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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
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.
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 $\tilde{f}$ by $\tilde{f}\left(z\right)=f\left(z\right)$ for all $z\in {B}_{r}^{\text{'}}\left(c\right),$ and $\tilde{f}\left(c\right)={lim}_{z\to c}f\left(z\right).$ Then
As in part (a) of [link] , define $F$ on ${B}_{r}\left(c\right)$ by
Then, by that exercise, $F$ is analytic on ${B}_{r}\left(c\right).$ We show next that $F\left(z\right)=\tilde{f}\left(z\right)$ 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 $\u03f5>0$ be given. Choose $\delta >0$ such that $\delta <|z-c|/2$ and such that $|\tilde{f}\left(\zeta \right)-\tilde{f}\left(c\right)|<\u03f5$ if $|\zeta -c|<\delta .$ Then, using part (c) of [link] , we have that
where the last equality holds because the function $\tilde{f}\left(c\right)/(\zeta -z)$ is an analytic function of $\zeta $ on the disk ${B}_{\delta}\left(c\right),$ and hence the integral is 0 by [link] . So,
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