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 :

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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:

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

Here is the proof of the lemma:

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

  • $\begingroup$ 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... ) $\endgroup$ – keoxkeox Oct 24 '18 at 7:00
  • $\begingroup$ If boundedness doesn't hold in the lemma, the singularity is no longer removable. $\endgroup$ – Berci Oct 24 '18 at 7:02
  • $\begingroup$ @Berci, sorry I don't understand the connection of it to the question.. $\endgroup$ – user231343 Oct 24 '18 at 7:03
  • $\begingroup$ @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 $z=0$. $\endgroup$ – eyeballfrog Oct 24 '18 at 7:09
  • $\begingroup$ 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.). $\endgroup$ – 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 $0 < |z-z_0| < \epsilon$.


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