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

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 $$f_0(z) = 1/z$$. [click image to enlarge]

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 $$2\pi i$$?

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.

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 $$0$$.  • I like to think the other way round. These things are hard to see geometrically or otherwise, and the beauty and power of complex analysis is that it makes them look so easy. – Giuseppe Negro Jan 21 '19 at 11:52
• @GiuseppeNegro: Of course, you have to take the time to see things geometrically. But if you do, it's not so hard - and things arrange in a pleasing manner. And of course, it's just an alternative way to see things. And the "true" beauty arises when it leads to the same results as on the analytic way. (The price for "looking so easy" in complex analysis is, that I don't have something visual in mind - just symbolic expressions.) – Hans-Peter Stricker Jan 21 '19 at 12:00
• I agree with Hans Stricker and the visual intuition is good to have even if it's not the simplest or most beautiful way to look at it (although with enough understanding I believe you could make it so). And other things in complex analysis can be visualised too – Benjamin Thoburn Jan 21 '19 at 18:04
• "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 $2πi$?" The pole doesn't force it. The pole allows it. – Arthur Jan 22 '19 at 15:37
• @Dirk: Please have a look at Addendum 2. It reveals what you said: The residue of $1/z^2$ at $z = 0$ is $0$ (because the numbers in the third picture sum up to $0$). – Hans-Peter Stricker Jan 29 '19 at 9:40

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 $$\int_C \frac{1}{z}dz=\text{Circluation}\bar{f}+i\cdot\text{Flux}\bar{f}=0+i\cdot 2\pi=2\pi i$$