In my research, I came across a 1-cocycle in the following group cohomology complex:

Let $\Lambda_\mathbb{Z}$ be a lattice (i.e. isomorphic to $\mathbb{Z}^n)$; let $\Lambda_\mathbb{C} = \Lambda_\mathbb{Z} \otimes_\mathbb{Z} \mathbb{C}$; let $\mathbb{C}(\Lambda_\mathbb{C})$ be the rational functions on $\Lambda_\mathbb{C}$, and let the $\times$ denote the multiplicative group.

Let $\Lambda_\mathbb{Z}$ act on $\mathbb{C}(\Lambda_\mathbb{C})$ by translation. Then the bar resolution takes the form $\mathbb{C}(\Lambda_\mathbb{C})^\times \rightarrow Fun(\Lambda_\mathbb{Z}, \mathbb{C}(\Lambda_\mathbb{C})^\times) \rightarrow Fun(\Lambda_\mathbb{Z}^2, \mathbb{C}(\Lambda_\mathbb{C})^\times) \rightarrow \ldots$

where $Fun$ refers to arbitrary functions.

Is anything known about the cohomology of this complex? I think I've figured out the first cohomology group (the group is not easy to explain, so I'd rather not type it out if there is an immediate answer), and the zeroth cohomology group is clearly $\mathbb{C}$ (as the 0-cocycles are constant functions).

For a more general question, is anything known about cohomology with coefficients in the multiplicative group of a field?

Edited to add: I didn't know Hilbert's theorem 90, which says that if we have a Galois extension $L/K$ with Galois group $G$, then $H^1(G, L^\times)$ is trivial. That doesn't seem to apply to my original question, but answers the more general one.

Edit 2: In the case $n = 1$, I have a proof that the 2nd cohomology group is trivial, which I think generalizes to show that all higher cohomology groups are trivial. In that case, the 1st cohomology group is $\mathbb{C}(\mathbb{C}^\times)^\times$.

Edit 3: One thing that helps here is decomposing $\mathbb{C}(\Lambda_\mathbb{C})^\times)$ into $\mathbb{Z}[\Lambda_\mathbb{Z}]$-modules; unique factorization gives us that it decomposes as $\oplus_{O} \oplus_{p \in O} \mathbb{Z}p$ where $p$ ranges over irreducible polynomials, and $O$ are the orbits of irreducibles; the summands of the decomposition are $\oplus_{p \in O} \mathbb{Z}p$.

Let $C_p = \{c \in \Lambda_\mathbb{C}|T_c(p) = p\}$; because $p$ is a polynomial and polynomials aren't periodic, $C_p$ is a subspace. Therefore $C_p \cap \Lambda_\mathbb{Z}$ is a summand of $\Lambda_\mathbb{Z}$. We can set $\Lambda_\mathbb{Z} = \mathbb{Z}^n \supset C_p \cap \Lambda_\mathbb{Z} = \mathbb{Z}^m \oplus \{0\}$; then $\oplus_{p \in O} \mathbb{Z}p \simeq \mathbb{Z}^n/(\mathbb{Z}^m \oplus \{0\})$

I think I have a finite-step free resolution, which should then allow determining the group cohomology on each of the individual representations easily.

Edit 4: I think the group homology may be informative; is there a universal coefficient theorem for group cohomology that works with an action of the group on the coefficients? I've found one where the action is trivial, but not generally.