Cauchy's Theorem by Differential Geometry

Question

Is there a prove of Cauchy's theorem footing on the topology of the complex plane (homotopy, differential forms, etc.)?

More specific consider a differentiable Banach space valued complex function. Prove that the integral along every closed differentiable curve vanishes: $$\oint\omega=0\text{ for all }\omega\in\Omega(\mathbb{C})$$

(Though it is precisely what analysis is about I just don't like the usual proof by shrinking triangles since it is kind of technical so I hope there might be one showing why complex differentiable is so much more than just its real analogue.)

Answer 1

Given a holomorphic function: $f:A\to\mathbb{C}$

Consider the real manifold: $M:=D\subseteq A\subseteq\mathbb{R}^2:\quad\partial D=S$
and the real one forms: $\omega:=udx-vdy,\tilde{\omega}:=ydx+xdy:\quad u=\mathrm{Re}f,v=\mathrm{Im}f$

Then we have: $$f\text{ holomorphic}\Rightarrow u,v\text{ C-R}\Rightarrow\omega,\tilde{\omega}\text{ closed}\Rightarrow\int_S\omega,\tilde{\omega}=0\Rightarrow\oint\omega,\tilde{\omega}=0\Rightarrow\oint f=0$$

...where the domain is simply connected so that any path is homotopic to eachother...
...and the manifold is chosen compact so that Stoke's theorem is applicable...
...and simply connected so its boundary is a closed curve.

Answer 2

Yes, if we're in the smooth category, holomorphicity of $f$ is equivalent to the fact that the $1$-form $\omega = f(z)\,dz$ is closed. And Cauchy's Theorem is just Green/Stokes's Theorem. Homotopy-invariance is just the general topological fact that if $f$ and $g$ are homotopic maps $S^1\to D$ (for some domain $D\subset\Bbb C$), then $\int_{S^1} f^*\omega = \int_{S^1} g^*\omega$ for any closed $1$-form $\omega$.

EDIT: The best way to see my first statement is to do the exercise of turning $$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$$ into $$df = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial\bar z}d\bar z$$ by using $dz = dx+i\,dy$, $d\bar z = dx - i\,dy$. You'll easily check that the Cauchy-Riemann equations are equivalent to $\dfrac{\partial f}{\partial\bar z} = 0$. Then, if we have the $1$-form $\omega = f(z)\,dz$, we see that $d\omega = \dfrac{\partial f}{\partial\bar z}d\bar z\wedge dz = 0$ if and only if $f$ is holomorphic.