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 $B(a;R)$”.