# About the definition of isolated singularity of a complex function

I'm learning the part Isolated Singularity Categorization, and there's a point in the definition of the isolated singularity which confused me a lot:

A function $f$ has an isolated singularity at a point $z = a$ if there is a number $R > 0$ such that $F$ is analytic on $\{z: 0 < |z-a| < R\}$.

So the part which bothered me is: If $a$ is an isolated singularity, then $f$ needs to be analytic on $\{z: 0 < |z-a| < R\}$, so what about $a$ itself? If $f$ is holomorphic at $a$, meaning that $f$ is analytic on $\{z:|z-a| < R\}$ for some $R$ (including $z$), then is $a$ called isolated singularity(which does not make much sense I think) in this case?. The definition says no clue about the analytic property at the point $a$ itself.

If at $a$, $f$ can be analytic or not, then there's some example in my book confused me, too. It says that the function $\sin(1 - {1 \over z})$ has only one isolated singularity, which is $z = 0$. Then it forces me to think that at isolated singularity, function should not be analytic, but then why don't the definition say it clear?

Sorry if it's a stupid question. Hope someone can clear it out for me. I really appreciate.

You're right, of course. A point where $f$ is analytic doesn't count as a singularity.

For example, in Conway's Functions of One Complex Variable I (p. 103) the definition reads “A function $f$ has an isolated singularity at $z=a$ if there is an $R>0$ such that $f$ is defined and analytic in $B(a;R)-\{a\}$ but not in $B(a;R)$”.

• Oh, which edition of Conway are you using? I'm reading that book, too. Actually, the definition I posted is from Conway, too, page 99. And your definition is different from the definition I read in my book. So maybe the edition you and I use are different. Apr 22 '16 at 8:47
• Second edition. Interesting that even he managed to get it wrong in the first edition. Nobody's perfect! Apr 22 '16 at 9:36
• Sorry, I have one more question: Based on your definition, is $f$ necessary to be DEFINED at $a$. It's clear that $f$ should not be analytic at $a$. But should it not be defined at $a$, too. For example, is $0$ a removable singularity of $f(x) = {\sin x \over x}$ if $x \neq 0$ and $f(0) = 2$ (this is an example from my teacher). Apr 22 '16 at 15:32
• For the definition of isolated singularity, it doesn't matter whether $f(a)$ is defined or not. It's a removable singularity if you can define $f(a)$ (or redefine, if $f(a)$ already was defined but with an “incorrect” value) so that $f$ becomes analytic in $B(a;R)$. Apr 22 '16 at 15:52
• I got it. Thanks a lot Hans Lundmark! Apr 22 '16 at 17:15