This is yet another question about the proof on page 340 of Tristan Needham's Visual Complex Analysis (this is the only thing I've read in the entire book as it's the only proof of the crossing rule for winding numbers I could find online). For reference, refer to this image: extract from page 340 of Tristan Needham's Visual Complex Analysis

This proof is only intuitional imo. Sure, it's not very hard to write it rigurously if you have the right setting, but this is exactly my problem.

So, suppose we want to prove this statement of the theorem:

Having $\gamma : [0, 1] \to \mathbb{C}$, a continuous closed curve and $z_1, z_2 \in \mathbb{C}\setminus\gamma^*$ two points laying in adjacent connected components of $\mathbb{C}\setminus\gamma^*$, that is, there is a path $\delta : [0, 1] \to \mathbb{C}$ with $\delta(0) = z_1$ and $\delta(1) = z_2$ with the property that $\gamma$ intersects it exactly once, ie. $\exists!(s,t) \in [0, 1]^2:\ \gamma(t) = \delta(s)$,

Then if the orientation of $\gamma$ is such that $z_2$ is on the left of the curve wrt motion down the curve, $I(\gamma, z_1) = I(\gamma, z_2) - 1$. Otherwise, we apply the theorem with $z_1$ and $z_2$ switched and we get that $I(\gamma, z_1) = I(\gamma, z_2) + 1$.

[Notation: $\gamma^* = \gamma([0, 1])$ and $I(\gamma, z)$ is the winding number of $\gamma$ around $z$]

Now, in order for us to be able to use the proof in the book, we need the following setup: the curve $\gamma$ looks like a straight line near the two points $z_1$ and $z_2$. I know that we can "refine" the curve by an homotopy and preserve the winding numbers around the two points, but how can we make sure that they remain in adjacent connected components after this transformation of the curve? In other words, I want to find an explicit way of changing the curve such that after the transformation we can still explicitly find a path linking the two points that $\gamma$ passes through exactly once. Then we can bring the two points as close to the curve as we need along this new path in order to make sure that there is enough space for a circular loop around $z_2$ in its connected component (by using the fact that connected components of complements of paths are open).

Another thing we could do is first transform the curve and then find new points, but then we have the same problem: how do we prove that the winding numbers around our new points (say $z_1', z_2'$) are the same as the ones around the old ones, ie. $I(\gamma, z_1') = I(\gamma, z_1)$ and $I(\gamma, z_2') = I(\gamma, z_2)$.

I've been scratching my head at this for days and I couldn't find a way to make it work. Any help would be appreciated.


2 Answers 2


I think I finally made it work. It does use topology theory and I admit that I should have added the tag [topology] as well because my question falls a bit outside the area of interest of complex anaysis. (However, I don't have proper knowledge of topological spaces as per se, but topological results I will be using are valid and I have studied in the context of metric spaces, which are just a special case of topological spaces).

First, let $z_1, z_2$ points situated in different connected components of $\mathbb{C}\setminus\gamma^*$ such that there exists a path $\delta$ connecting then and $\exists!(t_0,s_0) \in [0,1)^2:\ \gamma(t_0) = \delta(s_0)=: z_0$. We can assume wlog that $t_0 \ne 0$. Otherwise, we reparametrize the curve such that it starts at some other point $\gamma(t_1)$, $t_1 \in (0, 1)$. Note that such a reparametrization (using the term a bit loosely) still preserves winding numbers.

$z_1 \ne z_0 \ne z_2 \implies |z_1 - z_0| \ne 0$ and $|z_2 - z_0| \ne 0$. Let $0 < R < \min(|z_1 - z_0|, |z_2-z_0|)$. $\gamma$ is continous so $\exists(0 < \eta < \min(t_0, 1-t_0)):\ |t-t_0| < \eta \rightarrow |\gamma(t)-z_0| < R$. Let $a, b$ such that $0 < t_0-\eta < a < t_0 < b < t_0+\eta < 1$. We have that $z_1, z_2 \notin \gamma([a, b]) \subset B(z_0, R)$.
$\gamma([0, a])$ and $\gamma([b, 1])$ are the direct images of compact intervals through a continuous function, hence compact sets in $\mathbb{C}$. From the Heine-Borel Theorem we then have that they are closed sets in $\mathbb{C}$. Their union, which we'll denote by $\gamma^\times$ is therefore closed. Then, $\gamma^\times$ contains all of its accumulation points. Whence, $\gamma(t_0)$ is not an accumulation point of $\gamma^\times$ (remember that we defined $t_0$ such that $\gamma$ crosses it exactly once). This means that we can find a disk of radius $r > 0$ around $z_0$ such that $\overline{B(z_0, r)}\ \cap\ \gamma^\times=\varnothing$.

Next up, we bring the two points $z_1, z_2$ closer to $z_0$ along $\delta$. Since $\mathbb{C}\setminus\mathscr{C}(z_0, r)$ is disconnected, it is also path-disconnected, so $\delta$ must intersect it in order to move from $z_1$ or $z_2 \notin B(z_0, r)$ to $z_0 \in B(z_0, r)$. Let $s_1 = \min\{s \in (0, s_0)\ |\ |\delta(s) - z_0| = r\}$ (we know that this $\min$ exists because the set is the inverse image of a compact set through a continuous function, hence compact). Similarly, let $s_2 = \max\{ s \in (s_0, 1)\ |\ |\delta(s)-z_0|=r \}$. Denote $z_1' = \delta(s_1),\ z_2' = \delta(s_2)$. Then, $z_1', z_2' \in \mathscr C(z_0, r)$. Since $\delta|[0, s_1],\ \delta|[s_2, 1]$ link $z_1, z_1'$ and $z_2, z_2'$ without intersecting $\gamma$, they are respectively in the same connected components, so $I(\gamma, z_i') = I(\gamma, z_i),\ i\in\{1,2\}$.

Let $\delta_1$ be a parametrization of the straight line from $z_1'$ to $z_0$ and similarly $\delta_2$ from $z_0$ to $z_2'$. A disk is convex, so $\delta_1^*, \delta_2^* \in \overline{B(z_0, r)}$.

Define $L$ to be a segment centered at $z_0$ of the bisector line of the angle $\measuredangle z_1z_0z_2$ contained in $B(z_0, r)$. The two segments $\delta_1$ and $\delta_2$ separate $B(z_0, r)$ into 2 sectors. For the two arcs between $z_1'$ and $z_2'$ on $\mathscr C(z_0, r)$, $\gamma$ must intersect both of them because otherwise there would be a path on the circle connecting the two points without intersecting $\gamma$, which would mean that the points are in the same connected component of $\mathbb{C}\setminus{\gamma^*}$, which is false. Moreover, the two arcs without the points $z_1, z_2$ are closed sets in $\mathbb{C}\setminus\{z_1, z_2\}$ and the image of the continuous function $\gamma$ does not contain these two points, so the preimages of the arcs are closed sets in $[0, 1]$, thus bounded, thus from H-B, compact, hence, they admit a minimum and a maximum. For each of the two arcs, there's a first and a last time $\gamma$ crosses them, corresponding to entering and leaving the ball $B(z_0, r)$. I am convinced that $\gamma$ enters the ball through one sector and leaves through the other one, but I'm not quite able to prove this for now. So let $t_{\rm in}, t_{\rm out}$, $0 < t_{\rm in} < t_0 < t_{\rm out}< 1$ be the least and the greatest of the points where $\gamma$ intersects $\mathscr C(z_0, r)$.

Define $z_{\rm in}, z_{\rm out}$ as respectively the images of $t_{\rm in}, t_{\rm out}$ through $\gamma$. Denote the endpoints of $L$ as $l_{\rm in}$ and $l_{\rm out}$ according to which sector they are in ($t_{\rm in/out}$ in the same sector as $l_{\rm in/out}$).

For two points $x,y$ on the complex plane, we denote by $[x, y]$ a parametrisation of the line segment from $x$ to $y$ (while keeping the usual meaning for $x, y \in \mathbb{R}$). Also, we use $\cup$ for path concatenation. Now consider the closed curves:
$G_{\rm in} = \gamma\big|[a, t_{\rm in}] \cup [z_{\rm in}, l_{\rm in}] \cup [l_{\rm in}, z_0]$
$G_{\rm out} = [z_0, l_{\rm out}] \cup [l_{\rm out}, z_{\rm out}] \cup \gamma\big|[t_{\rm out}, b]$

$\gamma([a, t_0]), G_{\rm in}, \gamma([t_0, b]), G_{\rm out} \subset B(z_0, R) \not\ni z_1, z_2$ and $B(z_0, R)$ is simply-connected $\implies$ $\gamma([a, t_0]) \sim G_{\rm in}$ and $\gamma([t_0, b]) \sim G_{\rm out}$ are homotopic in $\mathbb C\setminus\{z_1, z_2\}$. We can replace the curve $\gamma$ with the curve $\overline\gamma := \gamma\big|[0, a] \cup G_{\rm in} \cup G_{\rm out} \cup \gamma\big|[b, 1]$. $\gamma \sim \overline\gamma$ in $\mathbb C\setminus\{z_1, z_2\}$, therefore $I(\overline\gamma, z_{1/2}) = I(\gamma, z_{1/2})$. Moreover, $\overline{\gamma}^* \setminus \overline{B(z_0, r)} \subseteq \gamma^* \setminus \overline{B(z_0, r)}$, so $\delta([0, s_0]), \delta([s_1, 1]) \notin \overline{\gamma}^*$. This means that $z_1', z_1$ and $z_2', z_2$ are still respectively in the same connected components of $\mathbb C\setminus \overline{\gamma}^*$. Then $I(\overline\gamma, z_{1/2}') = I(\overline\gamma, z_{1/2}) = I(\gamma, z_{1/2})$.

From this point onward we can use the method of proof presented in the refferenced book.

Via a geometric argument we see that there is a ball $B(z_0, r') \subset B(z_0, r)$ that is disjoint from $[l_{\rm in/out}, z_{\rm in/out}]$. Altenatively, $B(z_0, r)$ minus the two segments is an open set and $z_0$ is inside it. The boundary of this ball, $\mathscr C(z_0, r')$ is cut in half by $L$ and $z_1', z_2'$ are on either side of it. If we take the half that is intersected by $[z_0, z_2']$, we can replace the portion of $L$ that is inside $B(z_0, r')$ by it via an homotopy because both curves are inside $B(z_0, r)$ which is simply-connected and doesn't contain $z_1', z_2'$. Call this new curve $\widetilde{\gamma}$. This has the same winding numbers around $z_0, z_2'$ as $\overline\gamma$ around $z_1', z_2'$.

Choose a point $z_3$ on $[z_0, z_2']^* \cap B(z_0, r')$ close enough to $z_0$ that there is a circle around it, $\mathscr C(z_3, \rho)$, that is tangent to $L$ and is contained inside this half of $B(z_0, r')$. We have $I(\tilde\gamma, z_3) = I(\tilde\gamma, z_0) = I(\overline\gamma, z_1')$ and $I(\tilde\gamma, z_2') = I(\overline\gamma, z_2') = I(\overline\gamma, z_3)$. Construct the curve $\mathring\gamma$ by taking $\overline\gamma$ up to $z_0$ followed by the boundary of $B(z_3, \rho)$ traversed in opposite direction to $L$ (the opposite sense to the one of traversing $\mathscr C(z_0, r)$ going through $z_{\rm in}, z_1'$ and $z_{\rm out}$ in this order), then the rest of $\overline\gamma$. It is not very hard to see that $\tilde\gamma \sim \mathring\gamma$ in $\mathbb C \setminus \{z_3, z_2'\}$ (We can build the homotopy by considering segments from the intersection point of $\mathscr C(z_3, \rho)$ with $L$ to points on the semi-circle around it from $\tilde\gamma$ and dragging those points to the intersections of the segments with $\mathscr C(z_3, \rho)$ along the segments, with the same speed. At the same time, we extend either side of $L \setminus \overline{B(z_0, r')}$ up to where we drag the points $L \cap \mathscr C(z_0, r')$.)

This all means that $I(\gamma, z_1) = I(\mathring\gamma, z_3)$ and $I(\gamma, z_2) = I(\overline\gamma, z_3)$. Via a lemma that is easy to prove, as $\mathring\gamma$ is the concatenation of $\overline\gamma$ with $C = \partial B(z_3, \rho)$ under a change of starting point to $z_0$, $I(\mathring\gamma, z_3) = I(\overline\gamma, z_3) + I(C, z_3) \implies I(\gamma, z_1) = I(\gamma, z_2) + I(C, z3)$. Depending on the orientation of $C$, $I(C, z_3) = \pm 1$. If $\gamma$ is oriented such that $z_1$ is on its left, then $z_1'$ is on the left of $L$ (I'm not too sure whether "$z_1$/the connecetd component of $z_1$ is on the left side of $\gamma$" makes much sense generally, but in cases where it doesn't, I would argue that this works as a definition) which means that $z_{\rm in}, z_1'$ and $z_{\rm out}$ are traversed in this order clockwise, so $C$ is traversed counter-clockwise. This being said, we can finally conclude that $I(\gamma, z_1) = I(\gamma, z_2) - 1$.

Sidenote: I've gone through the last part of the proof (the one adapted from the proof in Tristan Needham's book) in a faster pace because, even though it is fundamental to the proof as a whole, my question was about the setup I got before that part, and also because I've been writing on this proof for more than a week now and I really want to call it quits and not waste any more time on it. If anyone could help me complete the missing detail from the first part, I'd be thankful.


I will try to answer your question and I expect this to be useful for you. There is a theorem called Jordan curve theorem which states that any closed and simple curve determines uniquely two different regions of the complex plane (one is the interior of the curve, and the other is an unbounded component), then it seems that the theorem is relating the winding number of two different points given on the different regions determined by the curve. The first thing I'll try to justify is that you can put the points and the curve as in the figure (actually, the curve will not be modified), I will only give an intuitive reason why this is true: remember that $\nu (K,r)= \int_K \frac{1}{z-r} \,dz$, now in this expression take $r$ as a variable that moves only in the region where $r$ was given. It is a consequence of the general Cauchy theorem that as long as you remain on this region the winding number remains the same, I'm sure that if you look in any good complex analysis book you'll find a proof (seek for homology), but I expect it is not difficult just to believe it for now. So we can get $r$ closer to an arc of $K$ staying always in the same original region that it was given, and since $K$ is smooth (or piecewise), by getting $r$ close enough to $K$ eventually the arc will seem to be like a straight line. Now do the same with $s$, get $s$ closer to $r$, since I am assuming that $K$ is simple then this is in fact possible (but I will not prove it). Then we have now another two points $r,s$ with the same winding numbers than the original ones and which also look like in the figure. So we can assume the original points and the curve look as we want to. Now focus only on the third and fourth images of the figure. Take the curve of the third image and start to bring closer the two sharp points in order to get the fourth curve in the limit. Let's call $M_n$ a sequence of figures like in the third image which in the limit converges to the figure $K+L$ in the fourth image. Since the function $\frac{1}{z-s}$ is uniformly continuous in any compact set not containing $s$, then if take a compact set containing $K+L$ in it's interior, it is possible to say that the integrals $\int_{M_n} \frac{dz}{z-s}$ converge to $\int_{K+L} \frac{dz}{z-s}$. But all the first integrals are simply equal to $\nu (K,r)$ which is constant. And the limit is equal to $\nu (K,s)+\nu (L,s)$, so we get the equality between these values.

  • $\begingroup$ I'm sorry but this is not useful at all to me. Please note that nowhere in the question did I say that the curve must be piecewise smooth. I am looking to formalise the proof for general continuous curves. Thus, the integral expression of the winding number is not valid here. Secondly, I explicity said in the title of the question that I need help in formalizing the proof, that is, I do understand it intuitively, I don't need an intuitive explanation. There are several more aspects that you don't take into account as well, like self-intersection of the curve near the crossing point. $\endgroup$ Aug 3, 2021 at 7:44
  • $\begingroup$ Ok, but actually almost in any complex analysis book line integrals, and therefore winding numbers, are defined for piecewise smooth curves, so if you want to see a proof for this fact with continuous assumption for the curve only then you must not search the answer in complex analysis, but in topology instead. Secondly, I believe that what I wrote is only intuitive in the sense that I'm using results that maybe you have not studied yet, but I mentioned them explicitly just for you to search and read them, and I you do so you would see the argument need no more "formal" justification. $\endgroup$
    – en3trix
    Aug 3, 2021 at 16:00
  • $\begingroup$ The last part of my answer, which was about justifying the integrals convergence, neither needs any more justification at all, since it gives the complete reasons why the integrals do converge, so this is not intuitive (yes, assuming always being dealing with a piecewise smooth curve). Finally, the result need no be true for non simple curves, but is easy to see that the assumption you made in your question reduce it to the case in which the curve is simple (since the hypothesis implies the points are given in adjacent components), for this you only have to study the Jordan curve theorem. $\endgroup$
    – en3trix
    Aug 3, 2021 at 16:02
  • $\begingroup$ In my vision, complex analysis and topology go hand-in-hand. It's really hard to study complex analysis without topology. The definition I use for the winding number is the variation of argument one if you know it. That seems to me like the most intuitively logical definition. For proofs of results, I almost never take things for granted. I don't like just understanding the theorem intuitively, I want to takle it down and settle the proof in stone. I think I found the answer to my question by myself and I will add it here as an answer. You can take a look when I'm done. $\endgroup$ Aug 3, 2021 at 16:07
  • $\begingroup$ It's true that topology is fundamental for complex analysis, though what I meant was that the problem for continuous curves is not of special concern of complex analysis but of topology and general homology theory. However, if you can "settle a proof in stone" by using complex analysis only, I would be glad to read it. $\endgroup$
    – en3trix
    Aug 3, 2021 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.