The unitary group can be seen as an algebraic group over a field $k$ with respect to a degree-2 separable field extension $K/k$ whose points over a $k$-algebra $R$ are given by $$\text{U}(n,K/k)(R):=\{A \in \text{GL}(n,K\otimes_kR):A^*A=I\}$$ where conjugation is given by the unique $k$-automorphism of $K$ which is an involution and fixes $k$.
When we set $k = \mathbb{R}$ and $K = \mathbb{C}$, this says that the real and complex points are given by $$\begin{aligned}\text{U}(n,\mathbb{C}/\mathbb{R})(\mathbb{R})&=\text{U}(n) \\ \text{U}(n,\mathbb{C}/\mathbb{R})(\mathbb{C})&=\text{GL}(n,\mathbb{C})\end{aligned}$$ (see the Wikipedia article for Unitary group).
It's obvious that the real points are indeed given by $\text{U}(n)$, but how can I see that the complex points are given by $\text{GL}(n,\mathbb{C})$? In other words, what is the isomorphism $$\{A \in \text{GL}(n,\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}):A^*A=I\} \longleftrightarrow \text{GL}(n,\mathbb{C})$$ supposed to be?
Very confusing. I think they meant to consider the algebraic variety defined over $\Bbb{R}$ $$V = \{ (C,D)\in M_n\times M_n, C^\top C + D^\top D = I, C^\top D-D^\top C = 0\}$$ with the algebraic group law $$(A,B)(C,D) = (AC-BD,AD+BC)$$ and to show that over $\Bbb{C}$ this is isomorphic to $GL_n$, the isomorphism being $$(C,D)\mapsto C+iD$$ the inverse being $$A \mapsto (\frac{A+A^{-\top}}2, \frac{A-A^{-\top}}{2i})$$