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The Winding Number of a Curve: A Comprehensive Guide

Silence is a source of great strength, Lao Tzu

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Introduction

A complex function $f(z)$ maps $z = x + iy \in \mathbb{C}$ to another complex number. For example: $f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy, f(z) = \frac{1}{z}, f(z) = \sqrt{z^2 + 7}$.

A contour is a continuous, piecewise-smooth curve defined parametrically as: $z(t) = x(t) + iy(t), \quad a \leq t \leq b$.

Definition (Smooth Contour Integral). Let ᵞ be a smooth contour (a continuously differentiable path in the complex plane), $\gamma: [a, b] \to \mathbb{C}$. Let $f: \gamma^* \to \mathbb{C}$ be a continuous complex-valued function defined on the trace $\gamma^*$ of the contour (i.e. along the image of $\gamma$). Then, the contour integral of f along $\gamma$ is defined as $\int_{\gamma} f(z)dz := \int_{a}^{b} f(\gamma(t)) \gamma^{'}(t)dt$.

Properties

Cauchy Integral Formula

  1. For a simple pole at a point $z_k$ (i.e., the denominator has a simple root) is calculated using the formula $Res(f, z_k) = \lim_{z \to z_k}(z-z_k)f(z)$.
  2. For a pole of order m, $Res(f, z_k) = \frac{1}{(m-1)!}\lim_{z \to z_k} \frac{d^{m-1}}{dz^{m-1}}[(z-z_k)^mf(z)]$
  1. Reversal of orientation. $\int_{-\gamma} f(z)dz = -\int_{\gamma} f(z)dz$ where $-\gamma:[a,b] \to \mathbb{C}$ is the reverse of the contour (traversing the same path in the opposite direction) defined by $(- \gamma)(t)=\gamma(a+b-t)$. This property states that reversing the direction of a contour changes or flips the sign of the integral. Intuitively, this is analogous to how reversing the limits of integration in real analysis changes the sign: $\int_a^b f(x)dx = -\int_b^a f(x)dx$. In complex analysis, the orientation of the contour matters because the integral depends on the direction in which we traverse the path.
  2. Additivity under subdivision. Suppose a < c < b, let split $\gamma$ into two sub-contours $\gamma_1 = \gamma|_{[a, \gamma_1]} \text{ and } \gamma_2 = \gamma|_{[c, b]}$, then $\int_{\gamma} f(z)dz = \int_{\gamma_1} f(z)dz + \int_{\gamma_2} f(z)dz$. This property states that integrating over the whole contour is the same as integrating over the pieces successively. This is the complex analogue of the additive property of definite integrals: $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx $ for a < c < b.
  3. Invariance of Contour Integrals Under Reparameterization. Let $\tilde{\gamma}$ be a contour, defined by a function $\tilde{\gamma}: [c, d] \to \mathbb{C}$. If $\tilde{\gamma}$ is another parameterization of the same oriented path $\gamma$ meaning there exists a one-to-one, continuously differentiable map $\psi: [c,d]\to[a,b]$ with a positive derivative $\psi'(t)>0$ such that $\tilde{\gamma}(t) = \gamma(\psi(t))$, then $\int_{\gamma} f(z)dz = \int_{\tilde{\gamma}} f(z)dz$
  1. The Curve itself ($\gamma$).
  2. The Interior (Int($\gamma$)) is the finite area enclosed by the curve (e.g., if you draw a circle on a piece of paper, the interior region would be everything inside the circle). It is a bounded, simply connected region.
  3. The Exterior (Ext($\gamma$)) is the infinite area outside the curve (e.g., using the same circle example, the exterior region would include all points on the paper that are not inside the circle). It is an unbounded region.

The Winding Number of a Curve

Lemma. If γ is a piecewise differentiable closed curve that does not pass through a point a in the complex plane, then the value of the integral $\oint_{\gamma} \frac{dz}{z-a}$ is an integer multiple of 2πi.

This Lemma proves that the integral $\oint_{\gamma} \frac{dz}{z-a}$ can only take on a discrete set of values: $\{ \dots, -2\pi i, 0, 2\pi i, 4\pi i, \dots \}$.

Proof.

Let the closed contour $\gamma$ be parameterized by $z = \gamma(t)$ for $t \in [\alpha, \beta]$. Since $\gamma$ is closed, the starting point $\gamma(\alpha)$ equals the final point $\gamma(\beta), \gamma(\alpha) = \gamma(\beta)$.

We define the function $h(t)$ as the value of the integral along the path $\gamma$ from the starting time $\alpha$ up to t: $h(t) = \int_{\alpha}^t \frac{\gamma'(u) du}{\gamma(u)-a} \text{ for } t \in [\alpha, \beta]$.

  1. Since $\gamma$ is piecewise differentiable (continuous everywhere on $[\alpha, \beta]$ except possibly at a finite number of points) and $\gamma(t) \neq a$ (the path doesn’t pass through the singularity a), the denominator $\gamma(u) - a$ is never zero on the entire integration path, ensuring the denominator is bounded away from zero ($\gamma$ is a continuous curve on a compact interval $[\alpha, \beta], \gamma(u) - a$ is a non-zero continuous function), hence the resulting integrand is piecewise continuous on the compact interval $[\alpha, \beta]$ and then, the integral that defines h(t) is a proper Riemann integral (or a standard contour integral), which means it exists and is unique for every t, i.e., h(t) is well-defined. $h(t)$ is also a continuous function on $[\alpha, \beta]$.

    $h(t) = \int_{\alpha}^t f(u) du \quad \text{where } f(u) = \frac{\gamma'(u)}{\gamma(u)-a}$. A fundamental theorem of calculus states that if $f(u)$ is piecewise continuous on a closed interval $[\alpha, \beta]$, then the integral function $h(t)$ is continuous on that entire interval. Notice that h(t) represents the accumulated area as t increases. A piecewise continuous function f(u) can have a finite number of jumps. Integrating a finite jump results in a corner in the graph of the integral function h(t), but it does not result in a break or jump in h(t). For h(t) to be discontinuous, the integrand f(u) would have to contain an infinite singularity (which is ruled out because $\gamma(u) \neq a$).

  2. At the start of the path, $h(\alpha) = \int_{\alpha}^\alpha \frac{\gamma'(u) du}{\gamma(u)-a} = 0$.
  3. By the Fundamental Theorem of Calculus (for the real variable $t$), the derivative of $h(t)$ with respect to $t$ is the integrand evaluated at t: $h'(t) = \frac{\gamma'(t)}{\gamma(t)-a}$

Consider the auxiliary function $G(t) = e^{-h(t)} (\gamma(t)-a)$. We compute the derivative of $G(t)$ with respect to t.

$$ \begin{aligned} \frac{d}{dt}(e^{-h(t)} (\gamma(t)-a)) = & [\text{ Product rule: } \frac{d}{dt}(uv) = u'v + uv'] \\[2pt] &=e^{-h(t)}(-h'(t))(\gamma(t)-a) + e^{-h(t)}(\gamma'(t)) \\[2pt] &= [\text{Now, we substitute the expression for $h'(t)$ from (3)}] \\[2pt] &=e^{-h(t)}(-\frac{\gamma'(t)}{\gamma(t)-a})(\gamma(t)-a) + e^{-h(t)}(\gamma'(t)) \\[2pt] &=e^{-h(t)}(-\gamma'(t))+ e^{-h(t)}(\gamma'(t))\\[2pt] &=0. \end{aligned} $$

Since the derivative of the function $G(t)$ is identically zero on the interval $[\alpha, \beta]$, the function itself must be a constant, $e^{-h(t)} (\gamma(t)-a) = c$ where c is a constant.

This can be rearranged to express the path $\gamma(t)$ in terms of h(t): $\gamma(t)-a = ce^{h(t)}$. In particular, we evaluate the expression at $t = \beta, ce^{h(\beta)} = \gamma(\beta)-a = \gamma(\alpha) - a =[\text{Evaluating the expression at } t = \alpha] ce^{h(\alpha)} =[(2)] ce^0 = c$.

Since we know $h(\alpha) = 0$ (2), and $e^0 = 1:c(1) = \gamma(\alpha) - a \implies c = \gamma(\alpha) - a$. Since the curve does not pass through a, $c = \gamma(\alpha) - a \neq 0$.

Therefore, $ce^{h(\beta)} = c \leadsto[c \neq 0, \text{ we can divide both sides by c}] e^{h(\beta)} = 1 \leadsto h(\beta) = 2\pi i n$ for some integer n.

The solution to the equation $e^w = 1$ in the complex plane is $w = 2\pi i n$, where n is any integer ($\dots, -2, -1, 0, 1, 2, \dots$).

Finally, we observe that the value $h(\beta)$ is the desired integral: $\oint_{\gamma} \frac{dz}{z-a} = \int_{\alpha}^{\beta} \frac{\gamma'(t) dt}{\gamma(t)-a} = h(\beta) = 2\pi i n$

Definition. Suppose we are given a closed, oriented curve in the xy plane. Then, the winding number of the curve is equal to the total number of counterclockwise turns that the object makes around the origin.

When counting the total number of turns, counterclockwise motion counts as positive (it contributes +1 per full loop), while clockwise motion counts as negative (it contributes −1 per full loo).

Formal definition. Let $\gamma: [α, β] → \mathbb{C}$ be a piecewise smooth, closed curve (so γ(α) = γ(β)), the winding number of γ about $z_0$, a complex point not on the curve $z_0$ ∉ γ([α, β]), also known as the index of $z_0$ with respect to γ, is defined as $n(\gamma, z_0) = Ind_{\gamma}(z_0) = \frac{1}{2\pi i} \oint_{\gamma} \frac{dz}{z - z_0} = \frac{1}{2\pi i} \int_{\alpha}^{\beta} \frac{\gamma'(t)}{\gamma(t) - z_0}dt$.

Figure 4 illustrates indices −2, −1, 0, 1, 2, 3 for concentric loops with various orientations; the multi-loop picture shows different points $z_0, z_1, z_2$ with different winding counts. image info

Key insight

To understand why this formula works, consider the polar representation. Write: $z-z_0 = re^{i\theta} \leadsto dz = e^{i\theta}dr + ire^{i\theta}d\theta \leadsto \frac{dz}{z-z_0} = \frac{e^{i\theta}dr + ire^{i\theta}d\theta}{re^{i\theta}} = \frac{dr}{r}+id\theta$

$\oint_{\gamma} \frac{dz}{z - z_0} = \int_a^b \frac{dr}{r}+id\theta = ln(r)\bigg|_{a}^{b} + i×\text{total change in θ}$

The term $\frac{dr}{r}$ integrates to zero because ln(r) returns to its initial value (closed curve).

$\oint_{\gamma} \frac{dz}{z - z_0} = i×\text{total change in θ} = i(2\pi k)$ where k is the net number of revolutions. Dividing by 2πi gives k, the winding number.

Relation to Jordan Curve Theorem

For a simple closed curve γ:

Examples

$n(\gamma, 0) = \frac{1}{2\pi i} \int_{0}^{1} \frac{1}{e^{i2\pi t}} \cdot \frac{d}{dt}(e^{i2\pi t}) dt = \frac{1}{2\pi i} \int_{0}^{1} \frac{i2\pi e^{i2\pi t}}{e^{i2\pi t}} dt = \frac{1}{2\pi i} \cdot i2\pi = 1$

A circle centered at the origin with a point inside has a winding number of +1 (counterclockwise) or -1 (clockwise).

For any $z_0$ with $|z_0| \lt R, \mathbb{n}(\gamma; z_0) = 1.$ This can be quickly verified using Cauchy’s Integral Formula. Let f(z)=1. The formula states: $\oint_\gamma \frac{f(z)}{z-z_0} dz = 2\pi i \cdot f(z_0)$. Substituting into the winding number definition: $\mathbb{n}(\gamma;z_0) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-z_0} dz = \frac{1}{2\pi i} (2\pi i \cdot f(z_0)) = f(z_0) = 1.$

For any $z_0$ with $|z_0| > R, \mathbb{n}(\gamma;z_0) = 0.$ This follows from Cauchy’s Theorem. If $z_0$ is outside the circle, the function f(z) = $\frac{1}{z-z_0}$ is analytic at all points inside and on the contour γ. Therefore, its integral over the closed contour is zero: $\mathbb{n}(\gamma;z_0) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-z_0} dz = \frac{1}{2\pi i} \cdot 0 = 0.$

Hence, $\mathbb{Int}(\gamma) = \{ z \in \mathbb{C} | \mathbb{n}(\gamma; z_0) = 1 \} = \{ z \in \mathbb{C} | |z| < R \} \text{ and } \mathbb{Ext}(\gamma) = = \{ z \in \mathbb{C} | \mathbb{n}(\gamma; z_0) = 0 \} = \{ z \in \mathbb{C} | |z| > R \}.$

The total winding number of the combined path γ = γ₁​+γ₂ is the sum of the individual winding numbers: n(γ, 0) = n(γ₁, 0) + n(γ₂, 0) = 1 + (-1) = 0.

The two loops cancel each other out. The combined path γ winds around the origin zero times

Case 1. $z_0 \in Int(\gamma)$

  1. Approximate the Curve with a Polygon. A Jordan curve γ can be approximated arbitrarily well by a simple polygon P that lies entirely within its interior. We can construct P to be positively oriented and “close enough” to γ so that it also contains $z_0$. Because the winding number integral is continuous with respect to the path, this approximation doesn’t change the result: $n(\gamma, z_0) = n(P, z_0)$

  2. Triangulate the Polygon. Any simple polygon can be divided into a finite number of triangles ($T_1, T_2, \cdots T_k$). The integral over the boundary of the polygon is the sum of the integrals over the boundaries of these triangles. This is because all the internal edges are traversed twice in opposite directions, and their integrals cancel each other out, leaving only the integral over the exterior boundary: $n(P, z_0) = \sum_{j=1}^k n(∂T_j,z_0)$.

  3. Isolate the Key Triangle. The point $z_0$ must lie in the interior of exactly one of these triangles, let’s call it $T_1$. For any other triangle $T_j, j \ne 1$, the point $z_0$ is in its exterior. The function $\frac{1}{z-z_0}$ is analytic inside and on the boundary of $T_j$. By Cauchy’s Integral Theorem, the integral is zero. Thus, $n(∂T_j,z_0) = 0$ for j ≠ 1.

    This means our sum collapses to a single term: $n(P, z_0) = \sum_{j=1}^k n(∂T_j,z_0) = n(∂T_1,z_0)$

  4. Deform the Triangle to a Circle (Homotopy). Let C be a small, positively oriented circle centered at $z_0$ that lies completely inside the triangle $T_1$. The function $\frac{1}{z-z_0}$ is analytic in the region between the triangle’s boundary and the circle. By the Deformation Principle (homotopy invariance), we can continuously deform the triangular path $∂T_1$ into the circular path C without changing the value of the integral: $n(∂T_1,z_0) = n(C, z_0)$. The winding number of a positively oriented circle around its center is known to be 1, so $n(\gamma, z_0) = n(P, z_0) = \sum_{j=1}^k n(∂T_j,z_0) = n(∂T_1,z_0) = n(C, z_0) = 1.$

Case 2. $z_0 \in Ext(\gamma)$

If $z_0$ is in the unbounded exterior region, the function $f(z) = \frac{1}{z-z_0}$ is analytic at all points inside and on the contour γ. The singularity at $z_0$ is outside the domain enclosed by the curve. By Cauchy’s Integral Theorem, the integral of an analytic function over a simple closed contour is zero: $\oint_{\gamma} \frac{1}{z-z_0} = 0$. Therefore, the winding number is: $n(\gamma; z_0) = \frac{1}{2\pi i} \oint_{\gamma} \frac{1}{z-z_0} = \frac{1}{2\pi i} \cdot 0 = 0.$

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