I think I have some misunderstandings regarding some basic concepts.
First, the question I'm dealing with is the following:
Let $f$ be analytic in $\{z ;|z|>1 \}$, and $\int_{|z|=2}f(z)dz=0$. Show that $f$ has a primitive in $\{z ;|z|>1 \}$
I know that if $f$ is analytic on this domain, then it has a Laurent expansion which is integrable term-by-term (where $|z| > 1$) hence a primitive exists. What do I need the other condition $\int_{|z|=2}f(z)dz=0$ for?
$\textbf{Claim}$: Suppose $f$ is analytic in a domain $\Omega$. Then, $f$ has a primitive in $\Omega$ iff $\int\limits_{\gamma}{}f=0$ for every simple closed curve $\gamma \in \Omega$.
In your case, every simple closed curve $\gamma$ in $\Omega=\{z:|z|>1\}$ is homotopic to either a contractible loop or a loop around $0$, which we can assume to be the circle $|z|=2$. Now, check that the claim holds.