Prove a function has a removable singularity at $z=0$.

Question

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?

Answer 1

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.

Answer 2

Hints:

Suppose we develope a Laurent series for $\;f\;$ in $\;0<|z|<r\;,\;\;r\in\Bbb R^+\;$ :

$$f(z)=\ldots+\frac{a_{-n-1}}{z^{n+1}}+\frac{a_{-n}}{z^n}+\mathcal O(z^{-n+1})\implies z^{n-1}f(z)=\ldots\frac{a_{-n-1}}{z^2}+\frac{a_{-n}}z+\mathcal O(1)$$

Now just remember that

$$\oint\limits_{|z|=1|}\frac{dz}{z^n}=\begin{cases}2\pi i&,\;\;n=1\\{}\\0&,\;\;n\neq 1\end{cases}$$