7.7 Isolated singularities, and the residue theorem  (Page 8/4)

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

1.   $\tilde{f}$ is analytic on the entire open disk $B_r\left(c\right),$ whence
   $$f\left(z\right)=\sum _{k=0}^{\infty }{c}_k{(z-c)}^k$$
   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 $c\in S^0,$
   $${\int }_{C_S}f\left(\zeta \right)d\zeta =0.$$

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 $ϵ>0$ be given. Choose $\delta >0$ such that $\delta <|z-c|/2$ and such that $|\tilde{f}\left(\zeta \right)-\tilde{f}\left(c\right)|<ϵ$ 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,