# Are winding number and index of a not smooth closed curve the same?

Let $$\gamma:[0,1] \longrightarrow \mathbf{C} \backslash \{0\}$$ be a closed curve (continuous and of bounded variation). We call $$\operatorname{Ind}_\gamma(0) \overset{\mathrm{def}}{=} \frac{1}{2 \pi i}\int_\gamma \frac{1}{z}\ dz.$$ $$\textbf{the index of \gamma around 0.}$$ As for the winding number, we can take a pair of continuous real valued functions $$r$$ and $$\theta$$ that satisfies $$r:[0,1] \longrightarrow (0,\infty),$$ $$\theta:[0,1] \longrightarrow \mathbf{R},$$ $$\forall t \in [0,1]\, \left[\, \gamma(t) = r(t) \cdot e^{i \cdot \theta(t)}\, \right],$$ and $$-\pi < \theta(0) \leq \pi.$$ (The proof of this can be found in chapter 7 of A. F. Beardon, Complex Analysis: The Argument Principle in Analysis and Topology, Wiley-Interscience publication 1979). We call $$\operatorname{Wnd}_\gamma(0) \overset{\mathrm{def}}{=}\frac{\theta(1) - \theta(0)}{2\pi}$$ $$\textbf{the winding number of \gamma around 0}$$.

It is already known that if $$\gamma$$ is $$C^1$$-curve, then $$\operatorname{Ind}_\gamma(0) = \operatorname{Wnd}_\gamma(0)$$ (see definition of winding number, have doubt in definition.).

My question: When $$\gamma$$ is not $$C^1$$ path, $$\operatorname{Ind}_\gamma(0) = \operatorname{Wnd}_\gamma(0)?$$

Beardon's book is sadly out of print, but detailed proofs of results concerning the winding number, or the index [incidentally, I don't think that one can distinguish these two terms, at least not in the way attempted in the question, although the same number associated with a plane curve is defined in the literature in two different ways, and the question is therefore an interesting one], can also be found in Chapter 3 of Ash and Novinger, Complex Variables (2004), which is downloadable here.

Because of the importance of the concept, I think it is worth giving a fairly short and straightforward proof from first principles, rather than appealing to a series of general theorems (although that is admittedly more efficient, and it would probably also be worth giving a direct proof of the "fundamental theorem of calculus" for rectifiable paths).

Digression: This is especially so as it is not easy even to find books that bother to treat the case of rectifiable curves. The best-known exception may be Conway, Functions of One Complex Variable I (2nd ed. 1978). Another exception is Garling, A Course in Mathematical Analysis, Vol. III (2014). But even these two books prove the equality in question by using piecewise continuously differentiable approximations to rectifiable curves; and so it is hard, for me at least, to see the wood for the trees. Rudin, in Principles of Mathematical Analysis (3rd ed. 1976), remarks of the concept of rectifiability: "The case $$k = 2$$ (i.e., the case of plane curves) is of considerable importance in the study of analytic functions of a complex variable." In contrast, in Real and Complex Analysis (3rd ed. 1987), the same author writes: "In this section the required integration theory will be developed; we shall keep it as simple as possible, and shall regard it merely as a useful tool in the investigation of properties of holomorphic functions. [...] A path is a piecewise continuously differentiable curve in the plane." It the latter, more utilitarian view of curves that seems to dominate the literature on complex analysis. (There are some remarks about this in Burckel, An Introduction to Classical Complex Analysis, Vol. 1; but I haven't got that book yet, so I can't say any more about it.) (End of amateur lecture.) :)

Let $$[a, b]$$ be a compact interval of $$\mathbb{R}$$, and $$\gamma \colon [a, b] \to \mathbb{C}$$ a continuous function. The image set $$[\gamma] = \gamma([a, b])$$ is a compact subset of $$\mathbb{C}$$, therefore it is closed. Suppose $$0 \notin [\gamma]$$. Because $$[\gamma]$$ is closed, there exists $$R > 0$$ such that the open disc $$\mathbb{C}(0, R)$$, with centre $$0$$ and radius $$R$$, does not intersect $$[\gamma]$$.

Because $$\gamma$$ is a continuous function on a compact set, it is uniformly continuous. So there exists a partition of $$[a, b]$$, $$$$\tag{1}\label{eq:3257640:part:2} a = t_0 < t_1 < \cdots < t_{n-1} < t_n = b,$$$$ such that $$|\gamma(t) - \gamma(t')| < R \quad (t_{j-1} \leqslant t < t' \leqslant t_j, \ 1 \leqslant j \leqslant n).$$ Therefore, each of the sets $$\gamma([t_{j-1}, t_j])$$ is contained in an open disc (of radius $$R$$) not containing the origin. Because there exists a continuous choice of logarithm on any such disc, a simple inductive proof shows that there exists a continuous function $$\lambda \colon [a, b] \to \mathbb{C}$$ such that $$\gamma(t) = e^{\lambda(t)}$$ for all $$t \in [a, b]$$.

Recall that $$e^z = 1 + z + \rho(z)$$, where $$\rho(z)/z \to 0$$ as $$z \to 0$$.

Take $$\epsilon > 0$$ such that $$|\rho(z)| < |z|/2$$ when $$0 < |z| < \epsilon$$. (For example, we could take $$\epsilon = \tfrac{3}{4}$$. Not that it matters, of course!) Because $$\lambda$$ is uniformly continuous, there exists $$\delta > 0$$ such that $$|\lambda(t) - \lambda(t')| < \epsilon$$ when $$|t - t'| < \delta$$. Therefore, \begin{align*} |\gamma(t) - \gamma(t')| & = |e^{\lambda(t)} - e^{\lambda(t')}| \\ & = |e^{\lambda(t')}||e^{\lambda(t) - \lambda(t')} - 1| \\ & = |\gamma(t')||\lambda(t) - \lambda(t') + \rho(\lambda(t) - \lambda(t'))| \\ & \geqslant \frac{R|\lambda(t) - \lambda(t')|}{2} \quad (|t - t'| < \delta). \end{align*} In a partition \eqref{eq:3257640:part:2} (not necessarily with the same values of $$t_j$$ as before, of course), if $$|t_j - t_{j-1}| < \delta$$ ($$j = 1, \ldots, n$$), then $$\sum_{j=1}^n|\lambda(t_j) - \lambda(t_{j-1})| \leqslant \frac{2}{R}\sum_{j=1}^n|\gamma(t_j) - \gamma(t_{j-1})|.$$ If $$\gamma$$ is rectifiable, and has length $$L \geqslant 0$$, it follows that $$\lambda$$ is rectifiable, and has length at most $$2L/R$$.

The path integral $$$$\tag{2}\label{eq:3257640:path:2} \int_\gamma \frac{dz}{z} = \int_a^b \frac{d\gamma(t)}{\gamma(t)}$$$$ exists whenever the complex Riemann-Stieltjes integral on the right is defined. (For a general definition of complex Riemann-Stieltjes integrals, see e.g. Apostol, Mathematical Analysis (2nd ed. 1974).)

Given a partition \eqref{eq:3257640:part:2} - again, it has no necessary relation to previous such partitions - "tagged", as they say, with values $$s_j \in [t_{j-1}, t_j]$$ ($$j = 1, \ldots, n$$), the Riemann-Stieltjes sum for the integral on the right of \eqref{eq:3257640:path:2} is: \begin{align} & \phantom{={}} \sum_{j=1}^n\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(s_j)} \notag \\ & = \sum_{j=1}^n\frac{e^{\lambda(t_j)} - e^{\lambda(t_{j-1})}} {e^{\lambda(s_j)}} \notag \\ & = \sum_{j=1}^n\left(e^{\lambda(t_j) - \lambda(s_j)} - e^{\lambda(t_{j-1}) - \lambda(s_j)}\right) \notag \\ & = \sum_{j=1}^n\left[\lambda(t_j) - \lambda(t_{j-1}) + \rho(\lambda(t_j) - \lambda(s_j)) - \rho(\lambda(t_{j-1}) - \lambda(s_j))\right] \notag \\ & = \tag{3}\label{eq:3257640:sum:log} [\lambda(b) - \lambda(a)] + \sum_{j=1}^n\rho(\lambda(t_j) - \lambda(s_j)) - \sum_{j=1}^n\rho(\lambda(t_{j-1}) - \lambda(s_j)). \end{align} To prove that the Riemann-Stieltjes integral \eqref{eq:3257640:path:2} exists, and that it is equal to $$\lambda(b) - \lambda(a)$$, it is enough (see e.g. Exercise 7.3 of Apostol) to prove that for every $$\epsilon' > 0$$ there exists $$\delta' > 0$$ such that if $$t_j - t_{j-1} < \delta'$$ ($$j = 1, \ldots, n$$), then $$$$\tag{4}\label{eq:3257640:integral} \left\lvert \sum_{j=1}^n\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(s_j)} - [\lambda(b) - \lambda(a)] \right\rvert < \epsilon'.$$$$

If $$L = 0$$, i.e. if $$\gamma$$ is constant (a point "curve"), then $$\lambda$$ is also constant, and \eqref{eq:3257640:integral} is trivially satisfied; so, suppose from now on that $$L > 0$$; then at least some of the "error" terms in \eqref{eq:3257640:sum:log} are non-zero. Let $$\eta > 0$$ be such that $$\frac{|\rho(z)|}{|z|} < \frac{R\epsilon'}{2L} \quad (0 < |z| < \eta).$$ Let $$\delta' > 0$$ be such that $$|\lambda(t) - \lambda(t')| < \eta \quad (|t - t'| < \delta').$$ Then, for any partition \eqref{eq:3257640:part:2} satisfying $$t_j - t_{j-1} < \delta'$$ ($$j = 1, \ldots, n$$), tagged with values $$s_j \in [t_{j-1}, t_j]$$ ($$j = 1, \ldots, n$$), the "error" in \eqref{eq:3257640:sum:log} is bounded above, thus: \begin{align*} & \phantom{={}} \left\lvert\sum_{j=1}^n\rho(\lambda(t_j) - \lambda(s_j)) - \sum_{j=1}^n\rho(\lambda(t_{j-1}) - \lambda(s_j))\right\rvert \\ & \leqslant \sum_{j=1}^n\left(|\rho(\lambda(t_j) - \lambda(s_j))| + |\rho(\lambda(t_{j-1}) - \lambda(s_j))|\right) \\ & < \frac{R\epsilon'}{2L} \sum_{j=1}^n\left(|\lambda(t_j) - \lambda(s_j)| + |\lambda(t_{j-1}) - \lambda(s_j)|\right) \\ & \leqslant \epsilon'. \end{align*}

We have thus proved $$\int_\gamma \frac{dz}{z} = \lambda(b) - \lambda(a),$$ on the assumption only that $$\gamma$$ is rectifiable. If also $$\gamma$$ is closed, i.e. $$\gamma(a) = \gamma(b)$$, then $$\operatorname{re}\lambda(a) = \operatorname{re}\lambda(b)$$, and $$\operatorname{im}\lambda(t)$$ is a choice of $$\operatorname{arg}\gamma(t)$$ for $$t \in [a, b]$$, therefore: $$\int_\gamma \frac{dz}{z} = 2\pi{i}n(\gamma, 0),$$ where I have used what seems to be the nearest thing to a standard notation for the index, or winding number, of $$\gamma$$ about $$0$$.

• A small point, not important for the result: given $\kappa \in (0, 1)$, choose $\epsilon > 0$ such that $|\rho(z)| < \kappa|z|$ when $0 < |z| < \epsilon$. Then we find that $\lambda$ has length at most $L/((1 - \kappa)R)$, and therefore at most $L/R$, because $\kappa$ was arbitrary. This bound is attained when $\gamma$ is an arc of a circle with centre $0$. The figure $2L$ can be replaced by $L$ in the subsequent calculations, not that it makes any difference. – Calum Gilhooley Jun 16 at 0:56
• Your answer is simple and sophisticated, thank you ! However, I am not sure that $\lambda$ is rectifiable, although rectifiability of $\lambda$ does not affect the other parts of the proof. What I mean is, $$\sup{\left\{\, \sum_{j=1}^n |\lambda(t_j) - \lambda(t_{j-1})|\, \mid \quad a=t_0 < t_1 < \cdots < t_n=b \mbox{ and } |t_j-t_{j-1}| < \delta \mbox{ for any j in \{1,2,\cdots,n\}}\, \right\}} \leq \frac{2L}{R}$$ is correct, but I am not sure that $$\mbox{total variation of \lambda} \leq \frac{2L}{R}.$$ – Sho Jun 16 at 2:56
• Oops my comment was so foolish, because sum of variations of subdivision is bigger. I am sorry. – Sho Jun 16 at 8:49
• I'm sure you'd feel better if you could see how many mistakes I made in the four days I spent trying to put this proof together! For example, at first I didn't even think that rectifiability was needed. Also I started by trying to prove the false statement that the integral was equal to $2\pi{i}n(\gamma,0)$ even when $\gamma$ was not closed. Then I thought I could only prove $\lambda$ rectifiable (if at all) by splitting it up awkwardly into its real and imaginary parts. It's good to have a critical eye on the thing. I'll keep checking it for howlers. (I'm not quite awake enough yet to do so.) – Calum Gilhooley Jun 16 at 11:42
• I'm feeling better now, thanks to your words! – Sho Jun 16 at 12:25

I give a $$\mathbf{\text{positive}}$$ answer to my own question, thanks to How to show that the integral $\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - a}$ is integer-valued when the curve $\gamma$ is not piecewise smooth?.

In the beginning, I post here some main statements which we will observe in detail later:

• statement 1 Let $$\epsilon$$ be an arbitrarily given positive number, then we can take a piecewise straight closed curve $$\eta$$ such that $$0 \notin \operatorname{ran}(\eta)\ (= \left\{\, \eta(t)\, \mid\quad t \in [0,1]\, \right\}),$$ $$\left|\operatorname{Ind}_{\gamma}(0) - \operatorname{Ind}_{\eta}(0)\right| < \epsilon,$$ and $$\operatorname{Wnd}_{\gamma}(0) = \operatorname{Wnd}_{\eta}(0).$$

• statement 2 As for the abave $$\eta$$, $$\operatorname{Ind}_{\eta}(0) = \operatorname{Wnd}_{\eta}(0).$$

If we gain the above statements, then $$\left|\operatorname{Ind}_{\gamma}(0) - \operatorname{Wnd}_{\gamma}(0)\right| < \epsilon$$ holds for any positive number $$\epsilon$$. Hence $$\operatorname{Ind}_{\gamma}(0) = \operatorname{Wnd}_{\gamma}(0)$$ follows.

First of all, we call $$\theta$$ a $$\mathbf{\text{continuous choice of arguments of \gamma}}$$, if $$\theta$$ is a real valued continuous function on $$[0,1]$$ and satisfies $$\forall t \in [0,1]\, \left[\, \theta(t) \in \arg{\gamma(t)}\, \right],$$ where $$\arg$$ is defined as a mapping $$\mathbf{C} \backslash \{0\} \ni z \longmapsto \left\{\, x \in \mathbf{R}\, \mid\quad z = |z|e^{i x}\, \right\}.$$

I defined $$\operatorname{Wnd}_{\gamma}(0)$$ as $$\frac{\theta(1) - \theta(0)}{2\pi}$$ where $$\theta$$ is a continuous choice of arguments of $$\gamma$$ such that $$-\pi < \theta(0) \leq \pi.$$ However, this value does not depend on the selection of continuous choice of arguments of $$\gamma$$. To see this, let $$\theta$$ and $$\phi$$ be a continuous choice of arguments of $$\gamma$$, then for each $$t$$ in $$[0,1]$$, $$\theta(t) \in \arg{\gamma(t)} \wedge \phi(t) \in \arg{\gamma(t)},$$ hence $$\frac{\theta(t) - \phi(t)}{2\pi}$$ is an integer. In addition, $$[0,1] \ni t \longmapsto \frac{\theta(t) - \phi(t)}{2\pi}$$ is continuous, hence is constant valued. Hence $$\frac{\theta(0) - \phi(0)}{2\pi} = \frac{\theta(1) - \phi(1)}{2\pi}.$$ Hence $$\frac{\theta(1) - \theta(0)}{2\pi} = \frac{\phi(1) - \phi(0)}{2\pi}.$$ (The above explanation can be found in chapter 7 of A. F. Beardon, Complex Analysis: The Argument Principle in Analysis and Topology, Wiley-Interscience publication 1979.)

## Sketch of the statement 1

Let $$\epsilon$$ be an arbitrarily given positive number, $$f$$ be a mapping defined as $$\mathbf{C} \backslash \{0\} \ni z \longmapsto \frac{1}{z},$$ and $$d$$ be the distance from $$0$$ to $$\operatorname{ran}(\gamma)$$, that is to say $$d = \inf_{t \in [0,1]}|\gamma(t)|.$$ Let $$N$$ be the $$d/2$$-neighborhood of $$\operatorname{ran}(\gamma)$$, that is to say $$N = \bigcup_{t \in [0,1]} D(\gamma(t);d/2)$$ where $$D(\gamma(t);d/2)$$ is an open disc centered at $$\gamma(t)$$ and radius $$d/2$$: $$D(\gamma(t);d/2) = \left\{\, z \in \mathbf{C}\, \mid\quad \left|z - \gamma(t)\right| < \frac{d}{2}\, \right\}$$ Since $$\operatorname{ran}(\gamma)$$ is compact, $$\overline{N}$$, closure of $$N$$, is also compact and $$0 \notin \overline{N}.$$ And since $$f$$ is continuous on $$\overline{N}$$, we can take a positive number $$\delta$$ that satisfies $$\delta < \frac{d}{2}$$ and $$\forall z,w \in N\, \left(\, |z-w| < \delta \Longrightarrow |f(z) - f(w)| < \epsilon\, \right).$$

Now we can take nodes \begin{align} 0 = t_0 < t_1 < \cdots < t_n = 1 \end{align} such that $$\forall t\, \left(\, t \in \left[t_k, t_{k+1}\right] \Longrightarrow |\gamma(t) - \gamma(t_k)| < \delta\, \right)$$ holds for each $$k$$ in $$\{0,1,2,\cdots,n-1\}$$.

We now define a piesewise straight continuous path $$\eta$$ on $$[0,1]$$ such that $$\left[t_k,t_{k+1}\right] \ni t \longmapsto \gamma(t_k) + \frac{t - t_k}{t_{k+1} - t_k} \left(\gamma(t_{k+1}) - \gamma(t_k)\right)$$ for each $$k$$ in $$\{0,1,2,\cdots,n-1\}$$. Since the range of $$\eta$$ lies in $$N$$, $$\eta$$ does not take $$0$$.

Then we have $$\left|\int_{\gamma} \frac{1}{z}\ dz - \int_{\eta} \frac{1}{z}\ dz \right| < 2 \epsilon V(\gamma),$$ where $$V(\gamma)$$ denotes the total variation of $$\gamma$$ on $$[0,1]$$. The proof of this inequality is in How to show that the integral $\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - a}$ is integer-valued when the curve $\gamma$ is not piecewise smooth?, so omit that here.

In addition, we have $$\operatorname{Wnd}_{\gamma}(0) = \operatorname{Wnd}_{\eta}(0).$$ Before I write a sketch of this, we admit the next theorem.

Theorem Let $$\alpha$$ be a real number, $$L_{\alpha}$$ be an incision on complex plane such that $$L_{\alpha} = \left\{\, t e^{i \alpha}\, \mid\quad t \leq 0\, \right\},$$ and $$\operatorname{Arg}_{\alpha}$$ be a mapping from $$\mathbf{C} \backslash L_{\alpha}$$ to $$\mathbf{R}$$ such that $$z \longmapsto \alpha + \operatorname{Arg}{(z e^{-i \alpha})}.$$

$$\operatorname{Arg}{z}$$ is the principal value of $$\arg{z}$$, and in this answer we say $$p$$ is the principal value of $$\arg{z}$$ if $$z = |z| e^{i p} \wedge -\pi < p \leq \pi.$$

Then $$\operatorname{Arg}_{\alpha}$$ is continuous on $$\mathbf{C} \backslash L_{\alpha}$$ and $$\frac{z}{|z|} = e^{i \operatorname{Arg}_{\alpha}(z)}$$ holds for any $$z$$ in $$\mathbf{C} \backslash L_{\alpha}$$ (, namely $$\operatorname{Arg}_{\alpha}(z)$$ is an argument of $$z$$).

A proof of this can be found in hapter 5 of A. F. Beardon, Complex Analysis, but here I show a quick sketch.

1. First, $$\mathbf{C} \ni z \longmapsto z e^{-i \alpha}$$is continuous, and $$\operatorname{Arg}$$ is continuous on $$\mathbf{C} \backslash L_{0}$$.

2. Second, suppose $$z$$ is an element of $$\mathbf{C} \backslash L_{\alpha}$$, then $$z e^{-i \alpha} \in \mathbf{C} \backslash L_{0}.$$ Hence $$\operatorname{Arg}_{\alpha}$$ is continuous at this $$z$$.

3. Third, if $$z$$ is an element of $$\mathbf{C} \backslash L_{\alpha}$$, then $$e^{i \operatorname{Arg}_{\alpha}(z)} = e^{i \alpha} e^{i \operatorname{Arg}{(z e^{-i \alpha})}} = e^{i \alpha} \frac{z e^{-i \alpha}}{\left|z e^{-i \alpha}\right|} = \frac{z}{|z|}.$$

For each $$k$$ in $$\{0,1,2,\cdots,n-1\}$$, we define a mapping such that $$\left[t_k,t_{k+1}\right] \ni t \longmapsto \operatorname{Arg}_{\operatorname{Arg}{\gamma(t_k)}}(\gamma(t))$$ as $$\theta^{\gamma}_k$$, and $$\left[t_k,t_{k+1}\right] \ni t \longmapsto \operatorname{Arg}_{\operatorname{Arg}{\gamma(t_k)}}(\eta(t))$$ as $$\theta^{\eta}_k$$. Since $$D(\gamma(t_k),\delta) \subset \mathbf{C} \backslash L_{\operatorname{Arg}{\gamma(t_k)}},$$ $$\theta^{\gamma}_k$$ and $$\theta^{\eta}_k$$ are continuous on the interval $$\left[t_k,t_{k+1}\right]$$.

We now define a continuous function $$\theta^{\gamma}$$ and $$\theta^{\eta}$$ on $$[0,1]$$ by $$[0,1] \ni t \longmapsto \begin{cases} \theta^{\gamma}_0 (t) & \mbox{if } t \in \left[t_{0},t_{1}\right] \\ \theta^{\gamma}_{k} (t) + \sum_{j=1}^{k}\left[\theta^{\gamma}_{j-1} (t_{j}) - \theta^{\gamma}_{j} (t_{j})\right] & \mbox{if } 1 \leq k \wedge t \in \left[t_{k},t_{k+1}\right] \end{cases}$$ and $$[0,1] \ni t \longmapsto \begin{cases} \theta^{\eta}_0(t) & \mbox{if } t \in \left[t_{0},t_{1}\right] \\ \theta^{\eta}_{k}(t) + \sum_{j=1}^{k}\left[\theta^{\eta}_{j-1} (t_{j}) - \theta^{\eta}_{j} (t_{j})\right] & \mbox{if } 1 \leq k \wedge t \in \left[t_{k},t_{k+1}\right] \end{cases}$$ then we can check that $$\gamma(t) = |\gamma(t)| e^{i \theta^{\gamma}(t)}$$ and $$\eta(t) = |\eta(t)| e^{i \theta^{\eta}(t)}$$ holds for every $$t$$ in $$[0,1]$$. In other words, $$\theta^{\gamma}$$ is a continuous choice of arguments of $$\gamma$$ and $$\theta^{\eta}$$ is a continuous choice of arguments of $$\eta$$.

Observe that for each $$k$$ in $$\{0,1,2,\cdots,n\}$$ $$\theta^{\gamma}(t_k) = \theta^{\eta}(t_k).$$ This is because $$\theta^{\gamma}_k(t_k) = \theta^{\eta}_k(t_k)$$ holds for each $$k$$ in $$\{0,1,2,\cdots,n\}$$.

Finally we have $$\operatorname{Wnd}_{\gamma}(0) = \frac{\theta^{\gamma}(1) - \theta^{\gamma}(0)}{2\pi} = \frac{\theta^{\eta}(1) - \theta^{\eta}(0)}{2\pi} = \operatorname{Wnd}_{\eta}(0).$$

## As for the statement 2

I wrote that the proof of this is in definition of winding number, have doubt in definition. in my question, but I am not sure that $$r$$ and $$\theta$$ in the top answer of definition of winding number, have doubt in definition. are differentiable.

However, this statement follows from theorem 9.5.2 of A. F. Beardon, Complex Analysis: The Argument Principle in Analysis and Topology, Wiley-Interscience publication 1979.