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 $\mathbb{C}(\mathbb{C}^\times)^\times$.