Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer. $$\frac{\sin ^{2} z}{2 z-\pi} \text { at } z=\pi / 2$$

Short Answer

Expert verified
The residue is \( \frac{1}{2} \).

Step by step solution

01

Identify the form of the function

The given function is \(\frac{\text{sin}^2(z)}{2z - \pi}\). The goal is to find the residue at \(z = \frac{\pi}{2}\).
02

Simplify the expression at the residue point

Rewrite \(2z - \pi\) as \(2(z - \frac{\pi}{2})\). The function then becomes \(\frac{\text{sin}^2(z)}{2(z - \frac{\pi}{2})}\).
03

Evaluate the numerator at the point of interest

Evaluate \(\text{sin}^2(z)\) at \(z = \frac{\pi}{2}\): \(\text{sin}^2(\frac{\pi}{2}) = 1\).
04

Determine the residue

Since \( \text{sin}^2(z) = 1 \) at \( z = \frac{\pi}{2} \), substitute into the function: \(\frac{1}{2(z - \frac{\pi}{2})}\). The residue is the coefficient of \(\frac{1}{z - z_0}\) which gives \(\frac{1}{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Analysis
Complex analysis, also known as the theory of functions of a complex variable, is a branch of mathematical analysis that investigates functions defined on the complex numbers. Here are some key concepts you should understand:
  • Complex Numbers: Extend the idea of the one-dimensional number line to the complex plane. A complex number has the form \(z = a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\).
  • Complex Functions: These are functions that take complex numbers as inputs and provide complex numbers as outputs. The most basic example is a polynomial function.
  • Analytic Functions: Functions that are locally given by convergent power series. These functions are also called holomorphic functions.
  • Singularities: Points where a complex function does not behave nicely (like not being differentiable). Singularities play an essential role in complex analysis.
Complex analysis is essential because it provides powerful techniques and theories for solving mathematical problems that are intractable by methods in real analysis alone.
Residue Theorem
The Residue Theorem is a central result in complex analysis that helps evaluate complex integrals. Let’s break it down:
  • Residue: The residue of a function at a particular point \(z_0\) is a special value that often helps simplify complex integration around that point. For an isolated singularity \(z_0\), the residue is the coefficient \(a_{-1}\) in the Laurent series expansion of the function around \(z_0\).
  • In Practice: To find the residue at a singularity for a function \(f(z)\), you often factor the denominator and then apply series expansion techniques or directly substitute the point.
  • Residue Theorem: If \(f(z)\) is analytic in and on a simple closed contour \(C\) except for isolated singularities, the integral of \(f(z)\) over \(C\) is given by \[ \text{∮}_C f(z) \, dz = 2 \pi i \sum \text{Res}(f, z_k) \] where \(\text{Res}(f, z_k)\) are the residues at the singularities inside \(C\).
This theorem is powerful because it simplifies the process of evaluating complex integrals by reducing it to the problem of finding residues.
Laurent Series
The Laurent series is an expansion of a complex function into an infinite sum that includes terms with negative powers. This is particularly useful for studying the behavior of functions near singularities. Key points include:
  • Regular Part: The part of the Laurent series that resembles a Taylor series, including non-negative powers of \((z - z_0)\).
  • Principal Part: This includes terms that have negative powers of \((z - z_0)\) and is crucial for identifying the residue.
  • Form of the Series: For a function \(f(z)\) around a point \(z_0\), the Laurent series is given by \[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n \] where the coefficients \(a_n\) can be computed using contour integrals.
  • Computing Residues: As mentioned, the residue is the coefficient of \(\frac{1}{z - z_0}\), often denoted as \(a_{-1}\) in the Laurent series.
The Laurent series is an invaluable tool in complex analysis, allowing for a deeper understanding of function behavior near singularities and greatly aiding in the application of residue calculus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The values of the following integrals are known and can be found in integral tables or by computer. Your goal in evaluating them is to learn about contour integration by applying the methods discussed in the examples above. Then check your answers by computer. $$\int_{0}^{\infty} \frac{\cos x d x}{\left(1+9 x^{2}\right)^{2}}$$

For each of the following functions find the first few terms of each of the Laurent series about the origin, that is, one series for each annular ring between singular points. Find the residue of each function at the origin. ( Warning: To find the residue, you must use the Laurent series which converges near the origin.) Hints: See Problem 2. Use partial fractions as in equations (4.5) and \((4.7) .\) Expand a term \(1 /(z-a)\) in powers of \(z\) to get a series convergent for \(|z|a\) \(\frac{z-1}{z^{3}(z-2)}\)

Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer. $$\frac{z-2}{z(1-z)} \text { at } z=0 ; \text { at } z=1$$

Let \(f(z)=u+i v\) be an analytic function, and let \(\mathbf{F}\) be the vector \(\mathbf{F}=v \mathbf{i}+u \mathbf{j}\). Show that the equations div \(\mathbf{F}=0\) and \(\operatorname{curl} \mathbf{F}=0\) are equivalent to the Cauchy-Riemann equations.

Liouville's theorem: Suppose \(f(z)\) is analytic for all \(z\) (except \(\infty\) ), and bounded [that is, \(|f(z)| \leq M\) for all \(z\) and some \(M\) ]. Prove that \(f(z)\) is a constant. Hints: If \(f^{\prime}(z)=0,\) then \(f(z)=\) const. To show this, write \(f^{\prime}(z)\) as in Problem 3.21 where \(C\) is a circle of radius \(R\) and center \(z\), that is, \(w=z+R e^{i \theta} .\) Show that \(\left|f^{\prime}(z)\right| \leq M / R\) and let \(R \rightarrow \infty\)

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free