So we know that Hodge duality works this way
$$⋆(dx^{i_1} \wedge ... \wedge dx^{i_p})= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$
where $p$ represents the $p$ in $p$-form and $n$ is the dimensional number.
My question is: How does hodge duality work on the imaginary number $i$ on one hand and on partial derivative like $\partial_x\alpha$ let us say on the other where $\alpha$ is a complex function.
If you are working on a complex manifold with a Hermitian metric, then the Hodge star operator should be taken antilinear: $\star(\alpha + i\beta) = \star\alpha - i\star\beta$. If you work on a real manifold without a complex structure by itself, and you extend the scalars to the complex numbers, it may be that there is no harm in taking it to be linear.
The Hodge star operation acts on differential forms. Numbers, real or complex, transform as 0-forms. The Hodge dual of a 0-form will result in something proportional to the volume form of the manifold. In detail, for a $d$-dimensional manifold,
$\star 1 = \text{vol}_d = \sqrt{|g|}dx^1 \wedge ... \wedge dx^d$,
and the Hodge operation commutes with multiplication by complex numbers.
A partial derivative like $\partial_x \alpha$ is also a complex number (in general), so the above applies. However, given a scalar function such as $\alpha$, a 1-form can be obtained as
$d\alpha = (\partial_i \alpha) dx^i$.
Then the Hodge star acts on $d\alpha$ as it would on any 1-form.
