# 7.7 Isolated singularities, and the residue theorem

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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
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I'm not sure why it wrote it the other way
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I got X =-6
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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.
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