7.7 Isolated singularities, and the residue theorem

 Page 1 / 4
this module covers the residue theorem, removable and isolated singularities, some new developments from Cauchy's theorem, and related exercises.

The first result we present in this section is a natural extension of [link] . However, as we shall see, its consequences for computing contour integrals can hardly be overstated.

Let $S$ be a piecewise smooth geometric set whose boundary ${C}_{S}$ has finite length. Suppose ${c}_{1},...,{c}_{n}$ are distinct points in the interior ${S}^{0}$ of $S,$ and that ${r}_{1},...,{r}_{n}$ are positive numbers such that the closed disks $\left\{{\overline{B}}_{{r}_{k}}\left({c}_{k}\right)\right\}$ are contained in ${S}^{0}$ and pairwise disjoint. Suppose $f$ is continuous on $S\setminus \cup {B}_{{r}_{k}}\left({c}_{k}\right),$ i.e., at each point of $S$ that is not in any of the open disks ${B}_{{r}_{k}}\left({c}_{k}\right),$ and that $f$ is differentiable on ${S}^{0}\setminus \cup {\overline{B}}_{{r}_{k}}\left({c}_{k}\right),$ i.e., at each point of ${S}^{0}$ that is not in any of the closed disks ${\overline{B}}_{{r}_{k}}\left({c}_{k}\right).$ Write ${C}_{k}$ for the circle that is the boundary of the closed disk ${\overline{B}}_{{r}_{k}}\left({c}_{k}\right).$ Then

${\int }_{{C}_{S}}f\left(\zeta \right)\phantom{\rule{0.166667em}{0ex}}d\zeta =\sum _{k=1}^{n}{\int }_{{C}_{k}}f\left(\zeta \right)\phantom{\rule{0.166667em}{0ex}}d\zeta .$

This is just a special case of part (d) of [link] .

Let $f$ be continuous on the punctured disk ${\overline{{B}^{\text{'}}}}_{r}\left(c\right),$ analytic at each point $z$ in ${B}_{r}^{\text{'}}\left(c\right),$ and suppose $f$ is undefined at the central point $c.$ Such points $c$ are called isolated singularities of $f,$ and we wish now to classify these kinds of points. Here is the first kind:

A complex number $c$ is called a removable singularity of an analytic function $f$ if there exists an $r>0$ such that $f$ is continuous on the punctured disk ${\overline{{B}^{\text{'}}}}_{r}\left(c\right),$ analytic at each point in ${B}_{r}^{\text{'}}\left(c\right),$ and ${lim}_{z\to c}f\left(z\right)$ exists.

1. Define $f\left(z\right)=sinz/z$ for all $z\ne 0.$ Show that 0 is a removable singularity of $f.$
2. For $z\ne c,$ define $f\left(z\right)=\left(1-cos\left(z-c\right)\right)/\left(z-c\right).$ Show that $c$ is a removable singularity of $f.$
3. For $z\ne c,$ define $f\left(z\right)=\left(1-cos\left(z-c\right)\right)/{\left(z-c\right)}^{2}.$ Show that $c$ is still a removable singularity of $f.$
4. Let $g$ be an analytic function on ${B}_{r}\left(c\right),$ and set $f\left(z\right)=\left(g\left(z\right)-g\left(c\right)\right)/\left(z-c\right)$ for all $z\in {B}_{r}^{\text{'}}\left(c\right).$ Show that $c$ is a removable singularity of $f.$

The following theorem provides a good explanation for the term “removable singularity.” The idea is that this is not a “true” singularity; it's just thatfor some reason the natural definition of $f$ at $c$ has not yet been made.

Let $f$ be continuous on the punctured disk ${\overline{B}}_{r}^{\text{'}}\left(c\right)$ and differentiable at each point of the open punctured disk ${B}_{r}^{\text{'}}\left(c\right),$ and assume that $c$ is a removable singularity of $f.$ Define $\stackrel{˜}{f}$ by $\stackrel{˜}{f}\left(z\right)=f\left(z\right)$ for all $z\in {B}_{r}^{\text{'}}\left(c\right),$ and $\stackrel{˜}{f}\left(c\right)={lim}_{z\to c}f\left(z\right).$ Then

1.   $\stackrel{˜}{f}$ is analytic on the entire open disk ${B}_{r}\left(c\right),$ whence
$f\left(z\right)=\sum _{k=0}^{\infty }{c}_{k}{\left(z-c\right)}^{k}$
for all $z\in {B}_{r}^{\text{'}}\left(c\right).$
2. For any piecewise smooth geometric set $S\subseteq {B}_{r}\left(c\right),$ whose boundary ${C}_{S}$ has finite length, and for which $c\in {S}^{0},$
${\int }_{{C}_{S}}f\left(\zeta \right)\phantom{\rule{0.166667em}{0ex}}d\zeta =0.$

As in part (a) of [link] , define $F$ on ${B}_{r}\left(c\right)$ by

$F\left(z\right)=\frac{1}{2\pi i}{\int }_{{C}_{r}}\frac{f\left(\zeta \right)}{\zeta -z}\phantom{\rule{0.166667em}{0ex}}d\zeta .$

Then, by that exercise, $F$ is analytic on ${B}_{r}\left(c\right).$ We show next that $F\left(z\right)=\stackrel{˜}{f}\left(z\right)$ on ${B}_{r}\left(c\right),$ and this will complete the proof of part (1).

Let $z$ be a point in ${B}_{r}\left(c\right)$ that is not equal to $c,$ and let $ϵ>0$ be given. Choose $\delta >0$ such that $\delta <|z-c|/2$ and such that $|\stackrel{˜}{f}\left(\zeta \right)-\stackrel{˜}{f}\left(c\right)|<ϵ$ if $|\zeta -c|<\delta .$ Then, using part (c) of [link] , we have that

$\begin{array}{cc}\hfill \stackrel{˜}{f}\left(z\right)& =f\left(z\right)\\ & =& \frac{1}{2\pi i}{\int }_{{C}_{r}}\frac{f\left(\zeta \right)}{\zeta -z}\phantom{\rule{0.166667em}{0ex}}d\zeta -\frac{1}{2\pi i}{\int }_{{C}_{\delta }}\frac{f\left(\zeta \right)}{\zeta -z}\phantom{\rule{0.166667em}{0ex}}d\zeta \hfill \\ & =& F\left(z\right)-\frac{1}{2\pi i}{\int }_{{C}_{\delta }}\frac{f\left(\zeta \right)-\stackrel{˜}{f}\left(c\right)}{\zeta -z}\phantom{\rule{0.166667em}{0ex}}d\zeta -\frac{1}{2\pi i}{\int }_{{C}_{\delta }}\frac{\stackrel{˜}{f}\left(c\right)}{\zeta -z}\phantom{\rule{0.166667em}{0ex}}d\zeta \hfill \\ & =& F\left(z\right)-\frac{1}{2\pi i}{\int }_{{C}_{\delta }}\frac{\stackrel{˜}{f}\left(\zeta \right)-\stackrel{˜}{f}\left(c\right)}{\zeta -z}\phantom{\rule{0.166667em}{0ex}}d\zeta ,\hfill \end{array}$

where the last equality holds because the function $\stackrel{˜}{f}\left(c\right)/\left(\zeta -z\right)$ is an analytic function of $\zeta$ on the disk ${B}_{\delta }\left(c\right),$ and hence the integral is 0 by [link] . So,

$\begin{array}{ccc}\hfill |\stackrel{˜}{f}\left(z\right)-F\left(z\right)|& =& |\frac{1}{2\pi i}{\int }_{{C}_{\delta }}\frac{\stackrel{˜}{f}\left(\zeta \right)-\stackrel{˜}{f}\left(c\right)}{\zeta -z}\phantom{\rule{0.166667em}{0ex}}d\zeta |\hfill \\ & \le & \frac{1}{2\pi }{\int }_{{C}_{\delta }}\frac{|\stackrel{˜}{f}\left(\zeta \right)-\stackrel{˜}{f}\left(c\right)|}{|\zeta -z|}\phantom{\rule{0.166667em}{0ex}}ds\hfill \\ & \le & \frac{1}{2\pi }{\int }_{{C}_{\delta }}\frac{ϵ}{\delta /2}\phantom{\rule{0.166667em}{0ex}}ds\hfill \\ & =& \frac{2ϵ}{\delta }×\delta \hfill \\ & =& 2ϵ.\hfill \end{array}$

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!