Chapter 14: Problem 19
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 \(\frac{1}{2}\).
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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.
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 \(C\).
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.