Meromorphic function with $\oint z^n f(z) dz=0$ is holomorphic? Let $f$ be a meromorphic function on $\mathbb C$ with finitely many poles, located in $|z|<R/2$ for some large fixed $R$. Furthermore, suppose that
$$\oint_{|z|=R} z^nf(z) dz=0$$
for all $n\in \mathbb Z$. What can we infer from this condition? I naively expect that $f$ is holomorphic. Is it true?
 A: Yes, it is true. We can formulate the statement slightly more general:

Let $f$ be meromorphic in the disk $B_r(0)$ without poles on the circle $|z| = R$, $0 < R < r$. If $\oint_{|z|=R} z^nf(z) \, dz=0 $ for all integers $n$ then $f$ is identically zero.

Proof: $f$ has only finitely many poles in $B_R(0)$. Let $p$ be a polynomial with zeros exactly at those poles, counted with multiplicity. Then
$$
 g(z) = p(z) f(z)
$$
is holomorphic in a neighborhood of $\overline{B_R(0)}$ and
$$
g^{(k)}(0) = \frac{k!}{2 \pi i} \oint_{|z|=R} \frac{g(z)}{z^{k+1}} \, dz = \frac{k!}{2 \pi i}\oint_{|z|=R} \frac{p(z)}{z^{k+1}}f(z)\,  dz=0
$$
for all $k \ge 0$. So $g$ is identically zero, and it follows that $f$ is identically zero as well.
A: Since $f$ has is meromorphic with all its poles inside of some disc, it means $f$ is holomorphic outside of that disc, i.e $f$ is holomorphic on an annulus, and thus it admits a Laurent expansion centered at the origin
\begin{align}
f(z)&=\sum_{m\in\Bbb{Z}}a_mz^m\quad\left(R/2<|z|<\infty\right).
\end{align}
(where the series converges uniformly and absolutely on compact subsets).
Now, for each $n\in\Bbb{Z}$ we have
\begin{align}
z^nf(z)&=\sum_{m\in\Bbb{Z}}a_mz^{m+n}=\frac{a_{-(n+1)}}{z}+\sum_{m\neq -(n+1)}a_mz^{m+n}
\end{align}
The second term has an explicit anti-derivative (just integrate term-by term), so its integral over a loop vanishes. Hence,
\begin{align}
\int_{|z|=R}z^nf(z)\,dz&=\int_{|z|=R}\frac{a_{-(n+1)}}{z}\,dz=2\pi i a_{-(n+1)}.
\end{align}
If this integral vanishes for all $n\in\Bbb{Z}$, then each coefficient $a_m$ of the Laurent expansion is zero, hence $f(z)=0$ for all $R/2<|z|<\infty$, and thus by uniqueness of analytic continuation, $f=0$.
If you assume that the integral vanishes for all $n\in\Bbb{Z}_{\geq 0}$, then it follows that $a_{-1}=a_{-2}=\cdots =0$, i.e the negative part of the Laurent expansion vanishes; so $f(z)=\sum_{m=0}^{\infty}a_mz^m$ for all $R/2<|z|<\infty$; this series thus also converges for all $z\in\Bbb{C}$, so by uniqueness of analytic continuation, we have $f(z)=\sum_{m=0}^{\infty}a_mz^m$, so $f$ is an entire function.
So, by specifying the $n$ for which the integral vanishes, you can characterize the terms in the Laurent expansion which vanish, and thus deduce properties of $f$.
