 4i is the conjugate of 3 − 4i; so the quotient is 2 + i 2 + i 3 + 4i (2 + i)(3 + 4i) 6 + 8i + 3i − 4 2 + 11i 2 11 = = = = = + i 3 − 4i 3 − 4i 3 + 4i (3 − 4i)(3 + 4i) 9 + 12i − 12i + 16 25 25 25 Trigonometric Representations. Let us recall the polar coordinates x = r cos θ and y = r sin θ: Using this representation, we have that z = x + iy = r cos θ + ir sin θ: The value r is the distance from the point (x; y) in the plane to the origin.
    [Show full text]
  • 1 Some Very Useful Theorems
    Yuval Advanced Complex Analysis Mathcamp 2017 1 Some very useful theorems The ultimate goal of this class is to understand the geometric properties of analytic functions, but before we can do that, there are quite a few theorems that we need to prove first. All of these theorems are suprising, beautiful, and insanely useful. 1.1 Isolated roots and unique continuations Definition. A function f : C ! C is said to have isolated roots if its roots (the places where it evaluates to 0) are \spaced out". Formally, this means that if z0 2 C is a root, namely f(z0) = 0, then there is some " > 0 such that if 0 < jz − z0j < ", then f(z) 6= 0. In other words, the roots of f are not arbitrarily close to one another. Having isolated roots is a very special property. For instance, the function g : C ! C defined by g(x + iy) = x − y is an extremely nice function, but it does not have isolated roots: the entire line y = x is composed of roots of g, so every root has other roots arbitrarily close to it. Nevertheless, we have the following result: Theorem. If f : C ! C is analytic and not the zero function, then f has isolated roots. Proof. Suppose not. Then we have a root z0 that has other roots arbitrarily close to it. This means that we can find a sequence of points z1; z2;::: 2 C with f(zk) = 0 and limk!1 zk = z0. We expand f as a Taylor series centered at z0, by writing 1 X n f(z) = an(z − z0) n=0 for some collection of complex numbers an.
    [Show full text]
  • Complex Integration
    Chapter 1 Complex integration 1.1 Complex number quiz 1 1. Simplify 3+4i . 1 2. Simplify | 3+4i |. 3. Find the cube roots of 1. 4. Here are some identities for complex conjugate. Which ones need correc- tion? z + w =z ¯+w ¯, z − w =z ¯− w¯, zw =z ¯w¯, z/w =z/ ¯ w¯. Make suitable corrections, perhaps changing an equality to an inequality. 5. Here are some identities for absolute value. Which ones need correction? |z + w| = |z| + |w|, |z − w| = |z| − |w|, |zw| = |z||w|, |z/w| = |z|/|w|. Make suitable corrections, perhaps changing an equality to an inequality. 6. Define log(z) so that −π < = log(z) ≤ π. Discuss the identities elog(z) = z and log(ew) = w. 7. Define zw = ew log z. Find ii. 8. What is the power series of log(1 + z) about z = 0? What is the radius of convergence of this power series? 9. What is the power series of cos(z) about z = 0? What is its radius of convergence? 10. Fix w. How many solutions are there of cos(z) = w with −π < <z ≤ π. 1 2 CHAPTER 1. COMPLEX INTEGRATION 1.2 Complex functions 1.2.1 Closed and exact forms In the following a region will refer to an open subset of the plane. A differential form p dx + q dy is said to be closed in a region R if throughout the region ∂q ∂p = . (1.1) ∂x ∂y It is said to be exact in a region R if there is a function h defined on the region with dh = p dx + q dy.
    [Show full text]