Proof without words of $\oint zdz = 0$ and $\oint dz/z = 2\pi i$

Solution 1:

I think your question can be somehow answered according to an interesting idea, Pólya vector field, which I think is a better and more reasonable construction than your proof because it preserves some important properties of the function.

Construct a vector field on the complex plane based on a complex-valued (analytic) function $f(z)=u+iv$: $$\left\langle \Re (f),-\Im (f) \right\rangle$$ (Your construction is actually $\left\langle \Re (f),\Im (f) \right\rangle$)This can be denoted by $\bar{f}$ for convenience. The most important property of $f$ preserved by this field is the analyticity of $f$, which is ensured by Cauchy-Riemann Equations (let $z=x+iy$): $$\begin{align} \frac{\partial u}{\partial x}&=\frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}&=-\frac{\partial v}{\partial x} \end{align}$$ By moving the fractions on the right-hand side to the left-hand side, one will get $$\nabla\times\bar{f}=\nabla\cdot\bar{f}=0$$

Now, if you write $f(z)d_nz=(u+iv)(d_nx+id_ny)=(ud_nx-vd_ny)+i(ud_ny+vd_nx)$, you will find that $$\sum_{k=0}^{n-1}f(z)d_nz=\sum_{k=0}^{n-1}(ud_nx-vd_ny)+i\sum_{k=0}^{n-1}(ud_ny+vd_nx)$$ Let $n\to\infty$, we have ($C$ is the set of $z_k$'s) $$\int_{C} f(z)dz=\int\bar{f}\cdot\mathbf{T}ds+\int\bar{f}\cdot\mathbf{N}ds=\text{Circluation}\bar{f}+i\cdot\text{Flux}\bar{f}$$ where $\mathbf{T}$ and $\mathbf{N}$ are unit tangent and unit norm respectively.

Here is the answer to your question: if you plot the Pólya vector field of $f(z)=\frac{1}{z}$, $$\bar{f}=\left\langle \frac{\Re(z)}{|z|^2},\frac{\Im(z)}{|z|^2}\right\rangle$$ you will find that it is a SOURCE VECTOR FIELD because there is a pole at $z=0$!!!

Easily, you can get the integral on a circular path around a source, which is $$\int_C \frac{1}{z}dz=\text{Circluation}\bar{f}+i\cdot\text{Flux}\bar{f}=0+i\cdot 2\pi=2\pi i$$