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According to the lecture notes that I am reading, $a$ is a singular point if it is a limit point of set of regular points but not a regular point itself.

$a$ is said to be a regular point of a function $f$ if there exists an open disc around $a$ on which $f$ is differentiable.

$a$ is said to be an isolated singularity if there exists an open disc about $a$ which has no other singularity.

By these definitions, how does it follow that "If $a$ is an isolated singularity of $f$, then $f$ is analytic in a punctured open disc about $a$".

$a$ being isolated will give me an open disc which has no other singularity other than $a$. But how does it follow that $f$ will be analytic at other points?

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  • $\begingroup$ That's the definition of $a$ is an isolated singularity, the function is analytic on $0<|z-a|<r$ for some $r$. What is your $f$, the context? $\endgroup$
    – reuns
    Apr 30 '21 at 19:09
  • $\begingroup$ Goursat's theorem: if $f$ is (complex) differentiable at all points of an open set, then it is analytic there. $\endgroup$ Apr 30 '21 at 19:19
  • $\begingroup$ @RobertIsrael..why is differentiable on a punctured disc? $\endgroup$
    – Gitika
    Apr 30 '21 at 20:07
  • $\begingroup$ @reuns..$f$ is any function defined on a domain $\endgroup$
    – Gitika
    Apr 30 '21 at 20:08
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In keeping with Robert's comment on Goursat's theorem ($f$ is analytic at a point iff it's differentiable on a neighborhood of that point), notice that if you have an punctured disc of radius $\epsilon$ about the isolated singularity $a$ wherein $f$ is differentiable, you can pick a point $z_0$ in that open disc where $\displaystyle 0 < |z_0 - a| < \frac\epsilon2$ and examine a neighborhood of points $z$ so that $|z-z_0| < |z_0-a|$ implies $f$ is differentiable in a neighborhood of $z$. Thus by Goursat's theorem $f$ is analytic at $z$.

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