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True or false: $f$ holomorphic in $A=\{z\in \mathbb{C}: 0\lt |z|\lt 2\}$ and $\int_{|z|=1} z^n f(z)dz=0 \ \forall n = 0, 1, 2, ...$ then $f$ has a removable singularity at $z=0$.

I' m not sure if this is true or false because I can only prove that $f$ is either analytic at zero or has a removable singularity.

$f$ holomorphic in $A$ means that we can express $f$ as a Laurent series of the form

$$f(z)= \sum_{n=1}^{\infty}\frac{b_n}{z^n} + \sum_{n=0}^{\infty}{a_n}{z^n}$$

with $$a_n = \frac{1}{2 \pi i}\int_{\gamma} \frac {f(z)}{z^{n+1}}dz$$

$$b_n = \frac{1}{2 \pi i}\int_{\gamma} {f(z)}{z^{n-1}}dz$$

$\int_{|z|=1} z^n f(z)dz=0 \ \forall n = 0, 1, 2, ...$ means that $b_n = 2 \pi i \cdot 0 = 0 \ \forall n$, so $f$ has as worst a removable singularity at $z=0$.

Now to prove that $f$ has indeed a removable singularity one must prove that $f$ is not analytic at $z=0$. But $f(z)=\sum_{n=0}^{\infty}{a_n}{z^n}$ at $A$, so for $f$ not to be analytic at $z=0$ we must have that $f(0) \neq a_0= \frac{1}{2 \pi i}\int_{\gamma} \frac {f(z)}{z}dz$, which I haven't been able to prove.

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  • $\begingroup$ I think any entire $f$ is a counterexample. $\endgroup$ – Nitin Uniyal Aug 22 '19 at 12:34
  • $\begingroup$ @user658409 Changing to $f\equiv 1$, you might just as well ask why $\frac zz$ isn't a counterexample. But that's an archetypal example of a function with a removable singularity. The answer is that the domain of $f$ has a puncture in it, and that puncture is a singularity. But there is a value you could assign to $f$ at that puncture to make the function analytic, and that makes the singularity removable. $\endgroup$ – Arthur Aug 22 '19 at 13:08
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    $\begingroup$ First, as already pointed out you're confused about the definition of "removable singularity". Less to the point, since $f$ is only defined for $0<|z|<2$ it's not analytic at the origin - it can't be, since $f(0)$ is undefined! It's an extension of $f$ that's analytic at $0$ (which of course is precisely what it means to say $f$ has a removable singularity...) $\endgroup$ – David C. Ullrich Aug 22 '19 at 15:25
  • $\begingroup$ @NitinUniyal What's the definition of "removable singularity"? $\endgroup$ – David C. Ullrich Aug 22 '19 at 16:30
  • $\begingroup$ @David C.Ulrich..A singularity at $z=a$ is removable if there is no negative powers of $(z-a)$ in Laurent's expansion of $f(z)$. $\endgroup$ – Nitin Uniyal Aug 22 '19 at 16:44
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Your interpretation of removable singularity is not the accepted one. You are not required to prove that $f$ is not analytic at $0$. For example $f(z)=0$ for all $z \neq 0$ has a removable singularity at $0$.

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To expand a bit on Kavi's answer: Say $A=\{z:0<z<2\}$ and assume $f\in H(A)$.

You're concerned that $f$ may be analytic at the origin. Of course Kavi is exactly right when he says "so what?". But the situation has a curious aspect.

First, speaking carefully,

(i) $f$ is analytic at the origin

is impossible, because strictly speaking $f(0)$ is undefined. The sensible, formally correct version of (i) is

(ii) It's possible to define $f(0)$ in such a way that $f$ becomes analytic at the origin.

And here's the reason I'm posting this: Not only is (ii) no problem regarding whether $f$ has a removable singularity, in fact (ii) is precisely the definition of "$f$ has a removable singularity"!

So when you say "I can only prove $f$ has a removable singularity or (i)" you're really saying "I can only prove $f$ has a removable singularity or (ii)", and by definition that means you're saying this:

I can only prove $f$ has a removable singularity or $f$ has a removable singularity.

Moral: You need to know the definitions.

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    $\begingroup$ for some reason I assumed $f$ defined in $A \cup \{ 0 \} $ and analytic in $A$ and thought that I could somehow infer $f(0)$ from the integral equality... so yeah, I should have paid more attention to what the question actually said. $\endgroup$ – Yagger Aug 22 '19 at 16:42
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You are right: we have that $b_n=0$ for all $n \ge 1$ and therefore

$f(z)=\sum_{n=0}^{\infty}{a_n}{z^n}$ on $A$. Now define $g(z):= \sum_{n=0}^{\infty}{a_n}{z^n}$ for $|z|<2.$ Then $g$ is holomorphic and $g$ is a holomorphic continuation of $f$ on the disc $|z|<2.$

Conclusion: $f$ has a removable singularity at $z=0.$

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  • $\begingroup$ @user658409 Huh? If you know what a removable singularity is then it's utterly obvious that the function $f=0$ has a removable singularity. If you don't know what a removable singularity is you should look it up. $\endgroup$ – David C. Ullrich Aug 22 '19 at 15:50
  • $\begingroup$ Of course this exactly correct, but it seems sort of beside the point. The OP appears to have got this far on his own; what you say here says nothing about what he's confused about, that is "Now to prove that $f$ has indeed a removable singularity one must prove that $f$ is not analytic at $z=0$. " $\endgroup$ – David C. Ullrich Aug 22 '19 at 16:14

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