Integral of meromorphic function.

Question

Let $f(z)$ be meromorphic function in $\Omega\subset\mathbb{C}$ with a pole $z=z_0$. Then, is it true that there is a number $N$ such that
$$\sum_{n=1}^{\infty} \int_C f(z)(z-z_0)^{n-1}dz=\sum_{n=1}^{N}\int_C f(z)(z-z_0)^{n-1}dz$$
where $C$ is closed contour containing $z_0$. This integral and summation are principal part of coefficient of Laurent series. I guess, for any meromorphic function $f$, $$f(z)=(z-z_0)^N g(z)$$
for some $N\in \mathbb{N}$ and holomorphic function $g(z)$ in $\Omega$. But, I am failing it.

Answer 1

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 $0$.