I found this visual "proof" of $\oint zdz = 0$ and $\oint dz/z = 2\pi i$ quite compelling and first want to share it with you. But I have a real question, too, which I will ask at the end of this post, so please stay tuned.

Consider the unit circle $C_1$ in $\mathbb{C}$ with $n$ equally distributed numbers $z_k = e^{i2\pi k/n}$. Draw from each $z_k$ an arrow to $z_k + f(z_k)$ for $f(z) = z$ resp. $f(z) = 1/z$. To make the connection between $z$ and $1/z$ more comprehensible I plot the graphs $f_p(z) = pz + (1-p)/z$ for some values of $0 \leq p \leq 1$. Note that $f_1(z) = z$ and $f_0(z) = 1/z$.

Note that these plots are essentially **stream plots** a) restricted to the unit circle and b) displaying not only the direction of the vector field but also its magnitude.

Note that by $f_p(-z) = -f_p(z)$ we have $\sum_{k=0}^{n-1}f_p(z_k) = 0$ for all $p$ which can more easily be seen when drawing $f_p(z)$ from the origin:

_{Note the difference in color between $f_1(z) = z$ and $f_0(z) = 1/z$ which is due to different phases $\varphi = 2\pi k/n$ of the arguments.}

Now let's consider the factor $d_nz$, which is the secant (for $n\rightarrow \infty$ the tangent) vector of the circle at point $z$. Multiplying $f(z)$ by $d_nz$ shrinks and rotates $f(z)$. Because $|d_nz|\rightarrow 0$ for $n\rightarrow\infty$, we would have $|f(z)d_nz| \rightarrow 0$, too, so to keep the vector $f(z)d_nz$ visible, we normalize $d_nz$ by division by $|d_nz| = 2\pi/n$. So, only the rotation of $f(z)$ is in the focus:

Now we are already done: For $f(z) = f_1(z) = z$ the vectors $f(z_k)dz$ still sum up to $0$, while for $f(z) = f_0(z) = 1/z$ we have $n$ vectors of length $1$ all pointing into the same direction (up), so there sum is just $n\cdot i$. Multiplying by the normalization factor $2\pi/n$ we get the desired result:

$$\sum_{k=0}^{n-1}f_0(z_k)d_nz = 2\pi i$$

which holds for arbitrary $n$.

Put algebraically, one finds with $\zeta_k = e^{i2\pi k/n}$ the $n$-th roots of unity, $d_k = \zeta_k e^{i\pi / 2}$, and $1/\zeta_k = \overline{\zeta_k}$:

$$\sum_{k=0}^{n-1}\zeta_k d_k = \sum_{k=0}^{n-1}\zeta_k^2 e^{i\pi / 2} = i \sum_{k=0}^{n-1}\zeta_k^2 = i \sum_{k=0}^{n-1}\zeta_k = 0$$

$$\sum_{k=0}^{n-1}\overline{\zeta_k} d_k = \sum_{k=0}^{n-1}\overline{\zeta_k}\zeta_k e^{i\pi / 2} = i \sum_{k=0}^{n-1}1 = i \cdot n$$

Again, to see this more clearly, we can draw $f(z)d_nz$ from the origin:

_{Note that the single point for $p=0$, i.e. $f_0(z) = 1/z$, represents in fact $n$ points, all at the same position.}

The proof sketched here was somehow "synthetic" or geometrical. It's interesting to compare it with the analytic proof:

$$\oint_{C_1} dz/z = \int _0 ^{2 \pi} \dfrac{i e^{i z}}{e^{i z}} dz = \int _0 ^{2 \pi}i dz = 2 \pi i $$

and the proof by Cauchy's residue theorem

$$\oint_{C_1} dz/z = 2\pi i \operatorname{I}(C_1,0)\operatorname{Res}(1/z,0)= 2 \pi i$$

with $\operatorname{I}(C_1,0) = 1$ the winding number of $C_1$ around $0$ and $\operatorname{Res}(1/z,0) = 1$ the residue of $f(z)=1/z$ at $0$.

What I **don't** see in the pictures above is, what the value of the integral has to do with the existence of a pole at $0$, so my question is:

By which kind of "long-range interaction" does the singularity at $0$ "force" the vectors $dz/z$ to rotate exactly the way they do, summing up to $2\pi i$?

## Addendum 2

Dirk asked for pictures for $f(z) = 1/z^2$. Here they are for $z \rightarrow z + f(z)$, $z \rightarrow z + f(z)d_nz$, , $0 \rightarrow f(z)d_nz$:

For the sake of comparison here are the plots for $f(z) = z^k$, $k = 1,2,3,4$ (upper rows) and $f(z) = 1/z^k$ (lower rows):

Note the special character of the bottom left plot in the second picture which is the one that relates to the residue of $z^{-1}$ at $z=0$ being $1$. All other residues of $z^{\pm k}$ at $z=0$ are $0$.

Note also how derivation acts as rotation.

## Addendum 1

For the sake of comparison: The same visual proof works – even though not so easy to see at a glance – for other closed curves, e.g. a cardioid $\gamma$. In the first table you'll see $z \rightarrow z + f(z)$, in the second $z \rightarrow z + f(z)d_nz$, in the third $0 \rightarrow f(z)d_nz$. With a little experience you will "see" that $\oint_\gamma zdz = 0$ and $\oint_\gamma dz/z = 2\pi i$

And here for the diamond curve: the tables for $z \rightarrow z + f(z)$, $z \rightarrow z + f(z)d_nz$, and $0 \rightarrow f(z)d_nz$

The rotation of $f(z)$ as induced by multiplication by $dz$ can be seen as a "process" in stop motion:

_{Note that in these two cases – based on the circle – $f(0)$ is rotated by $\pi/2$, $f(e^{i\pi/2})$ by $\pi$, $f(e^{i\pi})$ by $-\pi/2$, and $f(e^{i3\pi/2})$ by $0$.}