# Residues and existence of complex antiderivative I know that there already is a similar question ou

Residues and existence of complex antiderivative
I know that there already is a similar question out there but I am not really satisfied with that answer. I think it is too complicated. Assume that $G\subseteq \mathbb{C}$ is open and simply connected, $P\subseteq \mathbb{C}$ has finite cardinality and $f:G\setminus P\to \mathbb{C}$ is holomorphic. Then
f has an antiderivative on its domain $\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$ .
Can I argue the following way? The crucial part is the second implication.
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}$: Since f has an antiderivative, all integrals over closed rectifiable curves vanish. If I let $p\in P$ be arbitrary and $\epsilon >0$ so small that $\mathrm{\partial }{B}_{\epsilon }\left(p\right)\subseteq G\setminus P$, then according to the residue theorem:

$\phantom{\rule{thickmathspace}{0ex}}⟸\phantom{\rule{thickmathspace}{0ex}}$: For any closed rectifiable curve $\gamma$ we have:

Therefore the existence of a primitive.
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Alexia Hart
Step 1
Under the given conditions on G, P, and f, the following three statements are equivalent:
(a) f has an antiderivative in G.
(b) ${\mathrm{Res}}_{p}\left(f\right)=0$ for all $p\in P$.
(c) ${\int }_{\gamma }f\left(z\right)\phantom{\rule{thinmathspace}{0ex}}dz=0$ for all closed rectifiable curves $\gamma$ in G.
Step 2
What you have demonstrated is that $\left(a\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(b\right)$ and that $\left(b\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(c\right)$.
It remains to show that $\left(c\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(a\right)$, and that can be done by showing that $F\left(z\right)={\int }_{{z}_{0}}^{z}f\left(w\right)\phantom{\rule{thinmathspace}{0ex}}dw$ (which is well-defined under the condition (c)) satisfies ${F}^{\prime }=f$