Stokes' Theorem and path integrals in $\mathbb{C}$

Question

I have seen very short proofs of the Lemma of Goursat and Cauchy's Theorem using Stokes' Theorem. I have learnt Stokes' Theorem in the setting of (not complex) smooth manifolds as described in Wikipedia.

Looking at the proofs of these I always do not completely understand the right hand side of the following equality:

$$\int_{\partial D}f(z) \rm{dz}=\int_{\partial D} \rm{i} u(x,y) + v(x,y) (dx+\rm{i}dy),$$ where $D$ is a compact surface (such as a triangle or a circle,...) in $\mathbb{C}$ and $f$ is a holomorphic function with $u$ as imaginary and $v$ as the real part of the function (i.e. $f=\rm{i}u+v$) and defined on a open subset containing the topological closure of $D$.

Can one rewrite this as an integral of a differential form on a manifold (without appearance of the imaginary unit)?

Strictly speaking, I know the integration of real-valued function with domains in $\mathbb{R}^n$ and integrals of complex-valued functions, which are defined in the real setting (i.e. as $\int_{A} u(x,y) d(x,y) + i\int_A v(x,y) d(x,y)$ for a complex-valued function $f=u+\rm{i}v$ with domain $\mathbb{R}^n$ and might using the identification $\mathbb{R}^2 \cong \mathbb{C}$ to allow subsets of $\mathbb{C}$ as domains). As well as, the integration of differential forms on (not complex) manifolds.

In all this integrals there are (suddenly) no imaginary unit in integral expression. Indeed even the integral of complex-valued functions boils down to two integrals of real-valued functions. Therefore, the right hand side of the above equality seems to be heuristic expression.

Answer 1

The contour integral $\int_{\partial D} f(z) \, \mathrm{d}z$ when written out becomes $$\int_{\partial D} (u+iv) \cdot (\mathrm{d}x + i \mathrm{d}y) = \Big(\int_{\partial D} u \, \mathrm{d}x - v \, \mathrm{d}y\Big) + i \Big(\int_{\partial D} v \, \mathrm{d}x + u \, \mathrm{d}y\Big).$$ This is not heuristic - complex integrals are defined to make this true.

When $u$ and $v$ are defined inside $D$, and the boundary of $D$ is smooth, Green's theorem implies that both integrals are zero since $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}.$

Answer 2

Agree with @user343900's answer.

I would add that when one moves from $\Bbb{C}$ to $\Bbb{R}^2$, the term $\mathrm{i}u + v$ becomes the vector $(v,u)$ and and $\mathrm{d}x + \mathrm{i}\mathrm{d}y$ becomes the vector $\mathrm{d}s = (\mathrm{d}x,\mathrm{d}y)$, i.e. becomes a line element (as is required by the integral being along the boundary of the region). This isn't a "heuristic" -- this is one-to-one translation along the identification of $\Bbb{C}$ and $\Bbb{R}^2$.