Laurent polynomials are generated by $\{x_1,x_{1}^{-1},...,x_n,x_{n}^{-1}\}$. Can we view this as $\mathbf{C}[y_1,y_2,....,y_{2n}]$ where $y_1 = x_1,y_2 = x_{1}^{-1}$ etc.
I suppose we must also modify the corresponding $\mathbf{A}^{2n}$ by removing the $(a_1,a_2,...,a_{2n})$ where $a_i = 0$ for $i$ even, or simply not require that we can plug any $a \in \mathbf{A}^{2n}$ into $f \in \mathbf{C}[y_1,y_2,....,y_{2n}]$.
I am very new to this subject so please let me know if I have misunderstood something (such as the relation between $\mathbf{A}^{n}$ and $\mathbf{C}[x_1,x_2,....,x_{n}]$ in general.
For each $i \in \{1, \dots, n \}$, define $y_i = x_i^{-1}$, and observe that $x_i$ and $y_i$ obey the relation $x_i y_i = 1$.
So your ring of Laurent polynomials is isomorphic to $$\mathbb C [x_1, y_1, \dots, x_n, y_n ] / (x_1 y_1 - 1, \ \dots \ , x_n y_n - 1).$$
This is coordinate ring for the affine variety $$V(x_1 y_1 - 1, \ \dots \ , x_n y_n - 1) \subset \mathbb A^{2n},$$ which is isomorphic to the quasi-affine variety $$ \{ (y_1, \dots, y_n) \in \mathbb A^{n} \ | \ y_1 \neq 0, \ \dots \ , y_n \neq 0 \} \subset \mathbb A^n.$$