/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 In Problems 5 to 8, find the res... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 7

In Problems 5 to 8, find the residues of the given function at all poles. Take \(z=r e^{i \theta}\), \(0 \leq \theta<2 \pi\). $$ \frac{\ln z}{1+z^{2}} $$

Short Answer

Expert verified
The residues at the poles \( z = i \) and \( z = -i \) are both \( \frac{\pi}{4} \).

Step by step solution

01

Identify the poles

The poles occur where the denominator is zero. For the given function \( \frac{\ln z}{1+z^2} \), set \( 1+z^2 = 0 \). Solving for \( z \), we get:\[ z^2 = -1 \rightarrow z = \pm i \] Hence, the poles are at \( z = i \) and \( z = -i \).
02

Compute the residue at \( z = i \)

The residue of a function \( f(z) \) at a simple pole \( z_0 \) is given by \( \lim_{z \to z_0} (z - z_0) f(z) \). For \( z = i \):\[ \text{Residue at } z = i = \lim_{z \to i} (z - i) \left( \frac{\ln z}{1+z^2} \right) = \lim_{z \to i} \frac{(z - i) \ln z}{1 + z^2} \] Since \( 1 + z^2 = (z - i)(z + i) \), we get:\[ = \lim_{z \to i} \frac{(z - i) \ln z}{(z - i)(z + i)} = \lim_{z \to i} \frac{\ln z}{z + i} \] Plug in \( z = i \):\[ \text{Residue at } z = i = \frac{\ln i}{2i} \]
03

Compute the residue at \( z = -i \)

Similarly, for \( z = -i \):\[ \text{Residue at } z = -i = \lim_{z \to -i} (z + i) \left( \frac{\ln z}{1+z^2} \right) = \lim_{z \to -i} \frac{(z + i) \ln z}{1 + z^2} \] Simplifying using \( 1 + z^2 = (z - i)(z + i) \), we get:\[ = \lim_{z \to -i} \frac{(z + i) \ln z}{(z - i)(z + i)} = \lim_{z \to -i} \frac{\ln z}{z - i} \] Plug in \( z = -i \):\[ \text{Residue at } z = -i = \frac{\ln (-i)}{-2i} \]
04

Simplify the residues

To simplify:\[ \text{Residue at } z = i = \frac{\ln i}{2i} \] Recall that \( \ln i = i \frac{\pi}{2} \), so:\[ \text{Residue at } z = i = \frac{i \frac{\pi}{2}}{2i} = \frac{\pi}{4} \] Similarly, for \( z = -i \):\[ \text{Residue at } z = -i = \frac{\ln (-i)}{-2i} \] and \( \ln (-i) = -i \frac{\pi}{2} \) so:\[ \text{Residue at } z = -i = \frac{-i \frac{\pi}{2}}{-2i} = \frac{\pi}{4} \]

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 91影视!

Key Concepts

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

Residue Theorem
The Residue Theorem is a significant result in complex analysis that allows us to evaluate complex integrals easily. It relates integrals around closed curves to the sum of residues within that curve. A residue of a function at a point is essentially the coefficient of \(\frac{1}{z - a}\) in its Laurent series expansion around that point. For example, in the given problem, the function \(\frac{\text{ln z}}{1+z^2}\)\has poles at \(\text{z = i and z = -i}\). By calculating the residues at these poles, we can apply the residue theorem to evaluate integrals involving this function.
Poles of a Function
Poles are specific types of singularities in complex functions where the function heads towards infinity. A pole at \(z = z_0\) occurs when the function can be written as \(\frac{h(z)}{(z - z_0)^n}\), with \(h(z)\) holomorphic (complex differentiable) and \(n\) being a positive integer. In the given exercise, the poles occur where the denominator \(\text{1 + z}^2= 0\)\. This simplifies to \(\text{z}^2 = -1 \rightarrow z = \text{卤i}\). Steps to identify poles:
  • Set the denominator of the function to zero.
  • Simplify to solve for \(z\).
In this example, solving \(1 + z^2 = 0\) gives the poles at \(\text{z = i and z = -i}\).
Logarithmic Function
A logarithmic function in the complex plane, denoted as \(\text{ln} z\), is a multi-valued function because of the periodic nature of the complex exponential function. The principal branch of \(\text{ln} z\) is usually defined on the complex plane cut along the negative real axis:
  • For any complex number \(z = re^{i\theta}\), where \(\text{0 鈮� 胃 < 2蟺}\), \(\text{ln} z = \text{ln} r + i胃\).
  • Recall the Euler's formula: \(\text{e}^{i\theta} = \text{cos}胃 + i \text{sin}胃\).
In the given exercise, to compute the residues at poles, we use the logarithm of \(i\) and \(-i\), with \(\text{ln} i = i \frac{\text{蟺}}{2}\) and \(\text{ln}(-i) = -i \frac{\text{蟺}}{2}\).
Complex Numbers
A complex number is written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(\text{i}\) is the imaginary unit satisfying \(\text{i}^2 = -1\). Complex numbers extend the idea of one-dimensional real numbers to the two-dimensional complex plane by including \(i\). They are essential in various fields of mathematics and engineering because they simplify calculations involving oscillations.
  • When working with functions like \(\frac{\text{ln} z}{1 + z^2}\), it's crucial to understand both real and imaginary parts.
  • Powers and roots of complex numbers can be computed using Euler's formula, which links complex exponentiation with trigonometric functions.
  • In polar form, a complex number \(z = re^{i胃}\), where \(r\) is the magnitude and \(胃\) is the argument (angle).
For instance, the poles \(z = i\) and \(-i\) in the given exercise are points on the complex plane where the function heads towards infinity.

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

Find the Cauchy-Riemann equations in polar coordinates. Hint : \(z=r e^{i \theta}, f(z)=u(r, \theta)+i \tau(r, \theta) .\) Follow the method of equations \((2.3)\) and \((2.4) .\)

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. (Warming: 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 \(|z|a\). $$ \frac{30}{(1+z)(z-2)(3+z)} $$

In Problems 5 to 8, find the residues of the given function at all poles. Take \(z=r e^{i \theta}\), \(0 \leq \theta<2 \pi\). $$ \frac{z^{1 / 3}}{1+z^{2}} $$

For each of the following functions, say whether the indicated point is regular, an essential? singularity, or a pole, and if a pole of what order it is. (a) \(\frac{e^{z}-1}{z^{2}+4}, \quad z=2 i\) (b) \(\tan ^{2} z, \quad z=\pi / 2\) (c) \(\frac{1-\cos z}{z^{4}}, \quad z=0\) (d) \(\cos \left(\frac{\pi}{z-\pi}\right), z=\pi\)

Two long parallel cylinders form a capacitor. (Let their cross sections be the images of \(u=a\) and \(u=-a .\) ) If they are held at potentials \(V_{0}\) and \(-V_{0}\), find the potential \(V(x, y)\) at points between them. Given that the charge (per unit length) on a cylinder is \(q=\) \(V_{0} /(2 a)\), show that the capacitance (per unit length), that is, \(4 /\left(2 V_{0}\right)\), is given by \(1 /(4\) arc cosh \(d / 2 r)\), where \(d\) is the distance between the centers of the two cylinders, and their radii are \(r\).
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
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.