All complex vector bundles over the circle are trivial

Suppose first that $n>0$. Every $A\in {\mathrm{GL}}_n(\mathbb{C})$ has polar form $A=UP$ with $U$ unitary and $P$ positive definite. The path $U((1-t)P+tI)$ joins $A$ to $U$ through invertible matrices. By the finite-dimensional spectral theorem, $U=W\mathrm{diag}(e^{i{\theta }_1},\dots ,e^{i{\theta }_n})W^{\ast }$ for real angles ${\theta }_j$, and $W\mathrm{diag}(e^{i(1-t){\theta }_1},\dots ,e^{i(1-t){\theta }_n})W^{\ast }$ joins $U$ to $I$. Hence ${\mathrm{GL}}_n(\mathbb{C})$ is path connected.

The two values of $g$ can therefore be joined independently to $I$, producing a homotopy $S^0\times I\to {\mathrm{GL}}_n(\mathbb{C})$ from $g$ to the constant identity map. By [F1], $E$ is isomorphic to the identity-clutched bundle, which is $S^1\times {\mathbb{C}}^n$.

If $n=0$, then ${\mathrm{GL}}_0(\mathbb{C})$ is a point and the same conclusion is forced. The real argument fails at step 1.1 because ${\mathrm{GL}}_1(\mathbb{R})={\mathbb{R}}^{\times }$ has two components; the clutching values in different components give the Möbius line. No choice principle is used.