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 $\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ ${\mathrm{R}\mathrm{e}\mathrm{s}}_{p}(f)=0\text{}\mathrm{\forall}p\in P$.

Can I argue the following way? The crucial part is the second implication.

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}$: 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 $\mathrm{\partial}{B}_{\epsilon}(p)\subseteq G\setminus P$, then according to the residue theorem:

$0={\int}_{\mathrm{\partial}{B}_{\epsilon}(p)}f(z)\text{}\mathrm{d}z=2\pi i{\mathrm{R}\mathrm{e}\mathrm{s}}_{p}(f)$

$\phantom{\rule{thickmathspace}{0ex}}\u27f8\phantom{\rule{thickmathspace}{0ex}}$: For any closed rectifiable curve $\gamma $ we have:

${\int}_{\gamma}f(z)\text{}\mathrm{d}z=\sum _{p\in P}{I}_{p}(\gamma ){\mathrm{R}\mathrm{e}\mathrm{s}}_{p}(f)=0$

Therefore the existence of a primitive.

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 $\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ ${\mathrm{R}\mathrm{e}\mathrm{s}}_{p}(f)=0\text{}\mathrm{\forall}p\in P$.

Can I argue the following way? The crucial part is the second implication.

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}$: 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 $\mathrm{\partial}{B}_{\epsilon}(p)\subseteq G\setminus P$, then according to the residue theorem:

$0={\int}_{\mathrm{\partial}{B}_{\epsilon}(p)}f(z)\text{}\mathrm{d}z=2\pi i{\mathrm{R}\mathrm{e}\mathrm{s}}_{p}(f)$

$\phantom{\rule{thickmathspace}{0ex}}\u27f8\phantom{\rule{thickmathspace}{0ex}}$: For any closed rectifiable curve $\gamma $ we have:

${\int}_{\gamma}f(z)\text{}\mathrm{d}z=\sum _{p\in P}{I}_{p}(\gamma ){\mathrm{R}\mathrm{e}\mathrm{s}}_{p}(f)=0$

Therefore the existence of a primitive.