Suppose that $V$ is a finitely generated $\mathbb Z$-module. A Hodge structure of wight $k$ on $V$ is a decomposition of the complexification of $V$ into complex vector spaces $V^{p,q}$ such that $\overline{V^{p,q}}=V^{q,p}$. More precisely:
$$ V\otimes_\mathbb Z \mathbb C = \bigoplus_{p+q=k} V^{p,q} . $$
I'm struggling to see how it is possible that $\overline{V^{p,q}}=V^{q,p}$ without getting $V^{p,q}=V^{q,p}$.
My reasoning is the following: even if the $V^{p,q}$ are taken as external, abstract, vector spaces, hence having abstract complex-conujugations, then in the end they must be compatible with the canonical conjugation on $V\otimes \mathbb C$, which means (if I'm not wrong) that we can consider the $V^{p,q}$ directly as complex subspaces.
Now, in this case, I cannot imagine how $\overline{V^{p,q}}$ could be different from $V^{p,q}$ itself, for if $v\otimes z$ is an element in it, then $$ \overline{v\otimes z}=v\otimes\bar z = \frac{\bar z}{z}(v\otimes z ), $$ which means that both $v\otimes z$ and $\overline{v\otimes z}$ belong to the same (complex) vector space. Hence, as sets $V^{p,q}=\overline{V^{p,q}}$.
I'm aware of the related question Definition of complex conjugate in complex vector space and the ones mentioned on there, as well as the book Linear algebra and geometry by Kostrikin & Manin (1989) and the notes of Keith Conrad about complexification (availabe here).
PS: In the case of manifolds, the complex-conjugation used carries $dz$ into $d\bar z$ and, if I'm not mistaken, and there are no possibility of relating both $dz$ and $d\bar z$ algebraically, but with the above complex-conjugation there is, so this example would not be valid.
