complex-analysiswinding-number
I've been looking for the answer to the following question for a little while now:
Let $γ$ be a closed (C1-)curve whose image is contained in ${z: |z| < R}$ for some $R > 0$. Show that for any $z$ with $|z| > R$ we have $\operatorname{Ind}(γ,z) = 0$.
I think I am supposed to use the definition of the index of a winding number, but I have absolutely no idea of how to do it. To me if $z$ is outside the curve then the index is 0 by definition…
Any pointers would be greatly appreciated thanks!
You left out a vital piece of information: that little arrow on the left, pointing down. It tells you in which direction the loop is travelled (only once, I assume). Therefore, the winding number of each point of the three lobes is $1$ and in the central region it is equal to $2$. I suggest that you read this intuitive description of the winding number.
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 $[a, b]$, \begin{equation} \tag{1}\label{eq:3257640:part:2} a = t_0 < t_1 < \cdots < t_{n-1} < t_n = b, \end{equation} 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 \begin{equation} \tag{2}\label{eq:3257640:path:2} \int_\gamma \frac{dz}{z} = \int_a^b \frac{d\gamma(t)}{\gamma(t)} \end{equation} 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 \begin{equation} \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'. \end{equation}
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 $0$.
Best Answer
$\mathrm{Ind}(\gamma,z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{1}{z-z_0} \textrm{d}z$.
Since $\left|z_0\right| > R$, the function $z \mapsto \frac{1}{z-z_0}$ is holomorphic on $\{z : \left|z\right| < R\}$ and the result follows by Cauchy's Theorem.