🧮 Brain Teaser

The Winding Number Is Always an Integer

Let γ:[0,1]→C∖{0}\gamma: [0,1] \to \mathbb{C} \setminus \{0\} be a closed curve (i.e., γ(0)=γ(1)\gamma(0) = \gamma(1)) that avoids the origin. The winding number of γ\gamma around 00 is defined by

n(γ,0)=12πi∫γdzz.n(\gamma, 0) = \frac{1}{2\pi i} \int_\gamma \frac{dz}{z}.

Show that n(γ,0)n(\gamma, 0) is always an integer, without invoking any machinery about branches of the logarithm or homotopy theory. Use only the following raw ingredients:

  • Define ϕ(t)=∫0tγ′(s)γ(s) ds\phi(t) = \int_0^t \frac{\gamma'(s)}{\gamma(s)}\, ds.
  • Consider the function h(t)=e−ϕ(t)γ(t)h(t) = e^{-\phi(t)} \gamma(t).

Why must n(γ,0)n(\gamma, 0) be an integer?

winding numberclosed curveexponentiallog-derivativeintegrating factor

Answer: Winding Number Is Always an Integer

Key Idea / Intuition

The winding number counts how many times γ\gamma loops around the origin. The cleanest proof avoids all topology and branch-cut gymnastics by constructing a "lifted" function that tracks the running phase of γ(t)\gamma(t). If we form h(t)=e−ϕ(t)γ(t)h(t) = e^{-\phi(t)}\gamma(t), where ϕ\phi is the "log-derivative integral," then hh turns out to be constant. Since γ\gamma is a closed curve, this forces eϕ(1)=1e^{\phi(1)} = 1, which means ϕ(1)\phi(1) must be an integer multiple of 2πi2\pi i — and that integer is exactly the winding number.


Formal Proof / Solution

Step 1: Define the "running logarithm" ϕ\phi.

Set ϕ(t)=∫0tγ′(s)γ(s) ds,t∈[0,1].\phi(t) = \int_0^t \frac{\gamma'(s)}{\gamma(s)}\, ds, \qquad t \in [0,1].

This is well-defined since γ(s)≠0\gamma(s) \neq 0 for all ss, and γ\gamma is (piecewise) C1C^1.

Note immediately that ϕ(1)=∫01γ′(s)γ(s) ds=∫γdzz=2πi⋅n(γ,0).\phi(1) = \int_0^1 \frac{\gamma'(s)}{\gamma(s)}\,ds = \int_\gamma \frac{dz}{z} = 2\pi i \cdot n(\gamma, 0).

Step 2: Show h(t)=e−ϕ(t)γ(t)h(t) = e^{-\phi(t)}\gamma(t) is constant.

Differentiate: h′(t)=−ϕ′(t) e−ϕ(t)γ(t)+e−ϕ(t)γ′(t).h'(t) = -\phi'(t)\, e^{-\phi(t)}\gamma(t) + e^{-\phi(t)}\gamma'(t).

But ϕ′(t)=γ′(t)γ(t)\phi'(t) = \dfrac{\gamma'(t)}{\gamma(t)}, so

h′(t)=e−ϕ(t) ⁣[−γ′(t)γ(t)⋅γ(t)+γ′(t)]=e−ϕ(t) ⁣[−γ′(t)+γ′(t)]=0.h'(t) = e^{-\phi(t)}\!\left[-\frac{\gamma'(t)}{\gamma(t)}\cdot \gamma(t) + \gamma'(t)\right] = e^{-\phi(t)}\!\left[-\gamma'(t) + \gamma'(t)\right] = 0.

Hence h(t)h(t) is constant on [0,1][0,1].

Step 3: Evaluate hh at the endpoints.

Since hh is constant: h(0)=e−ϕ(0)γ(0)=e0⋅γ(0)=γ(0),h(0) = e^{-\phi(0)}\gamma(0) = e^0 \cdot \gamma(0) = \gamma(0), h(1)=e−ϕ(1)γ(1).h(1) = e^{-\phi(1)}\gamma(1).

But γ\gamma is closed: γ(1)=γ(0)\gamma(1) = \gamma(0). Therefore

γ(0)=e−ϕ(1)γ(0).\gamma(0) = e^{-\phi(1)}\gamma(0).

Since γ(0)≠0\gamma(0) \neq 0, we can cancel it:

e−ϕ(1)=1  ⟹  eϕ(1)=1.e^{-\phi(1)} = 1 \implies e^{\phi(1)} = 1.

Step 4: Conclude integrality.

The equation ew=1e^{w} = 1 (for w∈Cw \in \mathbb{C}) holds if and only if w=2πikw = 2\pi i k for some k∈Zk \in \mathbb{Z}.

Therefore ϕ(1)=2πikfor some k∈Z,\phi(1) = 2\pi i k \quad \text{for some } k \in \mathbb{Z},

and so n(γ,0)=ϕ(1)2πi=k∈Z.■n(\gamma, 0) = \frac{\phi(1)}{2\pi i} = k \in \mathbb{Z}. \qquad \blacksquare

Summary of the trick: The auxiliary function h(t)=e−ϕ(t)γ(t)h(t) = e^{-\phi(t)}\gamma(t) is the "exponential integrating factor" that kills the ϕ′\phi' term, making h′≡0h' \equiv 0. The closedness of γ\gamma then forces eϕ(1)=1e^{\phi(1)}=1, which is the algebraic reason the winding number is an integer.

Source: Complex Analysis, Stein & Shakarchi, Chapter 1; mathematical folklore

Type: Complex AnalysisSource: Complex Analysis, Stein & Shakarchi, Chapter 1; mathematical folkloreEdit on GitHub ↗