Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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 3

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?

share|improve this answer
    
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).

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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