kuniga.me > NP-Incompleteness > The Winding Number

09 May 2024

This is our fourth post in the series with my notes on complex integration, corresponding to *Chapter 4* in Ahlfors’ Complex Analysis.

In this one, we’ll explore the concept of the winding number of a curve with respect to a point, which can be interpreted how many times a curve winds around that point. It has an interesting relationship with the Cauchy integral theorem that we learned about in a prior post.

The previous posts from the series:

In the previous post, Cauchy Integral Theorem [4], we concluded with *Theorem 4* saying that:

if $\gamma$ is contained in a disk $\Delta$ and that $f(z)$ is holomorphic in $\Delta$. And this holds even if $f(z)$ is not holomorphic in a finite set of points in $\Delta$, as long as each of these points $\xi$ satisfy:

\[(1) \quad \lim_{z \rightarrow \xi} (z - \xi) f(z) = 0\]Now let’s look at a specific function, $f(z) = 1/(z - a)$. Assuming the point $a$ is in the disk $\Delta$, then $f(z)$ is not holomorphic at $a$, but maybe it satisfies $(1)$? We can find if that’s the case by replacing $a$ in $(1)$:

\[\lim_{z \rightarrow a} (z - a) \frac{1}{z - a} = 1\]So no, we can’t use *Theorem 4* from [4] to conclude

equals 0. Still, it’s worth exploring this integral further. What does its value represent? In which case it is 0? This is what we’ll focus on this post.

Let $\gamma$ be a piecewise differentiable closed curve and $a$ a point not on the curve. We define **the winding number** of $a$ with respect to $\gamma$, denoted by $n(\gamma, a)$ as:

The geometric interpretation of this value is how many times the curve $\gamma$ winds (i.e. completes a revolution) counter-clockwise around $a$. Wikipedia provides several examples:

Note that when the curve winds clockwise the value is negative.

Let’s build a geometric intuition on why the formula $(2)$ corresponds to the number of revolutions. Recall that the set of points in the circumference of a circle of radius $\rho$ and center $a$ can be written in polar form as:

\[z(\theta) = a + \rho e^{i\theta}\]For $0 \le \theta \lt 2 \pi$. We can generalize this idea for any closed curve $\gamma$ by picking any point $a$ not in it. The major difference is that the “radius” (i.e. the distance between a point in $\gamma$ and $a$) $\rho$ would not be fixed, but a function of $\theta$. We can parameterize both by some $0 \le t \le 1$ as:

\[z(t) = a + \rho(t) e^{i\theta(t)}\]Noting that $a$ doesn’t have to be inside the curve. We won’t prove it, but it’s possible to show that if $\gamma$ is differentiable so is $\rho(t)$ and $\theta(t)$. Now if we imagine $t$ is time and we place an observer at point $a$ rotating to follow the point $z(t)$ as we travel from $t = 0$ to $t = 1$, the amount of “angle displacement” this observer will perform can be found by adding up the delta angle between adjacent timestamps, $\Delta \theta = \theta(t_i) - \theta(t_{i-1})$.

Because it’s a closed curve, the observer must finish facing at the same direction they started, meaning they completed an integer number of revolutions and thus the total angle displacement should be a multiple of $2 \pi$.

If we allow $\theta(t)$ to go beyond $2\pi$ (when multiple revolutions occur), then the total angle displacement is: $\theta(1) - \theta(0)$ and the number of revolutions (and hence the winding number) is

\[(3) \quad n(\gamma, a) = \frac{\theta(1) - \theta(0)}{2\pi}\]With this intution in mind, *Lemma 1* formalizes the correspondence between $(2)$ and the number of revolutions:

**Lemma 1.** Let $\gamma$ be a piecewise differentiable closed curve and a point $a$ not in $\gamma$. The curve can parametrized with respect to $a$:

for $0 \le t \le 1$ and differentiable functions $\rho(t)$ and $\theta(t)$. Then:

\[(4) \quad \int_{\gamma} \frac{dz}{z - a} = i(\theta(1) - \theta(0))\]We can replace $(3)$ in $(4)$:

\[\int_{\gamma} \frac{dz}{z - a} = (2 \pi i) n(\gamma, a)\]Which gives us $(2)$.

The first property is that if we flip the direction of a curve $\gamma$, we negate the winding number, i.e.:

\[n(\gamma, a) = -n(-\gamma, a)\]This follows from the application of $(2)$ and using that

\[\int_\gamma f(z)dz = -\int_{-\gamma} f(z)dz\]Intuitively, if you’re an observer away from the curve, you don’t need to “turn around yourself” to follow a point along it, so the winding number of an external point is 0, as in the third example of *Figure 1*. How about the example from *Figure 2* (left): is it inside or outside?

We need a more precise way to define “outsideness”. There’s a topological formal definition, but we won’t go over it here. Instead, we can get an intuition by imagining the curve is a rubberband on the surface of a table. You then put your finger at point $a$. If you can remove the rubberband without lifting your finger, then point $a$ is “outside”.

In the case of *Figure 2*, the observer ends up going around thelselves, but they reverse back before finish, so in the end their winding number is still 0 (left image). It’s possible to show that the winding number is 0 if and only if a point is outside the curve, so we could use this alternative definition for “outsideness”.

We won’t prove this equivalence here, but rather a weaker result, in which the observer is “away” from the curve, i.e. not surrounded by it:

**Lemma 2.** Let $\gamma$ be a closed curve, $C$ be a circle enclosing it and $a$ a point outside the circle. Then $n(\gamma, a) = 0$.

We can always enclose any bounded curve with a circle if we choose a sufficiently large radius and still have $\infty$ outside it, which leads us to the corollary:

**Corollary 3.** Let $\gamma$ be a closed curve. Then $n(\gamma, \infty) = 0$

Generalizing the idea of inside/outside, we can consider the regions determined by a curve $\gamma$. The curve $\gamma$ is a bounded and closed set, so its complement is unbounded and open. The complement can be partitioned into connected components, exactly one of which is unbounded (the one containing infinity).

*Lemma 4* shows that any two points in the same region have the same winding number.

**Lema 4.** Let $\gamma$ be a closed curve and let $\curly{R_i}$ be the set of components determined by it. Let $a, b$ be points in the same region $R_i$. Then $n(\gamma, a) = n(\gamma, b)$.

Let's define $\Omega$ as the complement of $\overline{uv}$. We claim that for $z \in \Omega$ the function: $$(4.1) \quad f(z) = \frac{z - u}{z - v}$$ returns either a positive real value or a imaginary number, that is, never a negative real value. The key reason is that since $z$ doesn't belong to $\overline{uv}$. With this in mind, let us write $(z - u)$ and $(z - v)$ in polar form, say $r_u e^{i\theta_u}$ and $r_v e^{i\theta_v}$, so $$f(z) = \frac{r_u}{r_v} e^{i(\theta_u - \theta_v)}$$ We can interpret $(z - u)$ and $(z - v)$ geometrically as directed vectors, from $\overrightarrow{uz}$ and $\overrightarrow{vz}$ respectively. We have two cases: either $z$ lies on the line defined by $u$ and $v$ (but not between them!) or not. In the first case, the angle $\theta_u = \theta_v$, since the vectors point in the same direction, so $f(z) = r_u/r_v$, that is, a positive real.

In case 2, $z$ is not in the line defined by $u, v$, so it forms a triangle as depicted in

This property is important because we wish to compute $\log(f(z))$. The function $\log(z)$ is not holomorphic on the entire complex plane but is for $\mathbb{C} - \mathbb{R}_{\le 0}$ (i.e. complex numbers excluding the non-positive reals). So basically we're saying that for any $z \in \Omega$, the log of $\log(f(z))$ has a derivative, which can be show to be: $$(4.2) \quad g(z) = \frac{d\log(f(z))}{dz} = \frac{1}{z - u} - \frac{1}{z - v}$$ Since $g(z)$ is the derivative of a holomorphic function $\log(f(z))$, in $\Omega$, we can use

A corollary from *Lemma 4*, *Corollary 3* and that the unbounded region determined by $\gamma$ contains $\infty$:

**Corollary 5.** Let $\gamma$ be a closed curve and $R_0$ be the unbounded region determined by $\gamma$. Then if $a \in R_0$, $n(\gamma, a) = 0$.

This provides a stronger result than *Lemma 3*. We don’t need $a$ to be outside a circle enclosing $\gamma$. As long as there’s a path from $\infty$ to $a$ not crossing $\gamma$, the winding number of $a$ is 0, for example the one in *Figure 3*.

It’s still not a necessary condition for the winding number to be 0 though, since in *Figure 2* there still no way out but the winding number is still 0.

So far we’ve been consider conditions that lead to $n(\gamma, a) = 0$. We now consider conditions that lead to $n(\gamma, a) = 1$.

To simplify calculations, we’ll assume $a$ is at the origin. If we take the geometric interpretation of winding number, we can see it’s invariant with translation. In fact, in the equation $(2)$, the expression $z - a$ is essentially doing this normalization.

We also assume $a$ is surrounded by the curve $\gamma$, otherwise we already know its winding number is 0. Then, visualizing this on the complex plane, $\gamma$ has to exist in all four quadrants since it surrounds the origin, such as the curve in *Figure 4*.

We can now state a sufficient condition for $n(\gamma, a) = 1$:

**Lemma 6.** Let $\gamma$ be a curve around the origin. We can pick points $z_1$ and $z_2$ such that $z_1$ has a positive imaginary component and $z_2$ a negative one. Denote the part of the curve from $z_1$ to $z_2$ as $\gamma_1$ and from $z_2$ to $z_1$ as $\gamma_2$.

If $\gamma_1$ doesn’t cross the positive real axis and $\gamma_2$ doesn’t cross the negative real axis, then $n(\gamma, a) = 1$, where $a$ is the origin.

We now build two closed curves: $\sigma_1 = \gamma_1 + \delta_2 - C_1 - \delta_1$ and $\sigma_2 = \gamma_2 + \delta_1 - C_2 - \delta_2$. $\sigma_2$ is shown in

In the example of *Figure 4*, *Lemma 6* says that because the blue path doesn’t cross the positive $x$-axis and the red path doesn’t cross the negative one, then $n(\gamma, a) = 1$.

To recap, we started by analyzing an example function for which we can’t use *Theorem 4* in [4] and showed that it has a nice geometric interpretation, the number of revolutions around a point.

We considered some properties such as those sufficient for $n(\gamma, a) = 0$ and $n(\gamma, a) = 1$. The winding number is not just a geometric curiosity though, it will be needed as we progress in our study of complex integration.