If $\int_{|z|=1} z^n f(z)dz=0 \ \forall n = 0, 1, 2, ...$, then $f$ has a removable singularity at $z=0$

Question

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.

Answer 1

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 $0$.

Answer 2

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 $z=0.$

Answer 3

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.