1) The document provides solutions to homework problems from a complex analysis course. It solves problems involving properties of polynomials, including showing two polynomials are equal if they have the same roots, and investigating properties of roots.
2) It also analyzes singularities of complex functions, determining whether singularities are removable, poles, or essential singularities. Functions include rational, trigonometric, exponential and combined functions.
3) The solutions demonstrate techniques for analyzing complex functions at isolated singular points using principles like series expansions and the Casorati-Weierstrass theorem.
1. The document contains an assignment on mathematics involving ordinary and partial differential equations, matrices, vector calculus, and their applications. It includes solving various types of differential equations, finding eigenvectors and eigenvalues of matrices, evaluating line, surface and volume integrals using Green's theorem and Stokes' theorem. The assignment contains 20 problems spanning these topics.
2. The assignment covers key concepts in differential equations, linear algebra, and vector calculus including solving ordinary differential equations, partial differential equations, systems of linear equations, eigenproblems, line integrals, surface integrals, divergence, curl, gradient, Laplacian, and theorems like Green's theorem and Stokes' theorem.
3. Students are required to solve 20 problems involving these
The document discusses solving the two-dimensional Laplace equation to model steady heat flow problems. It presents:
1) The general boundary value problem (BVP) for the Laplace equation in a semi-infinite and finite lamina.
2) The separation of variables method to obtain solutions as a sum of products of ordinary differential equations.
3) Applying boundary conditions to determine constants and obtain the general solution for temperature distribution.
4) Examples of applying the method to specific BVPs for steady heat flow, including plates with various boundary temperature profiles and geometries.
1. The document provides 14 problems involving partial differential equations (PDEs). The problems involve forming PDEs by eliminating arbitrary constants from functions, finding complete integrals, and solving PDEs.
2. Methods used include taking partial derivatives, finding auxiliary equations, and making substitutions to isolate the PDE or solve it.
3. The document covers a range of techniques for working with PDEs, including eliminating constants, finding trial solutions, integrating subsidiary equations, and solving auxiliary equations to find complete integrals.
This document presents three theorems that further generalize previous results on the distribution of zeros of polynomials. Theorem 1 states that if the coefficients of a polynomial satisfy certain monotonicity conditions involving parameters λ, ρ, and μ, then the zeros of the polynomial lie within defined disks. Theorem 2 makes a similar statement about the zeros, but applies the conditions to the imaginary coefficients rather than the real coefficients. Theorem 3 gives disk regions containing the zeros when the coefficients satisfy different monotonicity conditions and the arguments of the coefficients are bounded. The theorems reduce to previous results as special cases when certain parameters are set to 1.
This document provides solutions to homework problems from a complex analysis course. It analyzes the differentiability of several complex functions by checking if they satisfy the Cauchy-Riemann equations. For differentiable functions, it computes their derivatives according to the Cauchy-Riemann theorem. The solutions involve defining real and imaginary parts of functions, taking partial derivatives, and applying limits.
The document discusses Fourier series and periodic functions. It provides:
- Definitions of Fourier series and periodic functions.
- Examples of periodic functions including trigonometric and other functions.
- Euler's formulae for calculating the coefficients of a Fourier series.
- Integration properties used to solve Fourier series problems.
- Two examples of determining the Fourier series for given periodic functions and using it to deduce mathematical results.
In this paper, restricting the coefficients of a polynomial to certain conditions, we locate a region containing all of its zeros. Our results generalize many known results in addition to some interesting results which can be obtained by choosing certain values of the parameters. Mathematics Subject Classification: 30C10, 30C15
This document provides solutions to problems from a complex analysis homework assignment. The solutions involve applying theorems related to contour integrals, Cauchy's integral formula, and derivatives to evaluate definite integrals over curves and find derivatives of functions. Key steps include using the Cauchy-Goursat theorem when functions are analytic over regions, and applying Cauchy's integral formula and its generalization to derivatives to evaluate integrals that involve singularities inside curves.
1. The document contains an assignment on mathematics involving ordinary and partial differential equations, matrices, vector calculus, and their applications. It includes solving various types of differential equations, finding eigenvectors and eigenvalues of matrices, evaluating line, surface and volume integrals using Green's theorem and Stokes' theorem. The assignment contains 20 problems spanning these topics.
2. The assignment covers key concepts in differential equations, linear algebra, and vector calculus including solving ordinary differential equations, partial differential equations, systems of linear equations, eigenproblems, line integrals, surface integrals, divergence, curl, gradient, Laplacian, and theorems like Green's theorem and Stokes' theorem.
3. Students are required to solve 20 problems involving these
The document discusses solving the two-dimensional Laplace equation to model steady heat flow problems. It presents:
1) The general boundary value problem (BVP) for the Laplace equation in a semi-infinite and finite lamina.
2) The separation of variables method to obtain solutions as a sum of products of ordinary differential equations.
3) Applying boundary conditions to determine constants and obtain the general solution for temperature distribution.
4) Examples of applying the method to specific BVPs for steady heat flow, including plates with various boundary temperature profiles and geometries.
1. The document provides 14 problems involving partial differential equations (PDEs). The problems involve forming PDEs by eliminating arbitrary constants from functions, finding complete integrals, and solving PDEs.
2. Methods used include taking partial derivatives, finding auxiliary equations, and making substitutions to isolate the PDE or solve it.
3. The document covers a range of techniques for working with PDEs, including eliminating constants, finding trial solutions, integrating subsidiary equations, and solving auxiliary equations to find complete integrals.
This document presents three theorems that further generalize previous results on the distribution of zeros of polynomials. Theorem 1 states that if the coefficients of a polynomial satisfy certain monotonicity conditions involving parameters λ, ρ, and μ, then the zeros of the polynomial lie within defined disks. Theorem 2 makes a similar statement about the zeros, but applies the conditions to the imaginary coefficients rather than the real coefficients. Theorem 3 gives disk regions containing the zeros when the coefficients satisfy different monotonicity conditions and the arguments of the coefficients are bounded. The theorems reduce to previous results as special cases when certain parameters are set to 1.
This document provides solutions to homework problems from a complex analysis course. It analyzes the differentiability of several complex functions by checking if they satisfy the Cauchy-Riemann equations. For differentiable functions, it computes their derivatives according to the Cauchy-Riemann theorem. The solutions involve defining real and imaginary parts of functions, taking partial derivatives, and applying limits.
The document discusses Fourier series and periodic functions. It provides:
- Definitions of Fourier series and periodic functions.
- Examples of periodic functions including trigonometric and other functions.
- Euler's formulae for calculating the coefficients of a Fourier series.
- Integration properties used to solve Fourier series problems.
- Two examples of determining the Fourier series for given periodic functions and using it to deduce mathematical results.
In this paper, restricting the coefficients of a polynomial to certain conditions, we locate a region containing all of its zeros. Our results generalize many known results in addition to some interesting results which can be obtained by choosing certain values of the parameters. Mathematics Subject Classification: 30C10, 30C15
This document provides solutions to problems from a complex analysis homework assignment. The solutions involve applying theorems related to contour integrals, Cauchy's integral formula, and derivatives to evaluate definite integrals over curves and find derivatives of functions. Key steps include using the Cauchy-Goursat theorem when functions are analytic over regions, and applying Cauchy's integral formula and its generalization to derivatives to evaluate integrals that involve singularities inside curves.
(i) The infinite series ∑(2x)n/n from n=1 to infinity is convergent for -1/2 ≤ x ≤ 1/2 and has an infinite sum of 2x.
(ii) The infinite series ∑((x-1)n)/2n from n=1 to infinity is convergent for -1 < x < 3 and has an infinite sum of x-1.
(iii) The infinite series ∑(n!xn-1) from n=1 to infinity has an interval of convergence of the entire real number line R.
The document discusses Fourier series and their applications. It begins by introducing how Fourier originally developed the technique to study heat transfer and how it can represent periodic functions as an infinite series of sine and cosine terms. It then provides the definition and examples of Fourier series representations. The key points are that Fourier series decompose a function into sinusoidal basis functions with coefficients determined by integrating the function against each basis function. The series may converge to the original function under certain conditions.
The document discusses partial differential equations and their solutions. It can be summarized as:
1) A partial differential equation involves a function of two or more variables and some of its partial derivatives, with one dependent variable and one or more independent variables. Standard notation is presented for partial derivatives.
2) Partial differential equations can be formed by eliminating arbitrary constants or arbitrary functions from an equation relating the dependent and independent variables. Examples of each method are provided.
3) Solutions to partial differential equations can be complete, containing the maximum number of arbitrary constants allowed, particular where the constants are given specific values, or singular where no constants are present. Methods for determining the general solution are described.
1. The chain rule describes how to take the derivative of a function composed of other functions. It states that if z = f(x, y) and x and y are functions of t, then the derivative of z with respect to t is the sum of the partial derivatives multiplied by the derivatives of x and y with respect to t.
2. The chain rule is applied to find the derivative of two example functions, w = xy and w = xy + z, with respect to t along given paths for x, y, and z in terms of t.
3. The chain rule is generalized to the case of a function u of n variables, where each variable is a function of m other variables
Location of Regions Containing all or Some Zeros of a Polynomialijceronline
In this paper we locate the regions which contain all or some of the zeros of a polynomial when the coefficients of the polynomial are restricted to certain conditions. Mathematics Subject Classification: 30C10, 30C15.
This document discusses partial differential equations (PDEs). It provides examples of how PDEs can be formed by eliminating constants or functions from relations involving multiple variables. It also discusses different types of first-order PDEs and methods for solving them. Several example problems are presented with step-by-step solutions showing how to derive and solve PDEs that model different physical situations. Standard forms and techniques for reducing PDEs to simpler forms are also outlined.
International Journal of Engineering Research and Development (IJERD)IJERD Editor
journal publishing, how to publish research paper, Call For research paper, international journal, publishing a paper, IJERD, journal of science and technology, how to get a research paper published, publishing a paper, publishing of journal, publishing of research paper, reserach and review articles, IJERD Journal, How to publish your research paper, publish research paper, open access engineering journal, Engineering journal, Mathemetics journal, Physics journal, Chemistry journal, Computer Engineering, Computer Science journal, how to submit your paper, peer reviw journal, indexed journal, reserach and review articles, engineering journal, www.ijerd.com, research journals,
yahoo journals, bing journals, International Journal of Engineering Research and Development, google journals, hard copy of journal
The document discusses the chain rule and how to use it to differentiate and integrate composite functions. The chain rule states that if h(x) = g(f(x)), then h'(x) = g'(f(x))f'(x). It provides examples of applying the chain rule to differentiate functions like sin(x2 - 4) and integrate functions like ∫(3x2 + 4)3 dx. It also discusses how to integrate functions of the form f'(x)g(f(x)) by recognizing them as derivatives of composite functions.
Engineering Mathematics - Total derivatives, chain rule and derivative of imp...Jayanshu Gundaniya
This document discusses the total derivative and methods for finding derivatives of functions with multiple variables.
The total derivative expresses the total differential of a function u with respect to time t as the sum of the partial derivatives of u with respect to each variable x1, x2,...xn, multiplied by the rate of change of that variable with respect to time.
The chain rule is used to take derivatives of composite functions, where the output of one function is an input to another. The derivative is expressed as the product of the partial derivatives of each nested function.
Derivatives can also be taken for implicit functions, where not all variables can be solved for explicitly. The derivative of one variable with respect to another in an
The document discusses Legendre polynomials, which are special functions that arise in solutions to Laplace's equation in spherical coordinates. Some key points:
1) Legendre polynomials Pn(cosθ) are a set of orthogonal polynomials that satisfy Legendre's differential equation.
2) Pn(cosθ) can be defined using a generating function or by taking partial derivatives of 1/r.
3) Important properties of Legendre polynomials include P0(t)=1, Pn(1)=1, Pn(-1)=(-1)n, and a recurrence relation involving Pn+1, Pn, and their derivatives.
The document discusses Legendre functions, which are solutions to Legendre's differential equation. Legendre functions arise when solving Laplace's equation in spherical coordinates. Legendre polynomials were first introduced by Adrien-Marie Legendre in 1785 as coefficients in an expansion of Newtonian potential. The document covers topics such as Legendre polynomials, Rodrigues' formula, orthogonality of Legendre polynomials, and associated Legendre functions.
Successive Differentiation is the process of differentiating a given function successively times and the results of such differentiation are called successive derivatives. The higher order differential coefficients are of utmost importance in scientific and engineering applications.
1) The document discusses solving partial differential equations using variable separation. It presents the one-dimensional wave equation and solves it using variable separation.
2) It derives three cases for the solution based on whether k is positive, negative, or zero. It then presents the general solution as a summation involving sines and cosines.
3) It applies the general solution to two example problems of a vibrating string, finding the displacement as a function of position and time by satisfying the boundary conditions.
This document provides lecture notes on complex analysis covering four units of content:
1) The index of a close curve, Cauchy's theorem, and entire functions.
2) Counting zeroes, meromorphic functions, and maximum principle.
3) Spaces of continuous and analytic functions, and behavior of functions.
4) Comparison of entire functions, analytic continuation, and harmonic functions.
It also provides definitions and theorems regarding integrals over rectifiable curves, winding numbers, and Cauchy's theorem. Exercises and proofs are included.
This document discusses Hermite polynomials and their properties. It begins by introducing the Hermite equation, which arises from the theory of the linear harmonic oscillator. The Hermite equation is then solved using a series solution approach. Explicit formulas for the Hermite polynomials Hn(x) are derived for n=0,1,2,3,... using properties of the Hermite equation like recursion relations. Key properties of the Hermite polynomials like generating functions, even/odd behavior, Rodrigue's formula, orthogonality, and recurrence relations are also proved.
The document defines proper and improper integrals, and discusses different types of improper integrals based on whether the limits are infinite or the function is unbounded. It provides tests to determine if improper integrals converge or diverge, including the T1 test involving exponential functions, the T2 test involving power functions, and comparison tests. Examples are worked through applying these tests to determine if various improper integrals converge or diverge. The key information is on defining improper integrals and tests to analyze their convergence.
This unit covers the formation and solutions of partial differential equations (PDEs). PDEs can be obtained by eliminating arbitrary constants or functions from relating equations. Standard methods are used to solve first order PDEs and higher order linear PDEs with constant coefficients. Various physical processes are modeled using PDEs including the wave equation, heat equation, and Laplace's equation.
This document provides summaries of common special functions and polynomials, including Legendre polynomials, associated Legendre functions, Bessel functions, spherical Bessel functions, Hermite polynomials, and Laguerre polynomials. It defines the differential equations that govern each function, provides generating functions, discusses orthogonality properties, and lists some important recurrence relations and examples. The document is intended as a short survey of essential properties of these important mathematical functions.
The document provides an introduction to partial differential equations (PDEs). Some key points:
- PDEs involve functions of two or more independent variables, and arise in physics/engineering problems.
- PDEs contain partial derivatives with respect to two or more independent variables. Examples of common PDEs are given, including the Laplace, wave, and heat equations.
- The order of a PDE is defined as the order of the highest derivative. Methods for solving PDEs through direct integration and using Lagrange's method are briefly outlined.
This document contains solutions to problems from a complex analysis homework assignment. Problem 21.2 involves determining the convergence of several series using tests from Chapter 21. Problem 21.4 examines the sum of a geometric series. Subsequent problems analyze the pointwise and uniform convergence of series and determine radii of convergence using tests such as the Ratio Test and Cauchy-Hadamard formula.
1) The document discusses analytic functions on simply connected domains not containing the origin. It states that the logarithm and exponential functions of z are analytic on such domains.
2) It considers a closed curve γ in a domain Ω that winds around 1 and -1. As you travel around γ and return to the starting point z0, the arguments of 1-z, 1+z, and the square root of 1-z2 change by multiples of 2π.
3) By applying Cauchy's theorem and a change of variables, the integral of 1/√(1-z2) along γ is computed to be 2πi.
(i) The infinite series ∑(2x)n/n from n=1 to infinity is convergent for -1/2 ≤ x ≤ 1/2 and has an infinite sum of 2x.
(ii) The infinite series ∑((x-1)n)/2n from n=1 to infinity is convergent for -1 < x < 3 and has an infinite sum of x-1.
(iii) The infinite series ∑(n!xn-1) from n=1 to infinity has an interval of convergence of the entire real number line R.
The document discusses Fourier series and their applications. It begins by introducing how Fourier originally developed the technique to study heat transfer and how it can represent periodic functions as an infinite series of sine and cosine terms. It then provides the definition and examples of Fourier series representations. The key points are that Fourier series decompose a function into sinusoidal basis functions with coefficients determined by integrating the function against each basis function. The series may converge to the original function under certain conditions.
The document discusses partial differential equations and their solutions. It can be summarized as:
1) A partial differential equation involves a function of two or more variables and some of its partial derivatives, with one dependent variable and one or more independent variables. Standard notation is presented for partial derivatives.
2) Partial differential equations can be formed by eliminating arbitrary constants or arbitrary functions from an equation relating the dependent and independent variables. Examples of each method are provided.
3) Solutions to partial differential equations can be complete, containing the maximum number of arbitrary constants allowed, particular where the constants are given specific values, or singular where no constants are present. Methods for determining the general solution are described.
1. The chain rule describes how to take the derivative of a function composed of other functions. It states that if z = f(x, y) and x and y are functions of t, then the derivative of z with respect to t is the sum of the partial derivatives multiplied by the derivatives of x and y with respect to t.
2. The chain rule is applied to find the derivative of two example functions, w = xy and w = xy + z, with respect to t along given paths for x, y, and z in terms of t.
3. The chain rule is generalized to the case of a function u of n variables, where each variable is a function of m other variables
Location of Regions Containing all or Some Zeros of a Polynomialijceronline
In this paper we locate the regions which contain all or some of the zeros of a polynomial when the coefficients of the polynomial are restricted to certain conditions. Mathematics Subject Classification: 30C10, 30C15.
This document discusses partial differential equations (PDEs). It provides examples of how PDEs can be formed by eliminating constants or functions from relations involving multiple variables. It also discusses different types of first-order PDEs and methods for solving them. Several example problems are presented with step-by-step solutions showing how to derive and solve PDEs that model different physical situations. Standard forms and techniques for reducing PDEs to simpler forms are also outlined.
International Journal of Engineering Research and Development (IJERD)IJERD Editor
journal publishing, how to publish research paper, Call For research paper, international journal, publishing a paper, IJERD, journal of science and technology, how to get a research paper published, publishing a paper, publishing of journal, publishing of research paper, reserach and review articles, IJERD Journal, How to publish your research paper, publish research paper, open access engineering journal, Engineering journal, Mathemetics journal, Physics journal, Chemistry journal, Computer Engineering, Computer Science journal, how to submit your paper, peer reviw journal, indexed journal, reserach and review articles, engineering journal, www.ijerd.com, research journals,
yahoo journals, bing journals, International Journal of Engineering Research and Development, google journals, hard copy of journal
The document discusses the chain rule and how to use it to differentiate and integrate composite functions. The chain rule states that if h(x) = g(f(x)), then h'(x) = g'(f(x))f'(x). It provides examples of applying the chain rule to differentiate functions like sin(x2 - 4) and integrate functions like ∫(3x2 + 4)3 dx. It also discusses how to integrate functions of the form f'(x)g(f(x)) by recognizing them as derivatives of composite functions.
Engineering Mathematics - Total derivatives, chain rule and derivative of imp...Jayanshu Gundaniya
This document discusses the total derivative and methods for finding derivatives of functions with multiple variables.
The total derivative expresses the total differential of a function u with respect to time t as the sum of the partial derivatives of u with respect to each variable x1, x2,...xn, multiplied by the rate of change of that variable with respect to time.
The chain rule is used to take derivatives of composite functions, where the output of one function is an input to another. The derivative is expressed as the product of the partial derivatives of each nested function.
Derivatives can also be taken for implicit functions, where not all variables can be solved for explicitly. The derivative of one variable with respect to another in an
The document discusses Legendre polynomials, which are special functions that arise in solutions to Laplace's equation in spherical coordinates. Some key points:
1) Legendre polynomials Pn(cosθ) are a set of orthogonal polynomials that satisfy Legendre's differential equation.
2) Pn(cosθ) can be defined using a generating function or by taking partial derivatives of 1/r.
3) Important properties of Legendre polynomials include P0(t)=1, Pn(1)=1, Pn(-1)=(-1)n, and a recurrence relation involving Pn+1, Pn, and their derivatives.
The document discusses Legendre functions, which are solutions to Legendre's differential equation. Legendre functions arise when solving Laplace's equation in spherical coordinates. Legendre polynomials were first introduced by Adrien-Marie Legendre in 1785 as coefficients in an expansion of Newtonian potential. The document covers topics such as Legendre polynomials, Rodrigues' formula, orthogonality of Legendre polynomials, and associated Legendre functions.
Successive Differentiation is the process of differentiating a given function successively times and the results of such differentiation are called successive derivatives. The higher order differential coefficients are of utmost importance in scientific and engineering applications.
1) The document discusses solving partial differential equations using variable separation. It presents the one-dimensional wave equation and solves it using variable separation.
2) It derives three cases for the solution based on whether k is positive, negative, or zero. It then presents the general solution as a summation involving sines and cosines.
3) It applies the general solution to two example problems of a vibrating string, finding the displacement as a function of position and time by satisfying the boundary conditions.
This document provides lecture notes on complex analysis covering four units of content:
1) The index of a close curve, Cauchy's theorem, and entire functions.
2) Counting zeroes, meromorphic functions, and maximum principle.
3) Spaces of continuous and analytic functions, and behavior of functions.
4) Comparison of entire functions, analytic continuation, and harmonic functions.
It also provides definitions and theorems regarding integrals over rectifiable curves, winding numbers, and Cauchy's theorem. Exercises and proofs are included.
This document discusses Hermite polynomials and their properties. It begins by introducing the Hermite equation, which arises from the theory of the linear harmonic oscillator. The Hermite equation is then solved using a series solution approach. Explicit formulas for the Hermite polynomials Hn(x) are derived for n=0,1,2,3,... using properties of the Hermite equation like recursion relations. Key properties of the Hermite polynomials like generating functions, even/odd behavior, Rodrigue's formula, orthogonality, and recurrence relations are also proved.
The document defines proper and improper integrals, and discusses different types of improper integrals based on whether the limits are infinite or the function is unbounded. It provides tests to determine if improper integrals converge or diverge, including the T1 test involving exponential functions, the T2 test involving power functions, and comparison tests. Examples are worked through applying these tests to determine if various improper integrals converge or diverge. The key information is on defining improper integrals and tests to analyze their convergence.
This unit covers the formation and solutions of partial differential equations (PDEs). PDEs can be obtained by eliminating arbitrary constants or functions from relating equations. Standard methods are used to solve first order PDEs and higher order linear PDEs with constant coefficients. Various physical processes are modeled using PDEs including the wave equation, heat equation, and Laplace's equation.
This document provides summaries of common special functions and polynomials, including Legendre polynomials, associated Legendre functions, Bessel functions, spherical Bessel functions, Hermite polynomials, and Laguerre polynomials. It defines the differential equations that govern each function, provides generating functions, discusses orthogonality properties, and lists some important recurrence relations and examples. The document is intended as a short survey of essential properties of these important mathematical functions.
The document provides an introduction to partial differential equations (PDEs). Some key points:
- PDEs involve functions of two or more independent variables, and arise in physics/engineering problems.
- PDEs contain partial derivatives with respect to two or more independent variables. Examples of common PDEs are given, including the Laplace, wave, and heat equations.
- The order of a PDE is defined as the order of the highest derivative. Methods for solving PDEs through direct integration and using Lagrange's method are briefly outlined.
This document contains solutions to problems from a complex analysis homework assignment. Problem 21.2 involves determining the convergence of several series using tests from Chapter 21. Problem 21.4 examines the sum of a geometric series. Subsequent problems analyze the pointwise and uniform convergence of series and determine radii of convergence using tests such as the Ratio Test and Cauchy-Hadamard formula.
1) The document discusses analytic functions on simply connected domains not containing the origin. It states that the logarithm and exponential functions of z are analytic on such domains.
2) It considers a closed curve γ in a domain Ω that winds around 1 and -1. As you travel around γ and return to the starting point z0, the arguments of 1-z, 1+z, and the square root of 1-z2 change by multiples of 2π.
3) By applying Cauchy's theorem and a change of variables, the integral of 1/√(1-z2) along γ is computed to be 2πi.
Let f(z) be a function continuous at a point z0.
To show that f(z) is also continuous at z0, we need to show:
limz→z0 f(z) = f(z0)
Since f(z) is given to be continuous at z0, by the definition of continuity:
limz→z0 f(z) = f(z0)
Therefore, if f(z) is continuous at a point z0, it automatically satisfies the condition for continuity at z0. Hence, f(z) is also continuous at z0.
So in summary, if a function f(z) is continuous at a
The document provides solutions to problems from an IIT-JEE 2004 mathematics exam. Problem 1 asks the student to find the center and radius of a circle defined by a complex number relation. The solution shows that the center is the midpoint of points dividing the join of the constants in the ratio k:1, and gives the radius. Problem 2 asks the student to prove an inequality relating dot products of four vectors satisfying certain conditions. The solution shows that the vectors must be parallel or antiparallel.
HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2
1. The document provides solutions to homework problems from a complex analysis class.
2. It shows the work to find harmonic conjugates and derivatives of complex functions, evaluate complex expressions, and take logarithms and exponents of complex numbers.
3. Key steps include using the Cauchy-Riemann equations to test if functions are analytic, decomposing complex expressions into polar form, and applying properties of logarithms and exponents to manipulate expressions.
This document contains solutions to homework problems from a complex analysis course. It solves problems involving logarithms, exponentials, trigonometric and hyperbolic functions of complex numbers. Key steps and solutions are shown for problems involving contour integrals of complex functions along curves in the complex plane. The length of one such curve is computed to be 8a. Several contour integrals are evaluated, with one found to be equal to 1 - i. Bounds on contour integrals are also determined using theorems.
The document provides solutions to questions from an IIT-JEE mathematics exam. It includes 8 questions worth 2 marks each, 8 questions worth 4 marks each, and 2 questions worth 6 marks each. The solutions solve problems related to probability, trigonometry, geometry, calculus, and loci. The summary focuses on the high-level structure and content of the document.
Advanced Engineering Mathematics Solutions Manual.pdfWhitney Anderson
This document contains 27 multi-part exercises involving differential equations. The exercises cover topics such as determining whether differential equations are linear or nonlinear, solving differential equations, and classifying differential equations by order.
Antwoorden fourier and laplace transforms, manual solutionsHildemar Alvarez
The document provides solutions to selected exercises from chapter 1 of a textbook. It first shows the steps to express the sum of two time-harmonic signals as another time-harmonic signal. It then defines several integrals involving trigonometric and exponential functions, and calculates their values. Finally, it summarizes properties of linear time-invariant systems and provides example responses to different input signals.
This document provides solutions to exercises from a complex analysis homework assignment. It includes:
1) Computing powers and roots of complex numbers using polar forms and identities
2) Sketching and describing properties of sets involving complex quantities like openness, boundedness, and being a domain
3) Finding and sketching images of complex mappings
This document provides solutions to homework problems from a complex analysis course. It solves problems involving properties of complex numbers, Cauchy's inequality, roots of unity, and proving that if the roots of a polynomial equation satisfy a certain condition, then the roots must all be real. The solutions demonstrate algebraic manipulations and reasoning about complex numbers and functions.
Cauchy's integral theorem, Cauchy's integral formula, Cauchy's integral formula for derivatives, Taylor's Series, Maclaurin’s Series,Laurent's Series,Singularities and zeros, Cauchy's Residue theorem,Evaluation various types of complex integrals.
Successful people replace words like "wish", "try", and "should" with "I will" in their speech and thinking. Ineffective people do not make this replacement and continue using weaker language. The document provides mathematical formulas and properties involving complex numbers, including formulas for roots of unity, sums of trigonometric series, representations of lines and circles using complex numbers, and other identities.
This document contains the solutions to 5 questions related to calculus concepts like integration, derivatives, series approximation, and geometry of curves and surfaces. Some of the key steps include:
- Using integration to find volumes, masses, and centroids
- Finding critical points and classifying extrema
- Approximating a series to evaluate an integral
- Solving a geometric series problem to find an initial height
- Analyzing motion problems using kinematic equations
- Finding equations of planes and tangent lines to surfaces
1. The document is a math assignment containing 5 Riccati differential equations to solve.
2. The solutions provided include expressing the equations in terms of p, then integrating and solving for the constants.
3. The key steps are rewriting the equations in terms of p, then taking the integral of terms involving p and solving for the constants.
The simplified electron and muon model, Oscillating Spacetime: The Foundation...RitikBhardwaj56
Discover the Simplified Electron and Muon Model: A New Wave-Based Approach to Understanding Particles delves into a groundbreaking theory that presents electrons and muons as rotating soliton waves within oscillating spacetime. Geared towards students, researchers, and science buffs, this book breaks down complex ideas into simple explanations. It covers topics such as electron waves, temporal dynamics, and the implications of this model on particle physics. With clear illustrations and easy-to-follow explanations, readers will gain a new outlook on the universe's fundamental nature.
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
How to Make a Field Mandatory in Odoo 17Celine George
In Odoo, making a field required can be done through both Python code and XML views. When you set the required attribute to True in Python code, it makes the field required across all views where it's used. Conversely, when you set the required attribute in XML views, it makes the field required only in the context of that particular view.
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
How to Build a Module in Odoo 17 Using the Scaffold MethodCeline George
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
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How to Add Chatter in the odoo 17 ERP ModuleCeline George
In Odoo, the chatter is like a chat tool that helps you work together on records. You can leave notes and track things, making it easier to talk with your team and partners. Inside chatter, all communication history, activity, and changes will be displayed.
1. NCU Math, Spring 2014: Complex Analysis Homework Solution 6
Text Book: An Introduction to Complex Analysis
Problem. 19.1
Sol:
Without loss of generality, we may assume that Pn(z) =
n
k=0 akzk
where ak ∈ R for all k = 0, · · · , n. Since z0
is a root of Pn(z), we have Pn(z0) = 0. By virtue of ak ∈ R for all k = 0, · · · , n and Pn(z0) = 0, we can get that
Pn(z0) =
n
k=0
ak(z0)k
=
n
k=0
ak(z0)k
=
n
k=0
ak · (z0)k
=
n
k=0
ak(z0)k
=
n
k=0
ak(z0)k
= Pn(z0)
= 0.
Thus z0 is also a root of Pn(z).
Problem. 19.2
Sol:
We dene f(z) = P(z) − Q(z). Then f(z) is a polynomial of degree at most n. Since P(z) and Q(z) agree on
n + 1 distinct points, we can get that P(zk) = Q(zk) for k = 1, · · · , n + 1 and zk = zl if k = l. So z1, · · · , zn+1
are distinct roots of f(z). But if P(z) ≡ Q(z), there are at most n distinct roots of f(z) by Corollary 19.1. This
implies that P(z) ≡ Q(z). Thus P(z) = Q(z) for all z ∈ C.
Problem. 19.3
Sol:
It is easy to see that z = 1 is not a root of Pn(z).
1
2. Now, we want to show that for all z ∈ C − {±1} with |z| ≤ 1 is not a root of Pn(z). To see this, we assume
that z0 ∈ C − {±1} with |z0| ≤ 1 is a root of Pn(z). If we can nd a contradiction, we have done what we want to
prove. Since z0 is a root of Pn(z), we have |(z0 − 1)Pn(z0)| = 0. But
(z0 − 1)Pn(z0)
=(z0 − 1)(anzn
0 + · · · + a1z0 + a0)
=anzn+1
0 + (an−1 − an)zn
0 + · · · + (a0 − a1)z0 − a0.
We can get that
0 = |(z0 − 1)Pn(z0)|
≥ |a0| − |anzn+1
0 + (an−1 − an)zn
0 + · · · + (a0 − a1)z0| .
This implies that a0 = |a0| = |anzn+1
0 + (an−1 − an)zn
0 + · · · + (a0 − a1)z0|. Because
a0 = anzn+1
0 + (an−1 − an)zn
0 + · · · + (a0 − a1)z0
≤an|z0|n+1
+ (an−1 − an)|z0|n
+ · · · + (a0 − a1)|z0|
≤an + (an−1 − an) + · · · + (a0 − a1)
=a0,
we have that all inequalities must be equalities. Thus z0 ∈ R from the rst equality and |z0| = 1 from the second
equality. We get that z0 = ±1. This leads a contradiction. Therefore we have showed what we want.
Problem. 30.1
Sol:
(a)
Since
f(z) =
ez2
− 1
z
=
∞
k=0
(z2
)k
k! − 1
z
=
∞
k=1
z2k
k!
z
=
∞
k=1
z2k−1
k!
= z
∞
k=0
z2k
(k + 1)!
→ 0 as z → 0,
we can get that z = 0 is a removable singularity of f(z) by Theorem 29.1. Here, we dene f(0) = 0 such that f is
analytic at z = 0.
2
3. (b)
Since
f(z) =
sin z − z
z2
=
∞
k=0
(−1)k
z2k+1
(2k+1)! − z
z2
=
∞
k=1
(−1)k
z2k+1
(2k+1)!
z2
=
∞
k=0
(−1)k+1
z2k+3
(2k+3)!
z2
=
∞
k=0
(−1)k+1
z2k+1
(2k + 3)!
→ 0 as z → 0,
we can get that z = 0 is a removable singularity of f(z) by Theorem 29.1. Here, we dene f(0) = 0 such that f is
analytic at z = 0.
(c)
Since
f(z) =
1 − 1
2 z2
− cos z
sin z2
=
1 − 1
2 z2
−
∞
k=0
(−1)k
z2k
(2k)!
∞
k=0
(−1)k(z2)2k+1
(2k+1)!
=
−
∞
k=2
(−1)k
z2k
(2k)!
∞
k=0
(−1)kz4k+2
(2k+1)!
=
−
∞
k=0
(−1)k+2
z2k+4
(2k+4)!
z2 ∞
k=0
(−1)kz4k
(2k+1)!
=
−
∞
k=0
(−1)k+2
z2k+2
(2k+4)!
∞
k=0
(−1)kz4k
(2k+1)!
→ 0 as z → 0,
we can get that z = 0 is a removable singularity of f(z) by Theorem 29.1. Here, we dene f(0) = 0 such that f is
analytic at z = 0.
Problem. 30.2
Sol:
3
4. (a)
For convenience, we dene
f(z) =
cos z
z2−π2
/4
when z = ±π
2 ,
− 1
π when z = ±π
2 .
It is easy to see that f is analytic if z = ±π
2 . Since
cos z
z2 − π2
/4
=
∞
k=0
dk(cos z)
dzk
z= π
2
k! (z − π
2 )k
z2 − π2
/4
=
∞
k=1
dk(cos z)
dzk
z= π
2
k! (z − π
2 )k
z2 − π2
/4
=
∞
k=1
dk(cos z)
dzk
z= π
2
k! (z − π
2 )k−1
z + π/2
→
−1
π
as z →
π
2
,
we can get that z = π
2 is a removable singularity of f(z) by Theorem 29.1 and f is analytic at z = π
2 if f(π
2 ) = −1
π .
So f is analytic at z = π
2 . Similarly,
cos z
z2 − π2
/4
=
∞
k=0
dk(cos z)
dzk
z=− π
2
k! (z + π
2 )k
z2 − π2
/4
=
∞
k=1
dk(cos z)
dzk
z=− π
2
k! (z + π
2 )k
z2 − π2
/4
=
∞
k=1
dk(cos z)
dzk
z=− π
2
k! (z + π
2 )k−1
z − π/2
→
−1
π
as z →
π
2
,
we have that z = −π
2 is a removable singularity of f(z) by Theorem 29.1 and f is analytic at z = −π
2 if f(−π
2 ) = −1
π .
Thus f is analytic at z = −π
2 .
Therefore, f is entire.
4
5. (b)
For convenience, we dene
g(z) =
ez
−e−z
2z when z = 0,
1 when z = 0.
It is easy to see that g is analytic if z = 0. Since
ez
− e−z
2z
=
∞
k=0
zk
k! −
∞
k=0
(−z)k
k!
2z
=
∞
k=0
zk
−(−1)k
zk
k!
2z
=
2
∞
k=0
z2k+1
(2k+1)!
2z
=
∞
k=0
z2k
(2k + 1)!
→1 as z → 0,
we can get that z = 0 is a removable singularity of g(z) by Theorem 29.1 and g is analytic at z = 0 if g(0) = 1. So
g is analytic at z = 0.
Therefore, g is entire.
Problem. 30.3
Sol:
(a)
It is easy to see that the isolated singularities are 0 and 1.
Since z3
+1
z2(z−1) =
z3+1/z−1
z2 , z3
+1
z−1 is analytic at z = 1, and limz→0
z3
+1
z−1 = −1 = 0, we have z3
+1
z2(z−1) has a pole of
order 2 at z = 0 by Theorem 29.2.
Because z3
+1
z2(z−1) =
z3+1/z2
z−1 , z3
+1
z2 is analytic at z = 0, and limz→1
z3
+1
z2 = 2 = 0, we have z3
+1
z2(z−1) has a pole of
order 1 at z = 1 by Theorem 29.2.
(b)
It is easy to see that the only isolated singularity is 0.
5
6. Since for z = 0,
z2
e
1/z
= z2
∞
k=0
1
k!
1
zk
= z2
k=0
−∞
1
(−k)!
zk
=
k=0
−∞
1
(−k)!
zk+2
=
k=2
−∞
1
(−k + 2)!
zk
,
we can get that z = 0 is an essential singularity of z2
e1/z
by the denition.
(c)
It is easy to see that the isolated singularities are (2k+1)
2 π for all k ∈ Z.
Since cos z = 0 if z = (2k+1)
2 π for all k ∈ Z and (cos z) = − sin z = (−1)k+1
= 0 if z = (2k+1)
2 π for all k ∈ Z,
we can get that cos z has a zero of order 1 at z = (2k+1)
2 π for all k ∈ Z. We also have z = 0 at z = (2k+1)
2 π for all
k ∈ Z. This implies that z
cos z has a pole of order 1 at z = (2k+1)
2 π for all k ∈ Z by Corollary 29.3.
(d)
It is easy to see that the only isolated singularity is 0.
Since z2
has a zero of order 2 at z = 0 and sin 3z =
∞
k=0
(−1)k
(3z)2k+1
(2k+1)! has a zero of order 1 at z = 0 by Theorem
26.1, we have sin 3z
z2 has a pole of order 1 at z = 0 by Corollary 29.3.
(e)
It is easy to see that the isolated singularities are 0 and ±i.
Since cos 1
z =
∞
k=0
(−1)k
z−2k
(2k)! =
k=0
−∞
(−1)−k
z2k
(−2k)! , we can get that cos 1
z has an essential singularity at z = 0. So
by Theorem 30.1 (Casorati-Weierstrass Theorem), we have
1. there exists a sequence {αn} such that αn → 0 and limn→∞ | cos 1
αn
| = ∞.
2. for any complex number ω, there exists a sequence βn (depending on ω) such that βn → 0 and limn→∞ cos 1
βn
=
−ω
sin 1 .
Because sin(z−1)
z2+1 → − sin 1 as z → 0, we can get that limn→∞ | cos 1
αn
sin(αn−1)
α2
n+1 | = ∞ and limn→∞ cos 1
βn
sin(βn−1)
β2
n+1 =
ω. This implies that cos 1
z
sin(z−1)
z2+1 has an essential singularity at z = 0 by Theorem 30.1 (Casorati-Weierstrass
Theorem).
Since cos 1
z
sin(z−1)
z2+1 =
cos 1
z
sin(z−1)/z+i
z−i , cos 1
z sin(z−1)/z+i is analytic if z = −i and z = 0, and cos 1
z sin(z−1)/z+i|z=i =
cos(−i) sin(i−1)
2i = 0, we can get that cos 1
z
sin(z−1)
z2+1 has a pole of order 1 at z = i by Theorem 29.2.
Similarly, from cos 1
z sin(z−1)/z−i is analytic if z = i and z = 0, cos 1
z sin(z−1)/z−i|z=−i = cos i sin(−i−1)
−2i = 0, and
Theorem 29.2, we have cos 1
z
sin(z−1)
z2+1 =
cos 1
z
sin(z−1)/z−i
z+i has a pole of order 1 at z = −i.
6
7. (f)
It is easy to see that the isolated singularities are 1
2 , ±i, and k ∈ Z.
Since (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
=
(z2−1) cos πz/(z+i)2(z+2)(2z−1) sin2 πz
(z−i)2 , (z2
−1) cos πz/(z+i)2
(z+2)(2z−1) sin2
πz is analytic at
z = i, and (z2
−1) cos πz/(z+i)2
(z+2)(2z−1) sin2
πz
z=i
= −2 cos(iπ)
−4(i+2)(2i−1) sin2(iπ)
= 0, we can get that (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
has a pole of order 2 at z = i by Theorem 29.2.
Similarly, from (z2
−1) cos πz/(z−i)2
(z+2)(2z−1) sin2
πz is analytic at z = −i, (z2
−1) cos πz/(z−i)2
(z+2)(2z−1) sin2
πz
z=−i
=
−2 cos(−iπ)
−4(−i+2)(−2i−1) sin2(−iπ)
= 0, and Theorem 29.2, we have (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
=
(z2−1) cos πz/(z−i)2(z+2)(2z−1) sin2 πz
(z+i)2
has a pole of order 2 at z = −i.
Because
(z −
1
2
)
(z2
− 1) cos πz
(z + 2)(2z − 1)(z2 + 1)2 sin2
πz
=
(z2
− 1) cos πz
2(z + 2)(z2 + 1)2 sin2
πz
→
(1
4 − 1) · 0
2(1
2 + 2)(1
4 + 1)2 sin2
(π
2 )
= 0 as z →
1
2
,
we can get that z = 1
2 is a removable singularity of (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
by Theorem 29.1.
From Example 26.1, we have sin kπ has a zero of order 1 at z = k ∈ Z. By virtue of Corollary 26.1, we can get
that sin2
πz has a zero of order 2 at z = k ∈ Z.
Since z + 2 has a zero of order 1 at z = −2 and sin2
πz has a zero of order 2 at z = k ∈ Z, we can get
that (z + 2) sin2
πz has a zero of order 3 at z = −2. We also have that (z2
−1) cos πz
(2z−1)(z2+1)2 is analytic at z = −2 and
(z2
−1) cos πz
(2z−1)(z2+1)2
z=−2
= (4−1) cos(−2π)
(−4−1)(4+1)2 = 3
−125 = 0. This implies that (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
=
(z2−1) cos πz/(2z−1)(z2+1)2
(z+2) sin2 πz
has a pole of order 3 at z = −2 by Corollary 29.3.
From (z+1) cos πz
(z+2)(2z−1)(z2+1)2
z=1
= (1+1) cos π
(1+2)(2−1)(1+1)2 = −2
12 = 0 and (z+1) cos πz
(z+2)(2z−1)(z2+1)2 is analytic at z = 1, we can get
that (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 = (z − 1) (z+1) cos πz
(z+2)(2z−1)(z2+1)2 has a zero of order 1 at z = 1. By above, sin2
πz has a zero of
order 2 at z = k ∈ Z, and Corollary 29.3, we have z = 1 is a pole of order 1 of (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
.
Similarly, we can get that (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 = (z + 1) (z−1) cos πz
(z+2)(2z−1)(z2+1)2 has a zero of order 1 at z = −1. So
z = −1 is a pole of order 1 of (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
by virtue of above, sin2
πz has a zero of order 2 at z = k ∈ Z,
and Corollary 29.3.
Since (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 is analytic at z = k ∈ Z−{±1, −2}, (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2
z=k
= 0 for all z ∈ Z−{±1, −2},
sin2
πz has a zero of order 2 at z = k ∈ Z, and Corollary 29.3, we have (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
has a pole of order
2 at z ∈ Z − {±1, −2}.
Problem. 30.4
Sol:
7
8. Since f(z) has a pole of order m at z0, we can nd R 0 such that for z ∈ C with 0 |z − z0| R,
f(z) =
∞
k=−m ak(z − z0)k
where a−m ∈ C − {0}. So we can get that for z ∈ C with 0 |z − z0| R,
f (z) =
∞
k=−m
kak(z − z0)k−1
=
∞
k=−m−1
(k + 1)ak+1(z − z0)k
.
Then we have f (z) has a pole of order m + 1 at z0 from the denition of singularity.
Problem. 31.1
Sol:
(a)
From Theorem 31.1, we have
R
z3
+ 1
z2(z − 1)
, 0 = lim
z→0
1
1!
d
dz
z2
·
z3
+ 1
z2(z − 1)
= lim
z→0
d
dz
z3
+ 1
z − 1
= lim
z→0
3z2
(z − 1) − (z3
+ 1) · 1
(z − 1)2
= −1.
By virtue of Corollary 31.1, we can get that
R
z3
+ 1
z2(z − 1)
, 1 = lim
z→1
(z − 1) ·
z3
+ 1
z2(z − 1)
= lim
z→1
z3
+ 1
z2
= 2.
(b)
Since for z = 0, z2
e1/z
=
k=2
−∞
1
(−k+2)! zk
, we can get that R z2
e1/z
, 0 = 1
3! = 1
6 by the denition.
(c)
By virtue of Corollary 31.1, we can get that for all k ∈ Z, R z
cos z , (2k+1)
2 π = limz→
(2k+1)
2 π
(z − (2k+1)
2 π) · z
cos z .
Since limz→
(2k+1)
2 π
cos z
z−
(2k+1)
2 π
= − sin z|z=
(2k+1)
2 π
= (−1)k+1
and limz→
(2k+1)
2 π
z = (2k+1)
2 π for all k ∈ Z, we have for
8
9. all k ∈ Z,
R
z
cos z
,
(2k + 1)
2
π = lim
z→
(2k+1)
2 π
(z −
(2k + 1)
2
π) ·
z
cos z
=
(−1)k+1
(2k + 1)
2
π.
(d)
By virtue of Corollary 31.1, we can get that
R
sin 3z
z2
, 0 = lim
z→0
z ·
sin 3z
z2
= lim
z→0
sin 3z
z
= 3 cos 3z|z=0
= 3.
Problem. 31.2
Sol:
(a)
From Theorem 29.2, z2n
(z+1)n has a pole of order n at z = −1. So by Theorem 31.1, we have
R
z2n
(z + 1)n
, −1 = lim
z→−1
1
(n − 1)!
dn−1
dzn−1
(z + 1)n
·
z2n
(z + 1)n
= lim
z→−1
1
(n − 1)!
dn−1
dzn−1
z2n
= lim
z→−1
2n · (2n − 1) · · · · · (2n − n + 2)
(n − 1)!
z2n−n+1
= C2n
n−1(−1)n+1
.
(b)
9
10. From Problem 30.3 (e) and Corollary 31.1, we have
R cos
1
z
sin(z − 1)
z2 + 1
, i = lim
z→i
(z − i) · cos
1
z
sin(z − 1)
z2 + 1
= lim
z→i
cos
1
z
sin(z − 1)
z + i
=
cos(−i) sin(i − 1)
2i
=
ei(−i)
+e−i(−i)
2 · ei(i−1)
−e−i(i−1)
2i
2i
=
(e + e−1
) · (e−1−i
− e1+i
)
−8
=
cosh 1 · sinh(1 + i)
2
.
(c)
From Theorem 29.2, cos z
z2(z−π)3 has a pole of order 2 at z = 0. So by Theorem 31.1, we have
R
cos z
z2(z − π)3
, 0 = lim
z→0
1
1!
d
dz
z2
·
cos z
z2(z − π)3
= lim
z→0
d
dz
cos z
(z − π)3
= lim
z→0
−(z − π)3
sin z − 3(z − π)2
cos z
(z − π)6
= −
3
π4
.
(d)
From Theorem 29.2, cos z
z2(z−π)3 has a pole of order 3 at z = π. So by Theorem 31.1, we have
R
cos z
z2(z − π)3
, π = lim
z→π
1
2!
d2
dz2
(z − π)3
·
cos z
z2(z − π)3
=
1
2
lim
z→π
d2
dz2
cos z
z2
=
1
2
lim
z→π
d
dz
−z2
sin z − 2z cos z
z4
=
1
2
lim
z→π
z4
(−2z sin z − z2
cos z − 2 cos z + 2z sin z) − 4z3
(−z2
sin z − 2z cos z)
z8
=
1
2
π4
(π2
+ 2) − 4π3
(2π)
π8
=
π2
− 6
2π4
.
10
11. (e)
Since
z3
cos
1
z − 2
= (z − 2)3
+ 8(z − 2)2
+ 12(z − 2) + 8 ·
k=0
−∞
(−1)−k
(z − 2)2k
(−2k)!
,
we can get that
R z3
cos
1
z − 2
, 2 = 1 ·
(−1)2
4!
+ 8 · 0 + 12 ·
(−1)1
2!
+ 8 · 0
= −
143
24
by virtue of the denition.
(f)
Since for 0 |z| 1,
e
1
z
z2 + 1
=
1
1 − (−z2)
· e
1
z
=
∞
k=0
(−1)k
z2k
·
∞
k=0
z−k
k!
=
∞
k=0
(−1)k
z2k
·
k=0
−∞
zk
(−k)!
,
we can get that
R
e
1
z
z2 + 1
, 0 =
∞
k=0
(−1)k 1
(2k + 1)!
= sin 1.
by virtue of the denition.
Problem. 31.4
Sol:
(a)
11
12. It is easy to see that z2
(z−1)2(z+2) has a pole of order 2 at z = 1 and a pole of order of 1 at z = −2. So
R
z2
(z − 1)2(z + 2)
, 1 = lim
z→1
1
1!
d
dz
(z − 1)2
·
z2
(z − 1)2(z + 2)
= lim
z→1
d
dz
z2
(z + 2)
= lim
z→1
2z(z + 2) − z2
(z + 2)2
=
5
9
and
R
z2
(z − 1)2(z + 2)
, −2 = lim
z→−2
(z + 2) ·
z2
(z − 1)2(z + 2)
= lim
z→−2
z2
(z − 1)2
=
4
9
.
From Theorem 31.2, we have
ˆ
γ
z2
(z − 1)2(z + 2)
dz = 2πi R
z2
(z − 1)2(z + 2)
, 1 + R
z2
(z − 1)2(z + 2)
, −2
= 2πi.
(b)
From Theorem 31.2 and z3
(z−2)(z−1−i) is analytic for all |z| 1.5 with z = 1 + i, we have
ˆ
γ
z3
(z − 2)(z − 1 − i)
dz = 2πi · R
z3
(z − 2)(z − 1 − i)
, 1 + i .
It is easy to see that z3
(z−2)(z−1−i) has a pole of order 1 at z = 1 + i. So
R
z3
(z − 2)(z − 1 − i)
, 1 + i = lim
z→1+i
(z − 1 − i) ·
z3
(z − 2)(z − 1 − i)
= lim
z→1+i
z3
(z − 2)
=
(1 + i)3
−1 + i
= 2.
This implies that
´
γ
z3
(z−2)(z−1−i) dz = 4πi.
(c)
12
13. From Theorem 31.2 and 1
z4−1 is analytic inside γ with z = 1, z = i, and z = −i, we have
ˆ
γ
1
z4 − 1
dz = 2πi · R
1
z4 − 1
, 1 + R
1
z4 − 1
, i + R
1
z4 − 1
, −i .
It is easy to see that 1
z4−1 has a pole of order 1 at z = 1, ±i. So
R
1
z4 − 1
, 1 = lim
z→1
(z − 1) ·
1
z4 − 1
= lim
z→1
1
(z + 1)(z + i)(z − i)
=
1
2(1 + i)(1 − i)
=
1
4
.
Similarly, we can get that R 1
z4−1 , i = 1
(i+1)(i−1)2i = i
4 and R 1
z4−1 , −i = 1
(−i+1)(−i−1)(−2i) = − i
4 . This implies
that
´
γ
1
z4−1 dz = πi
2 .
(d)
From Theorem 31.2 and z
(z2−1)2(z2+1) is analytic inside γ with z = 1, z = i, and z = −i, we have
ˆ
γ
z
(z2 − 1)2(z2 + 1)
dz = 2πi · R
z
(z2 − 1)2(z2 + 1)
, 1 + R
z
(z2 − 1)2(z2 + 1)
, i + R
z
(z2 − 1)2(z2 + 1)
, −i .
It is easy to see that z
(z2−1)2(z2+1) has a pole of order 1 at z = ±i and a pole of order 2 at z = 1. So
R
z
(z2 − 1)2(z2 + 1)
, 1 = lim
z→1
1
1!
d
dz
(z − 1)2
·
z
(z2 − 1)2(z2 + 1)
= lim
z→1
d
dz
z
(z + 1)2(z2 + 1)
= lim
z→1
(z + 1)2
(z2
+ 1) − z(2(z + 1)(z2
+ 1) + 2z(z + 1)2
)
(z + 1)4(z2 + 1)2
=
4 · 2 − (2 · 2 · 2 + 2 · 4)
16 · 4
= −
1
8
.
We can also get that
R
z
(z2 − 1)2(z2 + 1)
, i = lim
z→i
(z − i) ·
z
(z2 − 1)2(z2 + 1)
= lim
z→i
z
(z2 − 1)2(z + i)
=
i
4 · 2i
=
1
8
.
13
14. Similarly, R z
(z2−1)2(z2+1) , −i = −i
4·(−2i) = 1
8 . This implies that
´
γ
1
z4−1 dz = πi
4 .
(e)
Since for |z| 0, we have
cos(
1
z2
)e
1
z =
∞
k=0
(−1)k
(2k)!
1
z4k
·
∞
k=0
1
k!
1
zk
=
k=0
−∞
(−1)−k
(−2k)!
z−4k
·
k=0
−∞
1
(−k)!
z−k
.
Then we can get that
R cos(
1
z2
)e
1
z , 0 =
(−1)0
0!
·
1
1!
= 1
by the denition. So by Theorem 31.2 and cos( 1
z2 )e
1
z is analytic at z = 0, we have
ˆ
γ
cos(
1
z2
)e
1
z dz = 2πi · R cos(
1
z2
)e
1
z , 0
= 2πi.
(f)
Because tan z = sin z
cos z and cos z has a zero of order 1 at z = (2k+1)
2 π for all k ∈ Z, we can get that for all k ∈ Z,
R tan z,
(2k + 1)
2
π = lim
z→
(2k+1)
2 π
(z −
(2k + 1)
2
π)
sin z
cos z
= lim
z→
(2k+1)
2 π
sin z
cos z
(z−
(2k+1)
2 π)
.
Since limz→
(2k+1)
2 π
sin z = (−1)k
and limz→
(2k+1)
2 π
cos z
(z−
(2k+1)
2 π)
= (− sin z)|z=
(2k+1)
2 π
= (−1)k+1
, we have R tan z, (2k+1)
2 π =
−1 for all k ∈ Z.
From Theorem 31.2, we have
ˆ
γ
tan zdz = 2πi · R tan z,
−π
2
+ R tan z,
π
2
= −4πi.
Problem. 31.5
Sol:
14
15. Let γ be the circle |z + 1| = 2 that traversed once in the counterclockwise direction. So
´
γ
dz
(z2+z)2 =
−2
´
γ
dz
(z2+z)2 .
It is easy to see that 1
(z2+z)2 = 1
z2(z+1)2 has a pole of order 2 at z = 0 and z = −1. Then
R
1
(z2 + z)2
, 0 = lim
z→0
1
1!
d
dz
z2
·
1
(z2 + z)2
= lim
z→0
d
dz
1
(z + 1)2
= lim
z→0
−2
(z + 1)3
= −2
and
R
1
(z2 + z)2
, −1 = lim
z→−1
1
1!
d
dz
(z + 1)2
·
1
(z2 + z)2
= lim
z→−1
d
dz
1
z2
= lim
z→−1
−2
z3
= 2.
By virtue of Theorem 31.2, we can get that
ˆ
γ
1
(z2 + z)2
dz = 2πi · R
1
(z2 + z)2
, 0 + R
1
(z2 + z)2
, −1
= 0.
This implies that
´
γ
dz
(z2+z)2 = −2
´
γ
dz
(z2+z)2 = 0.
Problem. 31.6
Sol:
Since f has an isolated singularity at 0, we can nd R 0 and ak ∈ C for all k ∈ Z such that f(z) =
∞
k=−∞ akzk
for all 0 |z| R. Because f is an even function, we have a2j+1 = 0 for all j ∈ Z by Problen 25.7. This implies
that R[f, 0] = a−1 = 0 from the denition.
15