First de Rham cohomology group of $\mathbb C\setminus \mathbb Z$

Question

I'm wondering what $H^1 (\mathbb C\setminus \mathbb Z)$ could be. I have an idea but I did not succeeded in proving it properly. I think $$H^1 (\mathbb C\setminus \mathbb Z )\simeq \mathbb R \langle \mathbb Z \rangle=\{\displaystyle \sum_{i=0}^m r_i\cdot n_i | \ m\in\mathbb N,\ r_i\in\mathbb R,\ n_i\in \mathbb Z\ \forall i\}$$ but I can't show the map $[\omega]\in H^1 (\mathbb C\setminus \mathbb Z) \longmapsto (\int_{C_{\frac {1}{2}}(n)} \omega )_{n\in \mathbb Z} \in \mathbb R \langle \mathbb Z \rangle$ is well defined, where for $n\in \mathbb Z $, $C_\frac {1}{2} (n) $ is the circle centered in n, radius $\frac {1}{2} $, parameterized in a standard way

Answer 1

I think what you have there is the singular homology group $H_1(X,\Bbb R)$ with $X=\Bbb C\setminus \Bbb Z$. This is a real vector space $V$ of countable dimension. I think the de Rham cohomology (also singular cohomology) group will be the dual of this $H^1(X,\Bbb R )\cong\text{Hom}_{\Bbb R}(V,\Bbb R)$ which will be a vector space of uncountable dimension.