No try with $z_0=0$, $f(z)= \frac1{z(z-1)}$ and $C$ the circle of radius $2$. Then for all $n\ge 2$, $$ \int_C f(z)z^{n-1}dz= 2i\pi $$
If you meant that $0$ is the only pole of $f$ and $\Omega$ is simply connected then yes sure, for $n$ large enough $f(z)z^{n-1}$ is analytic on $\Omega$ and its integral on a closed-contour contained in $\Omega$ is $0$.
$f$ is homomorphic in $D = \Bbb C \setminus \{ z_1, z_2, z_3, \ldots \}$, where $(z_k)$ is a (finite or infinite) sequence of complex numbers without accumulation point in $\Bbb C$, and $f$ has a pole at each $z_k$.
We can assume that the $z_k$ are sorted with respect to increasing modulus:
$$
0 \le |z_1| \le |z_2| \le |z_3| \le \ldots
$$
In the case of infinite poles we necessarily have $\lim_{k \to \infty} z_k = \infty$.
If $|z_k| < |z_{k+1}|$ for some $k$ then we define $A_k$ as the annulus
$$
A_k = \{ z : |z_k| < |z| < |z_{k+1}| \} \, .
$$
In the case of finitely many poles $z_1, \ldots, z_N$ we additionally define
$$
A_N = \{ z : |z_N| < |z|\} \, .
$$
The function $f$ is holomorphic in the each annulus $A_k$
and therefore can be developed into a Laurent series
$$ \tag{1}
f(z) = \sum_{n=-\infty}^\infty a_n^{(k)} z^n \quad \text{for } z \in A_k \, .
$$
Now assume conversely that $f$ can be developed into a Laurent series
$$ \tag{2}
f(z) = \sum_{n=-\infty}^\infty b_n z^n
$$
which converges in some annulus $B = \{ z : r < |z| < R \}$. Then $f$ is holomorphic in $B$, i.e. it does not contain any of the poles. Let
$$
m = \max \{ k : |z_k| \le r \} \, .
$$
There are two possible cases: $f$ has finitely many poles and $m = N$, or $|z_m| \le r < R \le |z_{m+1}|$. In both cases is $B \subseteq A_m$. Then $(1)$ with $k=m$ is a Laurent series for $f$ in $B$ and since Laurent series are unique, we necessarily have
$$
b_n = a_n^{(m)} \quad \text{for } n=0, 1, 2, \ldots \, .
$$
This shows that
- $f$ has a Laurent series expansion in each annulus $A_k$ determined by two “consecutive” poles with different modulus (and, in the case of finitely many poles, a Laurent series expansion in the exterior $A_N$ of a disk), and
- Any Laurent series expansion of $f$ in an annulus with center at the origin is the restriction of one of these expansions in some $A_k$.
Best Answer
No. In fact there are various ways to show there exist polynomials $p_n$ with $$\sum p_n(s)=e^{1/s}\quad(\Re s>0).$$