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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.

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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.

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