# Lemma in deriving the winding number?

I'm self-studying complex analysis, and in my book there are starred exercises on complex integration I'm interested in understanding.

Lemma 1 of the text states

If the piecewise differentiable closed curve $\gamma$ does not pass through the point $a$, then the value of the integral $$\int_\gamma\frac{dz}{z-a}$$ is a multiple of $2\pi i$

in preparation for defining the winding number.

One exercise says, give an alternate proof of Lemma 1 by dividing $\gamma$ into a finite number of subarcs such that there exists a single-valued branch of $\text{arg}(z-a)$ on each subarc. Pay particular attention to the compactness argument needed to prove the existence of such a subdivision.

I thought about it a bit, and don't really know how to approach it. Is there a proof or possibly a sketch I could attempt to work through in the meantime? Thank you.

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Here's an intuitive description to help get you started. You can find solutions online using some google searches. Since $\gamma$ is compact (every open cover has a finite subcover) you can find the appropriate subarcs $\gamma_1, \ldots ,\gamma_n$. Your function "integrates" to $\ln|z-a|+arg(z-a)$ and since $\gamma$ is closed the $\ln|z-a|$ values of the $\gamma_i$ on the endpoints of the arcs all sum to $0$. All that is left is the sum of the arguments, which is some multiple of $2\pi i$. This is just for intuition, not a proof. – Parsa Feb 14 '12 at 12:03

## 3 Answers

Yes, there is (thought your statement should say integer multiple, to be precise :) ). This is a proof due to D.J. Newman.

Let $\gamma$ be a sufficiently nice curve (of the kind you've described), and let $D$ be a domain, and let $\varphi(t)$ be a parameterisation of $\gamma$ for $0 \leq t \leq 1$. Consider the function $$F(t) = \frac{1}{\varphi(t)-a} \exp\left(\int_0^t \frac{\varphi'(\tau)}{\varphi(t)-a} d\tau\right).$$

1. Determine $F'(t)$.
2. Note that $\gamma$ is closed, so $\varphi(0) = \varphi(1)$.
3. What can you say about $F(0)$ and $F(1)$, given 2.?
4. What can you conclude about $\exp(\int)$?

Deduce the result.

If you need any help, let me know! Which textbook are you using, by the way?

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Thanks snarksi. I'm using Ahlfor's Complex Analysis. I think this is more or less the proof he offers in the book. I'm particularly interested in seeing some proof that follows his alternative suggestion about dividing $\gamma$ into finite subarcs. Sorry if I did not emphasize it enough! – Dedede Feb 14 '12 at 23:58

Let $\gamma:[\alpha,\beta] \to \mathbb C$ be a piecewise differentiable closed curve, which doesn't pass through the point $a$. Denote by $\Gamma$ the point set $\gamma([\alpha,\beta])=\{\gamma(t):\alpha \leq t \leq \beta \}$, and the points where $\gamma'$ fails to be continuous by $\alpha \leq t_1^* <t_2^*< \dots t_N^* \leq \beta$ (only a finite number of these).

A continuous image of a compact space is compact, and therefore $\Gamma$ is compact. Also $a \notin \Gamma$, so there exists $\epsilon=\min_{z \in \Gamma} d(z,a)>0$.

In order for a single valued branch of $\arg (z-a)$ to exist on a subarc, we require that the maximal distance between any two points on the subarc is less than $2\epsilon$. (doing that forbids the subarc performing a complete circle - in fact even half a cycle is impossible that way).

The length of $\Gamma$,$L$ is finite, so there exists a natural number $M$ such that $\frac{L}{M} < 2\epsilon$

Set $P_0=\gamma(\alpha)$.

Having $P_n=\gamma(t_n)$ defined, we define the next point $P_{n+1}$ to be:

• If there is a singular point $t_j^*$ ahead, and it is close enough (meaning $d(P_n,\gamma(t_j^*)) \leq 2\epsilon$) we choose $P_{n+1}=\gamma(t_j^*)$ where $t_j^*$ is the next singular point.

• If there is a singular point $t_j^*$ ahead, and it is far away (meaning $d(P_n,\gamma(t_j^*)) > 2\epsilon$), we define $t_{n+1}$ according to $t_{n+1}=\sup \{t \in [t_n,t_j^*]:\int_{t_n}^t \| \gamma'(u) \| \mathrm{d}u \leq 2 \epsilon \}$, and $P_{n+1}=\gamma(t_{n+1})$

• If there are no singular points ahead we define $t_{n+1}=\sup \{t \in [t_n,\beta]:\int_{t_n}^t \| \gamma'(u) \| \mathrm{d}u \leq 2 \epsilon \}$ and $P_{n+1}=\gamma(t_{n+1})$

You should convince yourself that it takes a finite number of iterations (at most the number of singular points + M) to get to the last point $\gamma(\beta)$. Moreover the subarc between any two consecutive points $P_n,P_{n+1}$ admits a single valued branch of $\arg(z-a)$.

Denote the subarcs above by $\gamma_1,\gamma_2,\dots,\gamma_K$. We have $\int_\gamma \frac{dz}{z-a}=\sum_{j=1}^K \int_{\gamma_j} \frac{dz}{z-a}$. In every subarc the single valued branch of the argument corresponds to an analytic branch of the logarithm. Thus:

$$\int_{\gamma_j} \frac{dz}{z-a}=\int_{\gamma_j} \mathrm{d} \log(z-a)=\int_{\gamma_j} \mathrm{d} \ln |z-a|+i \int_{\gamma_j} \mathrm{d} \arg(z-a)$$

When summing over all $j$, the real parts form a telescoping sum which adds up to zero. And the imaginary part is the sum of all angle increments, which is an integer multiple of $2\pi i$ (because we end up on the same ray from $a$ as we started with).

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WLOG you can take $a = 0$, and let the arc $\gamma$ be given by $\gamma:[a, b] \to \mathbb{C}$. For $\gamma(t)$ on the arc, there is a half plane $H_t$ consisting of the points $z$ with $\displaystyle \Re \frac{z}{\gamma(t)} > 0$ . This the half plane "across from" $\gamma(t)$. For each $t$, find $\delta_t$ small enough so that $\gamma(s) \in H_t$ when $s \in (t-\delta_t, t + \delta_t)$. This is an open cover of $[a, b]$. Use compactness to get a finite subcover of intervals around $t_i$ for $i = 1 \dots k$ with $a \le t_1 \le \dots \le t_k \le b$. The image of each subinterval $[t_i, t_{i+1}]$ is contained in a slit plane. So you can integrate the sub-arcs defined on these subintervals using a branch of log.

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