I've been searching for this proof, here and on the web too, but it seems like the answer is harder to find than expected. There are many similar questions about this but they all (implicitly or explicitly) suppose continuous differentiability.

First, let me state the definition of winding number:

Let $\gamma : [0, 1] \to \mathbb{C^*}$ be a continuos function. We know that for such a path, there exists a continuous argument function, that is a function $\phi : [0, 1] \to \mathbb{R}$ such that $\gamma(t) = |\gamma(t)|e^{i\phi(t)}$, for all $t \in [0, 1]$. We define the variation of argument along $\gamma$: $${\rm vararg}(\gamma)=\phi(1)-\phi(0)$$ (We can also prove that all such argument functions are of the form $\phi + 2n\pi$, for any integer $n$, so the variation of argument is well-defined. Also, if $\gamma$ is closed, the variation of argument is a multiple of $2\pi$)

The winding number (index) of the closed curve $\gamma$ around $0$ is defined as: $${\rm I}(\gamma, 0) = \frac{\rm vararg(\gamma)}{2\pi}$$

For a closed path $\gamma : [0, 1] \to \mathbb{C}\setminus\{z_0\}$ for some $z_0 \in \mathbb{C}$, the winding number of $\gamma$ around $z_0$ is defined by translation as: $${\rm I}(\gamma, z_0) = {\rm I}(\gamma - z_0, 0)$$

I consider this as the best definition of winding number because it is intuitive, general and everything follows from it.

Now I want to prove that the winding number is constant on the connected components of the complement of the curve. This is not very hard to prove using the integral expression of the winding number, but that assumes that $\gamma$ is $C^1$.

I know that I should somehow prove that $\phi$ is continuous or just that it is constant in an open ball around any point included in the connected component.

Intuitively, I can prove it like this: consider a curve $\gamma$ that doesn't go through $0$, a point $z_0 \ne 0$ inside a ball around $0$ that is contained in the connected component of $0$. Imagine two segments from $0$ to the curve and from $z_0$ to the curve, as we move along it. Both segments follow the same point on the circle around $0$ and their movements along little circles around their respective start points have the same intervals of increase/decrease in angle (going counter-/clockwise). Therefore, whenever one completes a circle in one direction, the other does so too. Hence, the variation of the argument of $\gamma$ around $0$ and around $z_0$ is the same. By translation, we generalize the result for other points than $0$.

This argument is hard to write rigorously and gets a little too geometric, I think. I wonder if an easier take on this can be carried out.

Any proof or help is appreciated.

  • $\begingroup$ by the way, you should take a look at Henri Cartan's complex analysis text. There, he defines the integral $\int_{\gamma}\omega$ where $\omega$ is a locally-exact (he calls this closed) continuous $1$-form on an open set $U\subset \Bbb{C}$ and $\gamma$ is a continuous path in $U$. In your case, you should take $\omega=\frac{x\,dy-y\,dx}{x^2-y^2}$. There, he also proves a lemma like if $\gamma_1,\gamma_2$ are continuously homotopic loops (or paths with same endpoints), then $\int_{\gamma_1}\omega=\int_{\gamma_2}\omega$. Also, several properties of index are discussed (see pages 57-63). $\endgroup$
    – peek-a-boo
    Jul 25, 2021 at 13:00
  • $\begingroup$ I'm afraid such an approach is not for me yet, as I know virtually nothing about differential forms. $\endgroup$ Jul 25, 2021 at 16:12
  • $\begingroup$ @peek-a-boo, But wait, wouldn't $x$ and $y$ have to be $C^1$, therefore making $\gamma$ a $C^1$ curve, for $\int_{\gamma}\omega$ to make sense? $\endgroup$ Jul 25, 2021 at 16:23
  • $\begingroup$ assuming $\gamma$ is $C^1$ surely simplifies matters, but it's not necessary because $\omega$ is a closed differential form. You should really take a look at the book for the definitions and theorems (you don't really need the full force of differential forms, since at this level, complex analysis only deals with $1$ forms, i.e for line integrals, and this is all explained very nicely in his book). This question presents the definitions if you don't have access to the book. $\endgroup$
    – peek-a-boo
    Jul 25, 2021 at 16:50
  • $\begingroup$ You might like Chapter 7 of A. F. Beardon, Complex Analysis: The Argument Principle in Analysis and Topology (1979; republished by Dover, 2020). $\endgroup$ Jul 25, 2021 at 19:11

1 Answer 1


It’s a consequence of the following lemma: let $\gamma:[0,1] \rightarrow \mathbb{C}^*$ be a closed curve. There exists $\epsilon >0$ such that for every closed curve $\delta:[0,1] \rightarrow \mathbb{C}^*$ with $\|\gamma-\delta\|_{\infty} <\epsilon$, $I(\gamma,0)=I(\delta,0)$.

Indeed, you get your result by taking $\gamma$ and $\delta$ to be close translates of the same closed curve.

How to prove the lemma? Well, let $r>0$ be such that $|\gamma| \geq r$, and assume $\|\gamma-\delta\|_{\infty} < r$. Then consider $\alpha=\gamma/\delta$: then $\alpha([0,1]) \subset D=\mathbb{C} \backslash (-\infty,0]$. But it’s easy to see that there is a complex logarithm $L: D \rightarrow \mathbb{C}$, and then the imaginary part of $L \circ \alpha$ is an argument of $\alpha$. Therefore $\mathrm{vararg}(\alpha)=0$, and thus $\mathrm{vararg}(\gamma)=\mathrm{vararg}(\delta)$ so $I(\gamma,0)=I(\delta,0)$.

  • $\begingroup$ What does $\gamma/\delta$ mean? $\endgroup$ Jul 25, 2021 at 12:48
  • $\begingroup$ It’s the quotient. $\endgroup$
    – Aphelli
    Jul 25, 2021 at 12:50
  • $\begingroup$ Yeah, i get that it's a quotient. But like, division of complex numbers, or is there some concept of quotient of paths, like the quotient of a group by a normal subgroup? $\endgroup$ Jul 25, 2021 at 12:54
  • 1
    $\begingroup$ @StefanOctavian: You consider it pointwise. If $|\gamma(t)| \ge r$ and $|\gamma(t)-\delta(t)| < r$ for all $t$ then $|\delta(t)/\gamma(t)-1| < 1$, so that the curve $\alpha$ defined by $\alpha(t) = \delta(t)/\gamma(t)$ lies in the right-half plane, where the principal branch of the logarithm is defined. $\endgroup$
    – Martin R
    Jul 25, 2021 at 12:59
  • $\begingroup$ Could you explain why $\mathrm{argvar}(\alpha) = 0 \implies \mathrm{argvar}(\gamma) = \mathrm{argvar}(\delta)$? I am unable to see this. $\endgroup$ Jul 25, 2021 at 16:15

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