Assigning a value to a removable singularity

Question

I know that if $ f $ is analytic on $ \mathbb{C}\setminus\{0\} $, with an isolated singularity at zero and $$ \oint_{\left|z\right|=1}z^{n}f(z)\ dz=0\ \ \forall n\in\mathbb{N}\cup\{0\}, $$ then $ f $ has a removable singularity at zero. I have shown this to be true by the considering principal part of the corresponding Laurent series, but is it possible to find the value of $ f(0) $ which extends $ f $ to an entire function? Do we have enough information to do this? I have tried to assign a value to $ f(0) $ using the Cauchy Integral Formula but have not gotten far with this approach.

Answer 1

No, we don't have enough information. For instance, $f(z)$ could be $\frac zz$, or it could be $\frac{2z}z$, and there is no way for us to tell the difference from those integrals. One way to get singularity-lifting value of $f(0)$ is by calculating $$ \frac{1}{2\pi i}\oint_{|z|=1}\frac{f(z)}zdz$$assuming the integral goes counterclockwise.