When $n=1$, we see that $U(1)$ is defined by the equation $z\bar z=1$, hence $a^2+b^2=1$ for $z=a+bi$. Taking complex $a,b$ we see that the solutions are nonzero complex points, hence $U(1)$ is isomorphic to $GL(1)$ over $\mathbb C.$
When $n=2$ this is less straightforward but still believable from examining the equation $$\begin{pmatrix}u&v\\w&z\end{pmatrix}\begin{pmatrix}\bar u&\bar w\\\bar v&\bar z\end{pmatrix}=\begin{pmatrix}u\bar u+v\bar v&u\bar w+v\bar z\\\bar uw+\bar vz&w\bar w+z\bar z\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$ which leads to \begin{align} u_1^2+u_2^2+v_1^2+v_2=w_1^2+w_2^2+z_1^2+z_2^2=1\\ u_1w_1-u_2w_2+v_1z_1-v_2z_2=u_1w_2+u_2w_1+v_1z_2+v_2z_1=0 \end{align} As I understand it one can prove that $U(n)$ and $GL(n)$ are real forms for general $n$ by showing the the Lie algebras $\mathfrak u(n)$ and $\mathfrak{gl}(n)$ are real forms of each other. Is there a more direct proof of this, or is this the most 'natural' argument?