# If $"a"$ is an isolated singularity, then there exists a punctured disc around $a$ on which $f$ is analytic

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?

• 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? Apr 30 '21 at 19:09
• Goursat's theorem: if $f$ is (complex) differentiable at all points of an open set, then it is analytic there. Apr 30 '21 at 19:19
• @RobertIsrael..why is differentiable on a punctured disc? Apr 30 '21 at 20:07
• @reuns..$f$ is any function defined on a domain Apr 30 '21 at 20:08

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 $$z$$.