$$ \mathbf{n} = \frac{1}{\sqrt 2}(\mathbf u + i \mathbf v)$$ Here is a $\mathbf n$ is a complex vector ($\mathbf n \in \mathbb C^3$). $\mathbf v$ and $\mathbf u$ both are real $3$ vectors.
I want to compute the line integral $$\Im \oint \mathbf{n}^* \cdot d \mathbf n $$
Simplifying the integrand we find \begin{align} \Im (\mathbf n^* \cdot d\mathbf n) =\frac{1}{2}\Im[(\mathbf u - i \mathbf v)\cdot(d\mathbf u + i d\mathbf v)] \\ =\frac{1}{2} \Im(\mathbf u . d \mathbf u + \mathbf v \cdot d\mathbf v + i \mathbf u\cdot d\mathbf v -i \mathbf v \cdot d \mathbf u) \\ = \frac{1}{2}( \mathbf u\cdot d\mathbf v - \mathbf v \cdot d\mathbf u) \\ \end{align} The vectors $\mathbf v$ and $\mathbf u$ are linearly independent vectors, so we can write the differential of the last equation \begin{align} \frac{1}{2}d( \mathbf u\cdot d\mathbf v - \mathbf v \cdot d\mathbf u) = d\mathbf u \wedge d\mathbf v = d\mathbf n^* \wedge\cdot d\mathbf n \end{align} Using Stokes theorem $$ \Im \oint_{\partial C} \mathbf{n}^* \cdot d \mathbf n =\Im \iint_{C}d\mathbf n^* \wedge \cdot d\mathbf n $$ Here $C$ is a simply connected region with boundary $\partial C$.
I know very basic of exterior algebra. The book which gives a very brief review (Mathematical methods for physicists - Arfken) just tells how to deal with differentials and all the examples I worked out have 0-form coefficients. This particular problem got me thinking that the Stokes theorem should work out for cases when the coefficients of differentials are not a 0-form. So in this case, the 3 vectors are the coefficients of the differentials and differentials are also 3 vectors.
From that line of thought if we think $\mathbf n$ and $\mathbf n^*$ are associated with orthogonal directions in some space then the application of Stoke's theorem is just a one-liner.
Is this kind of proof permissible? or one needs to do more work to arrive at the proof.
I found this integral here -
Berry, M. V. (1989). The quantum phase, five years after. Geometric phases in physics, 7-28.