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.
Best Answer
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 $0$.