# Is a complex function analytic on a domain $D$ if $D$ contains an isolated singularity?

Most likely a stupid question, but the materials I have give, for me, somewhat contradictory information. For example in a theorem relating to Taylor series, it says let $$f$$ be analytic on $$S$$, and let $$z_0$$ be the closes isolated singularity to a point $$\alpha$$, where in the picture it shows that $$z_0$$ is in S. Thank you in advance, and English is not my primary language.

Yes. For instance$$\begin{array}{rccc}\iota\colon&\mathbb C\setminus\{0\}&\longrightarrow&\mathbb C\\&z&\mapsto&\frac1z\end{array}$$is an analytical function. That's so because, for each $$a\in\mathbb C\setminus\{0\}$$, you have$$\frac1a-\frac{z-a}{a^2}+\frac{(z-a)^2}{a^3}-\frac{(z-a)^3}{a^4}+\cdots$$near $$a$$ (when $$\lvert z-a\rvert<\lvert a\vert$$, to be more precise).
That's what being an analytical function means: for each $$a$$ in its domain, there is a power series centered at $$a$$ with positive radius conconvergence whose sum (near $$a$$) is $$f(z)$$. The fact that there is an isolated singularity (such as $$0$$, in my example) changes nothing.