Let $f$ be a holomorphic function on $\mathbb{C}\smallsetminus \{0\}$. Suppose $\int_{|z|=1}z^nf(z)\,dz=0$ for any $n=0,1,2,\ldots$. Prove that $f$ has a removable singularity at $z=0$. How to prove?
If $\lim_{z\to 0}z^nf(z)=0$ then I use the Laurent seris and solve it. But how to deal with the integral?
Best Answer
If a function $f$ is analytic in $\mathbb C\smallsetminus\{0\}$ then it is expressed as $$ f(z)=\sum_{n=-\infty}^\infty a_nz^n, $$ and it is readily proved that $$ a_n=\frac{1}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}. $$ Hence, in your example $$ a_n=0, \quad \text{for all $n<0$}, $$ and consequently, the singularity at $z=0$ is removable.