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?
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$\begingroup$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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