NPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
|
|
- Emery Dickerson
- 5 years ago
- Views:
Transcription
1 NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 28
2 Complex Analysis Module: 6: Residue Calculus Lecture: 5: s on Improper integrals A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 2 / 28
3 Evaluate the integral Γ 1 + z sin z dz where Γ is the square whose vertices are the points 4 + 4i, 4 + 4i, 4 4i, 4 4i. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 3 / 28
4 Solution: The poles of the function z Γ 1 + z sin z occur when sin z = 0, that is, when z = nπ, where n is an integer. It follows that the only poles which lie inside Γ are those points 0, π, π. These are all simple poles. The residue at these points are 1, 1 π and 1 + π respectively. The value of the integral is therefore 2πi[1 + ( 1 π) + ( 1 + π)] = 2πi. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 4 / 28
5 Evaluate the integral 0 t sin t t dt. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 5 / 28
6 Solution: Since t t sin t t 4 is an even function, we have + 4 t sin t 0 t dt = 1 t sin t 2 t dt. The singularities of the function z half-plane are simple poles. zeiz z 4 in the upper + 4 These are at the points 1 + i and 1 + i with residues ei(1+i) 8i e i( 1+i), respectively. 8i and A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 6 / 28
7 Therefore the sum of the residue is e i(1+i) 8i + ei( 1+i) 8i = e 1 (e i e 1 ) = 1 sin 1, 8i 4e A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 7 / 28
8 This gives t sin t t ( dt =Im 2πi. 1 ) 4e sin 1 = π sin 1. 4e A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 8 / 28
9 Evaluate t t 3 1 dt. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 9 / 28
10 Solution: The poles of z z z 3 1 occur at 1, 1 2 ( 1 + i 3) and at 1 2 ( 1 i 3). Since the point 1 2 ( 1 i 3) lies in the lower half plane, we consider the other two. The residue at 1 is 1 2 and at 1 2 ( 1 + i 3) is 1 6 ( 1 i 3). Hence the value of the integral is t t 3 1 dt =Re = π 3 3 ( 2πi. 1 6 ( 1 i 3) + πi. 1 ) 3 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 10 / 28
11 Compute z =1 z 1 dz. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 11 / 28
12 Solution: Let Then Hence z =1 z = e iθ, 0 θ 2π. dz = ie iθ dθ; d(z) = dθ; z 1 = e iθ 1. z 1 dz = = 0 2π 0 2π e iθ 1 dθ = 2 sin θ/2 dθ = 8. 2π 0 2(1 cos θ) dθ A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 12 / 28
13 Evaluate the integral π/2 0 dx a 2 + sin 2 x a > 1. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 13 / 28
14 Solution: π dθ Let I = 0 a 2 + sin 2 θ. By substituting 2θ = t, we arrive at an integral I = 2π 0 dt 2a cos t. Again by the substitution z = e it, dz = ie it dt the integral becomes 1 I = C 2a ( z + 1 ) dz iz 2 z where C is the circle z = 1. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 14 / 28
15 where I =2i =2i =2i f (z) = C C C dz (4a 2 + 2)z (z 2 + 1) dz z 2 2z(2a 2 + 1) + 1 f (z) dz 1 z 2 2z(2a 2 + 1) + 1. Poles of f (z) are given by α = 2a a a 2 + 1, β = 2a a a A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 15 / 28
16 Clearly α > 1 and β < 1, therefore z = β is the only pole inside C. The residue is given by Res z=β f (z) = lim z β (z β)f (z) (z β) = lim z β (z α)(z β) 1 = lim z β (z α) = 1 β α 1 = 4a a A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 16 / 28
17 By Cauchy Residue theorem, ( ) 1 f (z)dz =2πi 4a = a C πi 2a a Hence This means π 0 π/2 0 [ ] πi I = 2i 2a a dθ a 2 + sin 2 θ = π a a dθ a 2 + sin 2 θ = π 2a a A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 17 / 28
18 Evaluate x 2 x + 2 x x dx A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 18 / 28
19 Solution: Let us consider the contour integral f (z) = z2 z + 2 z z = z 2 z + 2 (z 2 + 1)(z 2 + 9). By Cauchy theorem f (z) dz = 2πi(sum of the residues inside C). The poles of f (z) are give by z = ±i, ±3i. C The residue at z = i and z = 3i are respectively 3 7i 48. 1(i + 1) 16 and A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 19 / 28
20 Hence the value of the integral is given by ( ) 10i f (z) dz = 2πi 48 C = 5π 12 This means R R f (x) dx + f (z) dz = 5π Γ 12 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 20 / 28
21 As R, we get f (x) dz = 5π 12 Γ f (z) dz 0 Therefore x 2 x + 2 x x dx = 5π 12. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 21 / 28
22 Evaluate the integral 0 x 1/3 1 + x 2 dx. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 22 / 28
23 Solution: Consider f (z) dz, where f (z) = z1/3 C 1 + z 2. Here z = 0 is a branch point and the poles are given by z = ±i. The principal branch is chosen such that Logz = 0 when z = 1. The contour C consists of the large semi circle z = R in the upper half plane and the real axis from R to R intended at z = 0, a small circle γ of radius r. The only pole lying within C is z = i and the residue is given by 1 2 (eiπ/2 ) 4/3. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 23 / 28
24 By Cauchy theorem f (z) dz = 2πi [ 12 ] (eiπ/2 ) 4/3 This means γ R C R f (x) dx + f (z) dz + f (x) dx + γ γ = 2πi. Γ f (z) dz [ 12 (eiπ/2 ) 4/3 ]. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 24 / 28
25 If lim z 0 z f (z) = A then f (z) dz =A(θ 2 θ 1 ), lim γ 0 γ z 1/3 lim z f (z) = lim z 0 z z 2 = 0 and Therefore lim f (z) dz = i(π 0) = 0 γ 0 γ A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 25 / 28
26 Also z 4/3 lim z f (z) = lim z z z = lim z z 2 4/3 (1 + 1 ) z 2 =0 Hence lim f (z) dz = 0 R A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 26 / 28
27 Thus by letting R, γ 0, we have 0 x 1/3 1 + x 2 dx + x 1/ x 2 dx = πi(eiπ/2 ) 4/3 Putting x = y, the above equation becomes, 0 πi4/3 y 1/3 e 1 + y 2 dy + x 1/3 1 + x 2 dx = πi(eiπ/2 ) 4/3 (1 e πi4/3 ) 0 0 x 1/3 1 + x 2 dx = πi(eiπ/2 ) 4/3 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 27 / 28
28 Equating real part on both sides we get (1 cos 4π/3) 0 0 x 1/3 1 + x 2 dx = π( 1) sin π π x 1/3 π sin 1 + x 2 dx = 2 sin 2 π = π 2 cosec π A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 28 / 28
NPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on omplex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 1 / 18 omplex Analysis Module: 6: Residue
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 18 Complex Analysis Module: 5:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 20 Complex Analysis Module: 10:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 17 Complex Analysis Module: 5:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 19 Complex Analysis Module: 8:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 20 Complex Analysis Module: 2:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 14 Complex Analysis Module: 1:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 16 Complex Analysis Module: 2:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 27 Complex Analysis Module: 2:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 29 Complex Analysis Module: 4:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 36 Complex Analysis Module: 7:
More informationChapter 6: Residue Theory. Introduction. The Residue Theorem. 6.1 The Residue Theorem. 6.2 Trigonometric Integrals Over (0, 2π) Li, Yongzhao
Outline Chapter 6: Residue Theory Li, Yongzhao State Key Laboratory of Integrated Services Networks, Xidian University June 7, 2009 Introduction The Residue Theorem In the previous chapters, we have seen
More information(1) Let f(z) be the principal branch of z 4i. (a) Find f(i). Solution. f(i) = exp(4i Log(i)) = exp(4i(π/2)) = e 2π. (b) Show that
Let fz be the principal branch of z 4i. a Find fi. Solution. fi = exp4i Logi = exp4iπ/2 = e 2π. b Show that fz fz 2 fz z 2 fz fz 2 = λfz z 2 for all z, z 2 0, where λ =, e 8π or e 8π. Proof. We have =
More informationSolution for Final Review Problems 1
Solution for Final Review Problems Final time and location: Dec. Gymnasium, Rows 23, 25 5, 2, Wednesday, 9-2am, Main ) Let fz) be the principal branch of z i. a) Find f + i). b) Show that fz )fz 2 ) λfz
More informationMATH 106 HOMEWORK 4 SOLUTIONS. sin(2z) = 2 sin z cos z. (e zi + e zi ) 2. = 2 (ezi e zi )
MATH 16 HOMEWORK 4 SOLUTIONS 1 Show directly from the definition that sin(z) = ezi e zi i sin(z) = sin z cos z = (ezi e zi ) i (e zi + e zi ) = sin z cos z Write the following complex numbers in standard
More informationEvaluation of integrals
Evaluation of certain contour integrals: Type I Type I: Integrals of the form 2π F (cos θ, sin θ) dθ If we take z = e iθ, then cos θ = 1 (z + 1 ), sin θ = 1 (z 1 dz ) and dθ = 2 z 2i z iz. Substituting
More informationPSI Lectures on Complex Analysis
PSI Lectures on Complex Analysis Tibra Ali August 14, 14 Lecture 4 1 Evaluating integrals using the residue theorem ecall the residue theorem. If f (z) has singularities at z 1, z,..., z k which are enclosed
More informationComplex Variables...Review Problems (Residue Calculus Comments)...Fall Initial Draft
Complex Variables........Review Problems Residue Calculus Comments)........Fall 22 Initial Draft ) Show that the singular point of fz) is a pole; determine its order m and its residue B: a) e 2z )/z 4,
More informationMTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106
Name (Last name, First name): MTH 02 omplex Variables Final Exam May, 207 :0pm-5:0pm, Skurla Hall, Room 06 Exam Instructions: You have hour & 50 minutes to complete the exam. There are a total of problems.
More informationEE2007 Tutorial 7 Complex Numbers, Complex Functions, Limits and Continuity
EE27 Tutorial 7 omplex Numbers, omplex Functions, Limits and ontinuity Exercise 1. These are elementary exercises designed as a self-test for you to determine if you if have the necessary pre-requisite
More informationSuggested Homework Solutions
Suggested Homework Solutions Chapter Fourteen Section #9: Real and Imaginary parts of /z: z = x + iy = x + iy x iy ( ) x iy = x #9: Real and Imaginary parts of ln z: + i ( y ) ln z = ln(re iθ ) = ln r
More informationMath Spring 2014 Solutions to Assignment # 8 Completion Date: Friday May 30, 2014
Math 3 - Spring 4 Solutions to Assignment # 8 ompletion Date: Friday May 3, 4 Question. [p 49, #] By finding an antiderivative, evaluate each of these integrals, where the path is any contour between the
More informationComplex Variables. Instructions Solve any eight of the following ten problems. Explain your reasoning in complete sentences to maximize credit.
Instructions Solve any eight of the following ten problems. Explain your reasoning in complete sentences to maximize credit. 1. The TI-89 calculator says, reasonably enough, that x 1) 1/3 1 ) 3 = 8. lim
More information1 Discussion on multi-valued functions
Week 3 notes, Math 7651 1 Discussion on multi-valued functions Log function : Note that if z is written in its polar representation: z = r e iθ, where r = z and θ = arg z, then log z log r + i θ + 2inπ
More informationMath Spring 2014 Solutions to Assignment # 12 Completion Date: Thursday June 12, 2014
Math 3 - Spring 4 Solutions to Assignment # Completion Date: Thursday June, 4 Question. [p 67, #] Use residues to evaluate the improper integral x + ). Ans: π/4. Solution: Let fz) = below. + z ), and for
More informationChapter 30 MSMYP1 Further Complex Variable Theory
Chapter 30 MSMYP Further Complex Variable Theory (30.) Multifunctions A multifunction is a function that may take many values at the same point. Clearly such functions are problematic for an analytic study,
More informationMath Final Exam.
Math 106 - Final Exam. This is a closed book exam. No calculators are allowed. The exam consists of 8 questions worth 100 points. Good luck! Name: Acknowledgment and acceptance of honor code: Signature:
More informationSolutions to practice problems for the final
Solutions to practice problems for the final Holomorphicity, Cauchy-Riemann equations, and Cauchy-Goursat theorem 1. (a) Show that there is a holomorphic function on Ω = {z z > 2} whose derivative is z
More informationSynopsis of Complex Analysis. Ryan D. Reece
Synopsis of Complex Analysis Ryan D. Reece December 7, 2006 Chapter Complex Numbers. The Parts of a Complex Number A complex number, z, is an ordered pair of real numbers similar to the points in the real
More informationMTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106
Name (Last name, First name): MTH 32 Complex Variables Final Exam May, 27 3:3pm-5:3pm, Skurla Hall, Room 6 Exam Instructions: You have hour & 5 minutes to complete the exam. There are a total of problems.
More informationComplex Variables & Integral Transforms
Complex Variables & Integral Transforms Notes taken by J.Pearson, from a S4 course at the U.Manchester. Lecture delivered by Dr.W.Parnell July 9, 007 Contents 1 Complex Variables 3 1.1 General Relations
More information6. Residue calculus. where C is any simple closed contour around z 0 and inside N ε.
6. Residue calculus Let z 0 be an isolated singularity of f(z), then there exists a certain deleted neighborhood N ε = {z : 0 < z z 0 < ε} such that f is analytic everywhere inside N ε. We define Res(f,
More informationResidues and Contour Integration Problems
Residues and ontour Integration Problems lassify the singularity of fz at the indicated point.. fz = cotz at z =. Ans. Simple pole. Solution. The test for a simple pole at z = is that lim z z cotz exists
More informationMa 416: Complex Variables Solutions to Homework Assignment 6
Ma 46: omplex Variables Solutions to Homework Assignment 6 Prof. Wickerhauser Due Thursday, October th, 2 Read R. P. Boas, nvitation to omplex Analysis, hapter 2, sections 9A.. Evaluate the definite integral
More informationMTH 3102 Complex Variables Solutions: Practice Exam 2 Mar. 26, 2017
Name Last name, First name): MTH 31 omplex Variables Solutions: Practice Exam Mar. 6, 17 Exam Instructions: You have 1 hour & 1 minutes to complete the exam. There are a total of 7 problems. You must show
More informationMTH3101 Spring 2017 HW Assignment 4: Sec. 26: #6,7; Sec. 33: #5,7; Sec. 38: #8; Sec. 40: #2 The due date for this assignment is 2/23/17.
MTH0 Spring 07 HW Assignment : Sec. 6: #6,7; Sec. : #5,7; Sec. 8: #8; Sec. 0: # The due date for this assignment is //7. Sec. 6: #6. Use results in Sec. to verify that the function g z = ln r + iθ r >
More informationFINAL EXAM MATH 220A, UCSD, AUTUMN 14. You have three hours.
FINAL EXAM MATH 220A, UCSD, AUTUMN 4 You have three hours. Problem Points Score There are 6 problems, and the total number of points is 00. Show all your work. Please make your work as clear and easy to
More informationCHAPTER 3 ELEMENTARY FUNCTIONS 28. THE EXPONENTIAL FUNCTION. Definition: The exponential function: The exponential function e z by writing
CHAPTER 3 ELEMENTARY FUNCTIONS We consider here various elementary functions studied in calculus and define corresponding functions of a complex variable. To be specific, we define analytic functions of
More informationMA 412 Complex Analysis Final Exam
MA 4 Complex Analysis Final Exam Summer II Session, August 9, 00.. Find all the values of ( 8i) /3. Sketch the solutions. Answer: We start by writing 8i in polar form and then we ll compute the cubic root:
More informationSecond Midterm Exam Name: Practice Problems March 10, 2015
Math 160 1. Treibergs Second Midterm Exam Name: Practice Problems March 10, 015 1. Determine the singular points of the function and state why the function is analytic everywhere else: z 1 fz) = z + 1)z
More information2 Write down the range of values of α (real) or β (complex) for which the following integrals converge. (i) e z2 dz where {γ : z = se iα, < s < }
Mathematical Tripos Part II Michaelmas term 2007 Further Complex Methods, Examples sheet Dr S.T.C. Siklos Comments and corrections: e-mail to stcs@cam. Sheet with commentary available for supervisors.
More informationcauchy s integral theorem: examples
Physics 4 Spring 17 cauchy s integral theorem: examples lecture notes, spring semester 17 http://www.phys.uconn.edu/ rozman/courses/p4_17s/ Last modified: April 6, 17 Cauchy s theorem states that if f
More informationMath 185 Fall 2015, Sample Final Exam Solutions
Math 185 Fall 2015, Sample Final Exam Solutions Nikhil Srivastava December 12, 2015 1. True or false: (a) If f is analytic in the annulus A = {z : 1 < z < 2} then there exist functions g and h such that
More informationExercises involving contour integrals and trig integrals
8::9::9:7 c M K Warby MA364 Complex variable methods applications Exercises involving contour integrals trig integrals Let = = { e it : π t π }, { e it π : t 3π } with the direction of both arcs corresponding
More informationComplex varibles:contour integration examples. cos(ax) x
1 Problem 1: sinx/x omplex varibles:ontour integration examples Integration of sin x/x from to is an interesting problem 1.1 Method 1 In the first method let us consider e iax x dx = cos(ax) dx+i x sin(ax)
More informationConformal maps. Lent 2019 COMPLEX METHODS G. Taylor. A star means optional and not necessarily harder.
Lent 29 COMPLEX METHODS G. Taylor A star means optional and not necessarily harder. Conformal maps. (i) Let f(z) = az + b, with ad bc. Where in C is f conformal? cz + d (ii) Let f(z) = z +. What are the
More informationThe Calculus of Residues
hapter 7 The alculus of Residues If fz) has a pole of order m at z = z, it can be written as Eq. 6.7), or fz) = φz) = a z z ) + a z z ) +... + a m z z ) m, 7.) where φz) is analytic in the neighborhood
More informationComplex varibles:contour integration examples
omple varibles:ontour integration eamples 1 Problem 1 onsider the problem d 2 + 1 If we take the substitution = tan θ then d = sec 2 θdθ, which leads to dθ = π sec 2 θ tan 2 θ + 1 dθ Net we consider the
More informationTypes of Real Integrals
Math B: Complex Variables Types of Real Integrals p(x) I. Integrals of the form P.V. dx where p(x) and q(x) are polynomials and q(x) q(x) has no eros (for < x < ) and evaluate its integral along the fol-
More information4.5 The Open and Inverse Mapping Theorem
4.5 The Open and Inverse Mapping Theorem Theorem 4.5.1. [Open Mapping Theorem] Let f be analytic in an open set U such that f is not constant in any open disc. Then f(u) = {f() : U} is open. Lemma 4.5.1.
More informationLecture 16 and 17 Application to Evaluation of Real Integrals. R a (f)η(γ; a)
Lecture 16 and 17 Application to Evaluation of Real Integrals Theorem 1 Residue theorem: Let Ω be a simply connected domain and A be an isolated subset of Ω. Suppose f : Ω\A C is a holomorphic function.
More informationSelected Solutions To Problems in Complex Analysis
Selected Solutions To Problems in Complex Analysis E. Chernysh November 3, 6 Contents Page 8 Problem................................... Problem 4................................... Problem 5...................................
More informationComplex Series (3A) Young Won Lim 8/17/13
Complex Series (3A) 8/7/3 Copyright (c) 202, 203 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version.2 or
More information1 Res z k+1 (z c), 0 =
32. COMPLEX ANALYSIS FOR APPLICATIONS Mock Final examination. (Monday June 7..am 2.pm) You may consult your handwritten notes, the book by Gamelin, and the solutions and handouts provided during the Quarter.
More informationMATH 452. SAMPLE 3 SOLUTIONS May 3, (10 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic.
MATH 45 SAMPLE 3 SOLUTIONS May 3, 06. (0 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic. Because f is holomorphic, u and v satisfy the Cauchy-Riemann equations:
More information18.04 Practice problems exam 2, Spring 2018 Solutions
8.04 Practice problems exam, Spring 08 Solutions Problem. Harmonic functions (a) Show u(x, y) = x 3 3xy + 3x 3y is harmonic and find a harmonic conjugate. It s easy to compute: u x = 3x 3y + 6x, u xx =
More informationComplex Function. Chapter Complex Number. Contents
Chapter 6 Complex Function Contents 6. Complex Number 3 6.2 Elementary Functions 6.3 Function of Complex Variables, Limit and Derivatives 3 6.4 Analytic Functions and Their Derivatives 8 6.5 Line Integral
More informationEEE 203 COMPLEX CALCULUS JANUARY 02, α 1. a t b
Comple Analysis Parametric interval Curve 0 t z(t) ) 0 t α 0 t ( t α z α ( ) t a a t a+α 0 t a α z α 0 t a α z = z(t) γ a t b z = z( t) γ b t a = (γ, γ,..., γ n, γ n ) a t b = ( γ n, γ n,..., γ, γ ) b
More information(a) To show f(z) is analytic, explicitly evaluate partials,, etc. and show. = 0. To find v, integrate u = v to get v = dy u =
Homework -5 Solutions Problems (a) z = + 0i, (b) z = 7 + 24i 2 f(z) = u(x, y) + iv(x, y) with u(x, y) = e 2y cos(2x) and v(x, y) = e 2y sin(2x) u (a) To show f(z) is analytic, explicitly evaluate partials,,
More information1. DO NOT LIFT THIS COVER PAGE UNTIL INSTRUCTED TO DO SO. Write your student number and name at the top of this page. This test has SIX pages.
Student Number Name (Printed in INK Mathematics 54 July th, 007 SIMON FRASER UNIVERSITY Department of Mathematics Faculty of Science Midterm Instructor: S. Pimentel 1. DO NOT LIFT THIS COVER PAGE UNTIL
More informationComplex Homework Summer 2014
omplex Homework Summer 24 Based on Brown hurchill 7th Edition June 2, 24 ontents hw, omplex Arithmetic, onjugates, Polar Form 2 2 hw2 nth roots, Domains, Functions 2 3 hw3 Images, Transformations 3 4 hw4
More informationPart IB. Complex Analysis. Year
Part IB Complex Analysis Year 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2018 Paper 1, Section I 2A Complex Analysis or Complex Methods 7 (a) Show that w = log(z) is a conformal
More informationMATH 311: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE
MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE Recall the Residue Theorem: Let be a simple closed loop, traversed counterclockwise. Let f be a function that is analytic on and meromorphic inside. Then
More informationQualifying Exam Complex Analysis (Math 530) January 2019
Qualifying Exam Complex Analysis (Math 53) January 219 1. Let D be a domain. A function f : D C is antiholomorphic if for every z D the limit f(z + h) f(z) lim h h exists. Write f(z) = f(x + iy) = u(x,
More informationPhysics 2400 Midterm I Sample March 2017
Physics 4 Midterm I Sample March 17 Question: 1 3 4 5 Total Points: 1 1 1 1 6 Gamma function. Leibniz s rule. 1. (1 points) Find positive x that minimizes the value of the following integral I(x) = x+1
More informationFunctions of a Complex Variable and Integral Transforms
Functions of a Complex Variable and Integral Transforms Department of Mathematics Zhou Lingjun Textbook Functions of Complex Analysis with Applications to Engineering and Science, 3rd Edition. A. D. Snider
More informationMath 120 A Midterm 2 Solutions
Math 2 A Midterm 2 Solutions Jim Agler. Find all solutions to the equations tan z = and tan z = i. Solution. Let α be a complex number. Since the equation tan z = α becomes tan z = sin z eiz e iz cos z
More informationProblem Set 7 Solution Set
Problem Set 7 Solution Set Anthony Varilly Math 3: Complex Analysis, Fall 22 Let P (z be a polynomial Prove there exists a real positive number ɛ with the following property: for all non-zero complex numbers
More information13 Definite integrals
3 Definite integrals Read: Boas h. 4. 3. Laurent series: Def.: Laurent series (LS). If f() is analytic in a region R, then converges in R, with a n = πi f() = a n ( ) n + n= n= f() ; b ( ) n+ n = πi b
More informationINTEGRATION WORKSHOP 2004 COMPLEX ANALYSIS EXERCISES
INTEGRATION WORKSHOP 2004 COMPLEX ANALYSIS EXERCISES PHILIP FOTH 1. Cauchy s Formula and Cauchy s Theorem 1. Suppose that γ is a piecewise smooth positively ( counterclockwise ) oriented simple closed
More informationMAT389 Fall 2016, Problem Set 11
MAT389 Fall 216, Problem Set 11 Improper integrals 11.1 In each of the following cases, establish the convergence of the given integral and calculate its value. i) x 2 x 2 + 1) 2 ii) x x 2 + 1)x 2 + 2x
More informationCERTAIN INTEGRALS ARISING FROM RAMANUJAN S NOTEBOOKS
CERTAIN INTEGRALS ARISING FROM RAMANUJAN S NOTEBOOKS BRUCE C. BERNDT AND ARMIN STRAUB Abstract. In his third noteboo, Ramanujan claims that log x dx + π 2 x 2 dx. + 1 In a following cryptic line, which
More informationINTEGRATION WORKSHOP 2003 COMPLEX ANALYSIS EXERCISES
INTEGRATION WORKSHOP 23 COMPLEX ANALYSIS EXERCISES DOUGLAS ULMER 1. Meromorphic functions on the Riemann sphere It s often useful to allow functions to take the value. This exercise outlines one way to
More information1 z n = 1. 9.(Problem) Evaluate each of the following, that is, express each in standard Cartesian form x + iy. (2 i) 3. ( 1 + i. 2 i.
. 5(b). (Problem) Show that z n = z n and z n = z n for n =,,... (b) Use polar form, i.e. let z = re iθ, then z n = r n = z n. Note e iθ = cos θ + i sin θ =. 9.(Problem) Evaluate each of the following,
More informationTaylor and Laurent Series
Chapter 4 Taylor and Laurent Series 4.. Taylor Series 4... Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor series of a given function f : R R is given by: f (x + f (x (x x + f (x
More informationEE2 Mathematics : Complex Variables
EE Mathematics : omplex Variables J. D. Gibbon (Professor J. D Gibbon 1, Dept of Mathematics) j.d.gibbon@ic.ac.uk http://www.imperial.ac.uk/ jdg These notes are not identical word-for-word with my lectures
More informationComplex Analysis Math 185A, Winter 2010 Final: Solutions
Complex Analysis Math 85A, Winter 200 Final: Solutions. [25 pts] The Jacobian of two real-valued functions u(x, y), v(x, y) of (x, y) is defined by the determinant (u, v) J = (x, y) = u x u y v x v y.
More informationMath Homework 2
Math 73 Homework Due: September 8, 6 Suppose that f is holomorphic in a region Ω, ie an open connected set Prove that in any of the following cases (a) R(f) is constant; (b) I(f) is constant; (c) f is
More informationMA3111S COMPLEX ANALYSIS I
MA3111S COMPLEX ANALYSIS I 1. The Algebra of Complex Numbers A complex number is an expression of the form a + ib, where a and b are real numbers. a is called the real part of a + ib and b the imaginary
More informationMATH FINAL SOLUTION
MATH 185-4 FINAL SOLUTION 1. (8 points) Determine whether the following statements are true of false, no justification is required. (1) (1 point) Let D be a domain and let u,v : D R be two harmonic functions,
More informationA REVIEW OF RESIDUES AND INTEGRATION A PROCEDURAL APPROACH
A REVIEW OF RESIDUES AND INTEGRATION A PROEDURAL APPROAH ANDREW ARHIBALD 1. Introduction When working with complex functions, it is best to understand exactly how they work. Of course, complex functions
More informationarxiv: v3 [math.ca] 14 Oct 2015
Symmetry, Integrability and Geometry: Methods and Applications Certain Integrals Arising from Ramanujan s Noteboos SIGMA 11 215), 83, 11 pages Bruce C. BERNDT and Armin STRAUB arxiv:159.886v3 [math.ca]
More informationLECTURE-13 : GENERALIZED CAUCHY S THEOREM
LECTURE-3 : GENERALIZED CAUCHY S THEOREM VED V. DATAR The aim of this lecture to prove a general form of Cauchy s theorem applicable to multiply connected domains. We end with computations of some real
More informationExercises for Part 1
MATH200 Complex Analysis. Exercises for Part Exercises for Part The following exercises are provided for you to revise complex numbers. Exercise. Write the following expressions in the form x+iy, x,y R:
More informationPart IB. Further Analysis. Year
Year 2004 2003 2002 2001 10 2004 2/I/4E Let τ be the topology on N consisting of the empty set and all sets X N such that N \ X is finite. Let σ be the usual topology on R, and let ρ be the topology on
More informationExercises for Part 1
MATH200 Complex Analysis. Exercises for Part Exercises for Part The following exercises are provided for you to revise complex numbers. Exercise. Write the following expressions in the form x + iy, x,y
More informationComplex Inversion Formula for Stieltjes and Widder Transforms with Applications
Int. J. Contemp. Math. Sciences, Vol. 3, 8, no. 16, 761-77 Complex Inversion Formula for Stieltjes and Widder Transforms with Applications A. Aghili and A. Ansari Department of Mathematics, Faculty of
More informationCHAPTER 4. Elementary Functions. Dr. Pulak Sahoo
CHAPTER 4 Elementary Functions BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India E-mail : sahoopulak1@gmail.com 1 Module-4: Multivalued Functions-II
More informationSOLUTION SET IV FOR FALL z 2 1
SOLUTION SET IV FOR 8.75 FALL 4.. Residues... Functions of a Complex Variable In the following, I use the notation Res zz f(z) Res(z ) Res[f(z), z ], where Res is the residue of f(z) at (the isolated singularity)
More informationMath 421 Midterm 2 review questions
Math 42 Midterm 2 review questions Paul Hacking November 7, 205 () Let U be an open set and f : U a continuous function. Let be a smooth curve contained in U, with endpoints α and β, oriented from α to
More informationComplex Analysis MATH 6300 Fall 2013 Homework 4
Complex Analysis MATH 6300 Fall 2013 Homework 4 Due Wednesday, December 11 at 5 PM Note that to get full credit on any problem in this class, you must solve the problems in an efficient and elegant manner,
More information= 2πi Res. z=0 z (1 z) z 5. z=0. = 2πi 4 5z
MTH30 Spring 07 HW Assignment 7: From [B4]: hap. 6: Sec. 77, #3, 7; Sec. 79, #, (a); Sec. 8, #, 3, 5, Sec. 83, #5,,. The due date for this assignment is 04/5/7. Sec. 77, #3. In the example in Sec. 76,
More informationPart IB. Complex Methods. Year
Part IB Year 218 217 216 215 214 213 212 211 21 29 28 27 26 25 24 23 22 21 218 Paper 1, Section I 2A Complex Analysis or 7 (a) Show that w = log(z) is a conformal mapping from the right half z-plane, Re(z)
More informationHere are brief notes about topics covered in class on complex numbers, focusing on what is not covered in the textbook.
Phys374, Spring 2008, Prof. Ted Jacobson Department of Physics, University of Maryland Complex numbers version 5/21/08 Here are brief notes about topics covered in class on complex numbers, focusing on
More informationMTH4101 CALCULUS II REVISION NOTES. 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) ax 2 + bx + c = 0. x = b ± b 2 4ac 2a. i = 1.
MTH4101 CALCULUS II REVISION NOTES 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) 1.1 Introduction Types of numbers (natural, integers, rationals, reals) The need to solve quadratic equations:
More informationPROBLEM SET 3 FYS3140
PROBLEM SET FYS40 Problem. (Cauchy s theorem and integral formula) Cauchy s integral formula f(a) = πi z a dz πi f(a) a in z a dz = 0 otherwise we solve the following problems by comparing the integrals
More informationMath 411, Complex Analysis Definitions, Formulas and Theorems Winter y = sinα
Math 411, Complex Analysis Definitions, Formulas and Theorems Winter 014 Trigonometric Functions of Special Angles α, degrees α, radians sin α cos α tan α 0 0 0 1 0 30 π 6 45 π 4 1 3 1 3 1 y = sinα π 90,
More informationEE2007: Engineering Mathematics II Complex Analysis
EE2007: Engineering Mathematics II omplex Analysis Ling KV School of EEE, NTU ekvling@ntu.edu.sg V4.2: Ling KV, August 6, 2006 V4.1: Ling KV, Jul 2005 EE2007 V4.0: Ling KV, Jan 2005, EE2007 V3.1: Ling
More informationMA424, S13 HW #6: Homework Problems 1. Answer the following, showing all work clearly and neatly. ONLY EXACT VALUES WILL BE ACCEPTED.
MA424, S13 HW #6: 44-47 Homework Problems 1 Answer the following, showing all work clearly and neatly. ONLY EXACT VALUES WILL BE ACCEPTED. NOTATION: Recall that C r (z) is the positively oriented circle
More informationlim when the limit on the right exists, the improper integral is said to converge to that limit.
hapter 7 Applications of esidues - evaluation of definite and improper integrals occurring in real analysis and applied math - finding inverse Laplace transform by the methods of summing residues. 6. Evaluation
More information