# Definitions

The argument of a non-zero complex number is only defined modulo $$2\pi.$$ A convenient way to describe mathematically this relationship is to associate to any such number the set of admissible values of its argument:

### Definition – The Argument Function.

‌ The set-valued (or multi-valued) function $$\mathrm{Arg},$$ defined on $$\mathbb{C}^{*}$$ by $\mathrm{Arg} \, z = \left\{ \theta \in \mathbb{R} \; \left| \; e^{i\theta} = \frac{z}{|z|} \right. \right\},$ is called the argument function.

If we need a classic single-valued function instead, we have for example:

### Definition – Principal Value of the Argument.

‌ The principal value of the argument is the unique continuous function $\mathrm{arg}: \mathbb{C} \setminus \mathbb{R}_- \to \mathbb{R}$ such that $\arg \, 1 = 0$ which is a choice of the argument on its domain: $\forall \, z \in \mathbb{C} \setminus \mathbb{R}_-, \; \mathrm{arg}\, z \in \mathrm{Arg} \, z.$

### Proof (existence and uniqueness).

‌ Define $$\mathrm{arg}$$ on $$\mathbb{C} \setminus \mathbb{R}_- \to \mathbb{R}$$ by: $\arg (x + i y) = \left| \begin{array}{cl} \arctan y/x & \mbox{if } x > 0, \\ +\pi/2 - \arctan x/y & \mbox{if } y > 0, \\ -\pi/2 - \arctan x/y & \mbox{if } y < 0. \\ \end{array} \right.$ This definition is non-ambiguous: if $$x>0$$ and $$y>0,$$ we have $\arctan x/y + \arctan y/x = \pi/2$ and a similar equality holds when $$x>0$$ and $$y<0.$$ As each of the three expressions used to define $$\mathrm{arg}$$ has an open domain and is continuous, the function itself is continuous. It is a choice of the argument thanks to the definition of $$\arctan$$: for example, if $$x>0,$$ with $$\theta=\arg(x+iy),$$ we have $\frac{\sin\theta}{\cos\theta}=\tan \theta = \tan (\arctan y/x) = \frac{y}{x},$ hence, as $$\cos \theta > 0$$ and $$x>0,$$ there is a $$\lambda > 0$$ such that $x + i y = \lambda (\cos \theta + i \sin \theta) = \lambda e^{i\theta},$ This equation yields $$\arg x+ i y \in \mathrm{Arg} \, x + iy.$$ The proof for the half-planes $$y>0$$ and $$y<0$$ is similar.

If $$f$$ is another continuous choice of the argument on $$\mathbb{C} \setminus \mathbb{R}_-$$ such that $$f(1) = 0,$$ the image of $$\mathbb{C} \setminus \mathbb{R}_-$$ by the difference $$f - \arg$$ is a subset of $$2\pi \mathbb{Z}$$ that contains $$0,$$ and it’s also path-connected as the image of a path-connected set by a continuous function. Consequently, it is the singleton $$\{0\}$$: $$f$$ and $$\arg$$ are equal. $$\blacksquare$$

We cannot avoid the introduction of a cut in the complex plane when we search for a continuous choice of the argument: there is no continuous choice of the argument on $$\mathbb{C}^*.$$ However, for a continuous choice of the argument along a path of $$\mathbb{C}^*,$$ there is no such restriction:

The following theorem is a special case of the path lifting property (in the context of covering spaces; refer to (Hatcher 2002) for details).

### Theorem – Continuous Choice of the Argument.

‌ Let $$a \in \mathbb{C}$$ and $$\gamma$$ be a path of $$\mathbb{C} \setminus \{a\}.$$ Let $$\theta_0 \in \mathbb{R}$$ be a value of the argument of $$\gamma(0) - a$$: $\theta_0 \in \mathrm{Arg}(\gamma(0) - a).$ There is a unique continous function $$\theta:[0,1] \mapsto \mathbb{R}$$ such that $$\theta(0) = \theta_0$$ which is a choice of $$z \mapsto \mathrm{Arg}(z-a)$$ on $$\gamma$$: $\forall \, t \in [0,1], \; \theta(t) \in \mathrm{Arg}(\gamma(t) - a).$

### Proof.

‌ Let $$(x(t), y(t))$$ be the cartesian coordinates of $$\gamma(t)$$ in the system with origin $$a$$ and basis $$(e^{i\theta_0}, i e^{i\theta_0}).$$ As long as $$x(t) > 0,$$ the function $t \mapsto \theta_0 + \arg(x(t)+iy(t))$ is a continuous choice of the argument of $$\gamma(t) - a.$$ Let $$d$$ be the distance between $$a$$ and $$\gamma([0,1])$$ and let $$n \in \mathbb{N}$$ such that $|t - s| \leq 2^{-n} \, \Rightarrow \, |\gamma(t) - \gamma(s)| < d.$ The condition $$x(t) > 0$$ is ensured for any $$t$$ in $$[0, 2^{-n}].$$ This construction of a continuous choice may be iterated locally on every interval $$[k2^{-n}, (k+1)2^{-n}]$$ with a new coordinate system to provide a global continuous choice of the argument on $$[0,1].$$

The uniqueness of a continuous choice is a consequence of the intermediate value theorem: if we assume that there are two such functions $$\theta_1$$ and $$\theta_2$$ with the same initial value $$\theta_0,$$ as $$\theta_1(0) - \theta_2(0)=0,$$ if $$\theta_1(t) - \theta_2(t) \neq 0$$ for some $$t\in[0,1],$$ then either $$|\theta_1(t) - \theta_2(t)| < \pi,$$ or there is a $$\tau \in \left]0,t\right[$$ such that $$\theta_1(\tau) - \theta_2(\tau) \neq 0$$ and $$|\theta_1(\tau) -\theta_2(\tau)| < \pi.$$ In any case, there is a contradiction since all values of the argument differ of a multiple of $$2\pi.$$ $$\blacksquare$$

### Definition – Variation of the Argument.

‌ Let $$a \in \mathbb{C}$$ and $$\gamma$$ be a path of $$\mathbb{C} \setminus \{a\}.$$ The variation of $$z \mapsto \mathrm{Arg}(z-a)$$ on $$\gamma$$ is defined as $[z \mapsto \mathrm{Arg}(z-a)]_{\gamma} = \theta(1) - \theta(0)$ where $$\theta$$ is a continous choice of $$z \mapsto \mathrm{Arg}(z-a)$$ on $$\gamma.$$

### Proof (unambiguous definition).

‌ If $$\theta_1$$ and $$\theta_2$$ are two continuous choices of $$z \mapsto \mathrm{Arg}(z-a)$$ on $$\gamma,$$ for any $$t \in [0,1],$$ they differ of a multiple of $$2\pi.$$ As the function $$\theta_1 - \theta_2$$ is continuous, by the intermediate value theorem, it is constant. Hence $(\theta_1 - \theta_2)(1) = (\theta_1 - \theta_2)(0),$ and $$\theta_1(1) - \theta_1(0) = \theta_2(1) - \theta_2(0).$$ $$\blacksquare$$

### Definition – Winding Number / Index.

‌ Let $$a \in \mathbb{C}$$ and $$\gamma$$ be a closed path of $$\mathbb{C} \setminus \{a\}.$$ The winding number – or index – of $$\gamma$$ around $$a$$ is the integer $\mathrm{ind}(\gamma,a) = \frac{1}{2\pi}[z\mapsto \mathrm{Arg}(z-a)]_{\gamma}.$

### Proof – The Winding Number is an Integer.

‌ Let $$\theta$$ be a continuous choice function of $$z \mapsto \mathrm{Arg}(z-a)$$ on $$\gamma;$$ as the path $$\gamma$$ is closed, $$\theta(0)$$ and $$\theta(1),$$ which are values of the argument of $$\gamma(0)-a = \gamma(1)-a,$$ are equal modulo $$2\pi,$$ hence $$(\theta(1) - \theta(0))/2\pi$$ is an integer. $$\blacksquare$$

### Definition – Path Exterior & Interior.

‌ The exterior and interior of a closed path $$\gamma$$ are the subsets of the complex plane defined by $\mathrm{Ext} \, \gamma = \{ z \in \mathbb{C} \setminus \gamma([0,1]) \; | \; \mathrm{ind}(\gamma, z) = 0 \}.$ and $\mathrm{Int} \, \gamma = \mathbb{C} \setminus (\gamma([0,1]) \cup \mathrm{Ext} \, \gamma) = \{ z \in \mathbb{C} \setminus \gamma([0,1]) \; | \; \mathrm{ind}(\gamma, z) \neq 0 \}.$

# Properties

### Theorem – The Winding Number is Locally Constant.

‌ Let $$a \in \mathbb{C}$$ and $$\gamma$$ be a closed path of $$\mathbb{C} \setminus \{a\}.$$ There is a $$\epsilon > 0$$ such that, for any $$b \in \mathbb{C}$$ and any closed path $$\beta,$$ if $|b - a| < \epsilon \; \mbox{ and } \; (\forall \, t \in [0,1], \; |\beta(t) - \gamma(t)| < \epsilon)$ then $$\beta$$ is a path of $$\mathbb{C} \setminus \{b\}$$ and $\mathrm{ind}(\gamma,a) = \mathrm{ind}(\beta,b).$

### Proof.

‌ Let $$\epsilon = d(a, \gamma([0,1])) /2.$$ If $$|b - a| < \epsilon$$ and for any $$t \in [0,1],$$ $$|\gamma(t) - \beta(t)| < \epsilon,$$ then clearly $$b \in \mathbb{C} \setminus \beta([0,1]).$$ Additionally, for any $$t \in [0,1]$$ there are values $$\theta_1$$ of $$\mathrm{Arg} (\gamma(t)-a)$$ and $$\theta_2$$ of $$\mathrm{Arg}(\beta(t)-b)$$ such that $$|\theta_1 - \theta_2| < \pi/2.$$ If we select some values $$\theta_{1,0}$$ of $$\mathrm{Arg} (\gamma(0) - a)$$ and $$\theta_{2,0}$$ of $$\mathrm{Arg} (\beta(0) -b)$$ such that $$|\theta_{1,0} - \theta_{2,0}| < \pi/2,$$ then the corresponding continuous choices $$\theta_1$$ et $$\theta_2$$ satisfy $$|\theta_1(t) - \theta_2(t)| < \pi/2$$ for any $$t \in [0,1]$$(1). Consequently $|\mathrm{ind}(\gamma,a) - \mathrm{ind}(\beta,b)| = \left| \frac{\theta_1(1) - \theta_1(0)}{2\pi} - \frac{\theta_2(1) - \theta_2(0)}{2\pi} \right| < \frac{1}{2}.$ As both winding numbers are integers, they are equal. $$\blacksquare$$

### Corollary – The Winding Number is Constant on Components.

‌ Let $$\gamma$$ be a closed path. The function $z \in \mathbb{C} \setminus \gamma([0,1]) \mapsto \mathrm{ind}(\gamma,z)$ is constant on each component of $$\mathbb{C} \setminus \gamma([0,1])$$. If additionally the component is unbounded, the value of the winding number is zero.

### Proof.

‌ The mapping $$z \mapsto \mathrm{ind}(\gamma, z)$$ is locally constant – and hence constant – on every connected component of $$\mathbb{C} \setminus \gamma([0,1]).$$ If $$a$$ belongs to some unbounded component of this set, there is a $$b$$ in the same component such that $$|b|> r = \max_{t \in [0,1]} |\gamma(t)|.$$ It is possible to connect $$b$$ to any point $$c$$ such that $$|c| = r$$ by a circular path in $$\mathbb{C} \setminus \gamma([0,1]),$$ thus we may assume that $$b \in \mathbb{R}_-.$$ The function $\theta: t \in [0,1] \mapsto \mathrm{arg}(\gamma(t) - b)$ is a continuous choice of $$z \mapsto \mathrm{Arg}(z-b)$$ along $$\gamma$$ and it satisfies $\forall \, t \in [0,1], \; |\theta(t)| = \arctan \frac{\mathrm{Im} (\gamma(t) - b)}{\mathrm{Re} (\gamma(t) -b )} < \arctan \frac{r}{|b| - r} < \frac{\pi}{2}.$ As $$\gamma$$ is a closed path, $$\theta(0)$$ and $$\theta(1)$$ – which are equal modulo $$2\pi$$ – are actually equal and $\mathrm{ind}(\gamma, a) = \mathrm{ind}(\gamma, b) = \frac{\theta(1) - \theta(0)}{2\pi} =0$ as expected. $$\blacksquare$$

# Simply Connected Sets

### Definition – Simply/Multiply Connected Set & Holes.‌

‌ Let $$\Omega$$ be an open subset of the plane. A hole of $$\Omega$$ is a bounded component of its complement $$\mathbb{C} \setminus \Omega.$$ The set $$\Omega$$ is simply connected if it has no hole (if every component of its complement is unbounded) and multiply connected otherwise.

### Examples.

1. The open set $$\Omega = \{(x,y) \in \mathbb{R}^2 \; | \; x < -1 \, \mbox{ or }\, x>1\}$$ is not connected but it is simply connected: its complement has a unique component which is unbounded, hence it has no holes.

2. The open set $$\Omega = \mathbb{C} \setminus \overline{\{2^{-n} \, | \, n \in \mathbb{N}\}}$$ is multiply connected: its holes are exactly the singletons of its complement.

Intuitively, we should be able to circle around any hole of $$\Omega$$ without leaving the set; this idea leads to an alternate characterization of simply connected sets.

### Theorem – Simply Connected Sets & The Winding Number.

‌ An open subset $$\Omega$$ of the complex plane is simply connected if and only if the interior of any closed path $$\gamma$$ of $$\Omega$$ is included in $$\Omega$$: $\forall \, z \in \mathbb{C} \setminus \gamma([0,1]),\; \mathrm{ind}(\gamma, z) \neq 0 \, \Rightarrow \, z \in \Omega,$ or equivalently, if the complement of $$\Omega$$ is included in the exterior of $$\gamma$$: $\forall \, z \in \mathbb{C} \setminus \Omega, \; \mathrm{ind}(\gamma, z) = 0.$

### Examples.

1. If $$\gamma$$ is a closed path of $$\Omega = \{(x,y) \in \mathbb{R}^2 \; | \; x < -1 \, \mbox{ or }\, x>1\}$$ and $$z \in \mathbb{C} \setminus \Omega$$, since $$\mathbb{C} \setminus \Omega$$ is connected and unbounded, $$z$$ belongs to an unbounded component of $$\mathbb{C} \setminus \gamma([0,1]).$$ Thus $$\mathrm{ind}(\gamma, z)=0$$ for any $$z \in \mathbb{C} \setminus \Omega.$$

2. The open set $$\Omega = \mathbb{C} \setminus \overline{\{2^{-n} \, | \, n \in \mathbb{N}\}}$$ is open and multiply connected: for example $$\gamma = 1 + 1/4 [\circlearrowleft]$$ is a path of $$\Omega$$, $$z=1$$ is a point of $$\mathbb{C} \setminus \Omega$$ and $$\mathrm{ind}(\gamma, 1) = 1.$$

### Remark.

‌Note that we may not always be able to encircle only one hole at a time. For example, in the case of the set $$\Omega = \mathbb{C} \setminus \overline{\{2^{-n} \, | \, n \in \mathbb{N}\}},$$ we can find a closed path $$\gamma$$ of $$\mathbb{C} \setminus \Omega$$ such that $$\mathrm{ind}(\gamma, 0)=1,$$ but then we also have $$\mathrm{ind}(\gamma, 2^{-n})=1$$ for $$n$$ large enough: we cannot encircle the hole $$\{0\}$$ of $$\Omega$$ unless we also encircle an infinity of extra holes.

### Lemma.

‌ If the compact set $$K$$ is a hole of the open set $$\Omega$$, there is a compact subset $$L$$ of $$\mathbb{C} \setminus \Omega$$ such that $$K\subset L$$ and $$d(L, (\mathbb{C} \setminus \Omega) \setminus L) > 0.$$

### Proof of the Lemma.

‌ The set $$C = \mathbb{C} \setminus \Omega$$ is closed in $$\mathbb{C}$$ which is locally compact, thus it is locally compact. Since $$\mathbb{C}$$ is Hausdorff, its subspace $$C$$ is also Hausdorff. By the Šura-Bura theorem (Remmert 1998, 304), $$K$$ has a neighbourhood base in $$C$$ consisting in open compact subsets of $$C$$. Since $$C$$ is a neighbourhood of $$K$$ in $$C$$, there is a compact set $$L$$ such that $$K \subset L \subset C$$ which is open in $$C$$. Thus, its complement $$(\mathbb{C} \setminus \Omega) \setminus L$$ is also closed in $$C$$. Since $$C$$ is closed, both sets are also closed in $$\mathbb{C}$$. They are also disjoint by construction, and thus $$d(L, (\mathbb{C} \setminus \Omega) \setminus L) > 0.$$ $$\blacksquare$$

### Proof – Simply Connected Sets & The Winding Number.

‌ Assume that $$\Omega$$ is simply connected and let $$\gamma$$ be a closed path of $$\Omega$$. Let $$z \in \mathbb{C} \setminus \Omega;$$ this point belongs to an unbounded connected component of $$\mathbb{C} \setminus \Omega$$ and therefore to an unbounded connected component of $$\mathbb{C} \setminus \gamma([0,1]),$$ thus $$\mathrm{ind}(\gamma, z)=0.$$

Conversely, if $$\Omega$$ is not simply connected, the set $$\mathbb{C} \setminus \Omega$$ has a hole $$K$$ which is contained in some compact subset $$L$$ of $$\mathbb{C} \setminus \Omega$$ such that the distance $$\epsilon$$ between $$L$$ and $$(\mathbb{C} \setminus \Omega) \setminus L$$ is positive. Let $$r < \epsilon/\sqrt{2};$$ Define for any pair $$(k, l)$$ of integers the node $$n_{k,l} = (k + i l)r$$ and $$S_{k,l}$$ as the closed square with vertices $$n_{k,l},$$ $$n_{k+1,l},$$ $$n_{k+1,l+1}$$ and $$n_{k,l+1}.$$ The (positively) oriented boundary of the square $$S_{k,l}$$ is the polyline $[n_{k,l} \to n_{k+1,l} \to n_{k+1,l+1} \to n_{k,l+1} \to n_{k,l}]$ The collection of squares that intersect $$L$$ is finite and covers $$L.$$ Additionally, all of its squares are included in $$\Omega \cup L.$$

For any square $$S$$ in the cover of $$L$$ and any interior point $$a$$ of $$S$$ if $$\gamma$$ is the oriented boundary of $$S,$$ then $$\mathrm{ind}(\gamma, a) = 1.$$ Additionally, $$\mathrm{ind}(\mu, a) = 0$$ for the oriented boundary $$\mu$$ of any other square in the collection. Consequently, if $$\Gamma$$ denotes the collection of oriented line segments that composes the oriented boundaries of all squares of the cover of $$L,$$ we have $\sum_{\gamma \in \Gamma}\frac{1}{2\pi} [z \mapsto \mathrm{Arg}(z-a)]_{\gamma} = 1.$ Now if the line segment $$\gamma$$ belongs to $$\Gamma$$ and $$\gamma([0,1]) \cap L \neq \varnothing,$$ then $$\gamma^{\leftarrow}$$ also belongs to $$\Gamma;$$ if we remove all such pairs from $$\Gamma,$$ the resulting collection $$\Gamma'$$ also satisfies $\sum_{\gamma \in \Gamma'}\frac{1}{2\pi} [z \mapsto \mathrm{Arg}(z-a)]_{\gamma} = 1.$ and by construction the image of any $$\gamma$$ in $$\Gamma'$$ is included in $$\Omega.$$ The original collection $$\Gamma$$ is balanced: for any square vertice $$n,$$ the number of line segments with $$n$$ as an initial point and with $$n$$ as a terminal point is the same. The collection $$\Gamma'$$ has the same property. Consequently, the line segments of $$\Gamma'$$ may be assembled in a finite sequence of closed paths $$\gamma_1,$$ $$\dots,$$ $$\gamma_n$$ and $\sum_{k=1}^n \mathrm{ind}(\gamma_k, a) = 1.$ Every point of $$L$$ is either an interior point of some square of the collection, or the limit of such point; anyway, that means that $\forall \, z \in L, \; \sum_{k=1}^n \mathrm{ind}(\gamma_k, z) = 1$ and thus that there is at least one path $$\gamma_k$$ such that $$\mathrm{ind}(\gamma_k, z) \neq 0.$$ $$\blacksquare$$

# A Complex Analytic Approach

If a closed path is rectifiable, we may compute its winding number as a line integral; to prove this, we need the:

### Lemma.

‌ Let $$a \in \mathbb{C}$$ and $$\gamma$$ be a rectifiable path of $$\mathbb{C} \setminus \{a\}.$$ For any $$t \in [0,1],$$ let $$\gamma_t$$ be the path such that for any $$s\in[0,1],$$ $$\gamma_t(s) = \gamma(t s).$$ The function $$\mu:[0,1]\to \mathbb{C},$$ defined by $\mu(t) = \int_{\gamma_t} \frac{dz}{z-a}$ satisfies $\exists \, \lambda \in \mathbb{C}^*, \, \forall \, t \in [0,1], \, e^{\mu(t)} = \lambda \times (\gamma(t) - a).$

### Proof.

‌ We only prove the lemma under the assumption that $$\gamma$$ is continuously differentiable; the rectifiable case is a straightforward extension.

We have for any $$t\in [0,1]$$ $\mu(t) = \int_{\gamma_t} \frac{dz}{z-a} = \int_0^1 \frac{\gamma'(ts) \times t}{\gamma(t s) - a} ds = \int_0^t \frac{\gamma'(s)}{\gamma(s) - a} ds,$ hence $\mu'(t) = \frac{\gamma'(t)}{\gamma(t) - a}$ and the derivative of the quotient $$\phi(t) = {e^{\mu(t)}}/{(\gamma(t)}-a)$$ satisfies $\phi'(t) = \mu'(t) \phi(t) - \frac{\gamma'(t)}{\gamma(t) - a} \phi(t) = 0$ which yields the result. $$\blacksquare$$

### Theorem – The Winding Number as a Line Integral.

‌ Let $$a \in \mathbb{C}$$ and $$\gamma$$ be a rectifiable path of $$\mathbb{C} \setminus \{a\}.$$ Then $[z \mapsto \mathrm{Arg} (z-a)]_{\gamma} = \mathrm{Im} \left( \int_{\gamma} \frac{dz}{z - a} \right).$ If the path $$\gamma$$ is closed, then $\mathrm{ind}(\gamma,a) = \frac{1}{i2\pi} \int_{\gamma} \frac{dz}{z - a}.$

### Proof.

‌We use the function $$\mu$$ of the previous lemma. Applying the modulus to both sides of the equation $$e^{\mu(t)} = \lambda \times (\gamma(t) - a)$$ provides $$e^{\mathrm{Re}(\mu(t))} = |\lambda| \times |\gamma(t) - a|,$$ hence $e^{i \mathrm{Im}(\mu(t))} = \frac{\lambda}{|\lambda|} \frac{\gamma(t) - a}{|\gamma(t) - a|}.$ The function $$t \in [0,1] \mapsto \mathrm{Im}(\mu(t))$$ is – up to a constant – a continuous choice of $$z \mapsto \mathrm{Arg}\, (z-a)$$ on $$\gamma.$$ Consequently, $[z \mapsto \mathrm{Arg} \, (z-a)]_{\gamma} = \mathrm{Im}(\mu(1)) - \mathrm{Im}(\mu(0)) = \mathrm{Im}(\mu(1)),$ which is the desired result.

If additionally $$\gamma$$ is a closed path, the equations $\gamma(0) = \gamma(1) \; \mbox{ and } \; e^{\mathrm{Re}(\mu(t))} = |\lambda| \times |\gamma(t) - a|$ yield $$e^{\mathrm{Re}(\mu(0))}= e^{\mathrm{Re}(\mu(1))}$$ and hence $$\mathrm{Re}(\mu(1)) = \mathrm{Re}(\mu(0)) = 0.$$ Thus, $\mathrm{ind}(\gamma,a) = \frac{1}{2\pi} \mathrm{Im}(\mu(1)) = \frac{1}{i2\pi} \mu(1),$ which concludes the proof. $$\blacksquare$$