# Analyticity and removable singularity

In Churchill's book of Complex Analysis there are two statements that I can't match them to be consistent: In one place it says that a function must be analytic at a removable singular point :

why "must"? Because after removing singularity it becomes a series of positive powers.

But the following lemma say that if the function is not analytic at $$z_0$$ then definitely it has a removable singularity there:

and so must be analytic; but by the lemma it is not analytic! Where am I wrong?

Here is the proof of the lemma:

I don't understand why it supposes not being analytic then it arrvies at a Taylor series which implies analyticity?

• I am sorry but I do not see it forcing it to be analytical at $z_0$ to have a removable singularity. It only says it must be analytic around a neighborhood of $z_0$ that is centered around $z_0$ but not containing it. (First part shows how to remove a removable singularity to obtain an analytical continuation by example and second is nothing but the definition of a removable singularity. If a singularity has infinitely many singularities that cannot be isolated, then that is a different story... ) – keoxkeox Oct 24 '18 at 7:00
• If boundedness doesn't hold in the lemma, the singularity is no longer removable. – Berci Oct 24 '18 at 7:02
• @Berci, sorry I don't understand the connection of it to the question.. – user231343 Oct 24 '18 at 7:03
• @Edi Unboundedness allows for a pole or essential singularity at that point. For example, $1/z$ is analytic on $0 < |z| < \epsilon$ for all $\epsilon > 0$. However, it is not bounded on these domains, and of course it is not analytic at $z=0$. – eyeballfrog Oct 24 '18 at 7:09
• Do Taylor series assume analyticity? It is just a series definition, whether it is divergent or not is something else. I started to believe that you are in fact questioning validity of analytical continuations (Some people accept sum of all positive integers to be -1/12 and some find it senseless.). – keoxkeox Oct 24 '18 at 7:10

A removable singularity of a function $$f$$ is a point $$z_0$$ where $$f(z_0)$$ is undefined, but there exists a value $$c$$ such that, if we define $$f(z_0) = c$$, then $$f$$ is analytic in a neighborhood of $$z_0$$. Note that $$f$$ is not actually analytic at $$z_0$$--it is undefined. It's just that there's a way to define its value at $$z_0$$ to make it analytic.
What the lemma is proving is that if a function $$f$$ is analytic and bounded on the set $$0 < |z-z_0| < \epsilon$$ for some positive $$\epsilon$$, then either $$f$$ is analytic at $$z_0$$, or $$f$$ has a removable singularity there (and thus could be made analytic through a suitable choice of $$f(z_0)$$). Bounded is needed because $$z_0$$ could otherwise be a pole or essential singularity, where no choice of $$f(z_0)$$ could make $$f$$ analytic there, but in both of those cases $$f$$ would be unbounded on $$0 < |z-z_0| < \epsilon$$.