Holomorphic functions don't necessarily have antiderivatives/primitives. For example, $\frac{1}{z}$ is holomorphic on the punctured plane $V:= \mathbb{C}- \{ 0\},$ but it does not have any antiderivative on this domain. My guess (intuition) is that every holomorphic function on the punctured plane $V$ has this "bad" part $\frac{1}{z}$, maybe a multiple of $\frac{1}{z}.$ If we can eliminate this bad portion from the given holomorphic function, the remaining portion might have antiderivative. Is my guess true? My question is rewritten as the following:
If $f: V \rightarrow \mathbb{C}$ is analytic, then does there exists an analytic function $g: V \rightarrow \mathbb{C}$ that has antiderivative such that $ f(z) - g(z) = \alpha \frac{1}{z}$ for some $\alpha \in \mathbb{C}$ ? This is just my guess. I might be totally wrong. Could you help me with some explnation or counter examples?
Yes, that is correct. By Laurent's theorem, there are numbers $(a_n)_{n\in\Bbb Z}$ such that, for every $z\in V$, $f(z)=\sum_{n=-\infty}^\infty a_nz^n$. And $f(z)-\frac{a_1}z$ has an antiderivative, which is$$\sum_{n\in\Bbb Z\setminus\{-1\}}\frac{a_n}{n+1}z^{n+1}.$$