Properties of Hodge Duality 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.
 A: 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.
A: 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.
