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I want to prove that given that $\gamma_0,\gamma_1$ are closed rectifiable curves in a region G with $\gamma_0 \sim \gamma_1\ $(i.e. homotopic) in G and $w \in \Bbb C- \{G\}$. Then $n(\gamma_0,w)=n(\gamma_1,w).$

Okay so here's my attempt :

First of all we are trying to show that $n(\gamma_0,w)=n(\gamma_1,w). \Rightarrow \tfrac{1}{2\pi i}\int_{\gamma_0}\tfrac{dz}{z-w}=\tfrac{1}{2\pi i}\int_{\gamma_1}\tfrac{dz}{z-w}$

As w lies outside of $\gamma_1,\gamma_0$ then $\tfrac{1}{z-w}$ is holomorphic on a disk which includes $\gamma_1,\gamma_0$ so by Cauchy's theorem

$$\int_{\gamma_1}\tfrac{dz}{z-w}=0=\int_{\gamma_0}\tfrac{dz}{z-w}$$

This implies that $n(\gamma_0,w)=n(\gamma_1,w).$

Here's my trouble :

I'm not sure I understand why the fact that the two curves are homotopic matters.Here's what I think the reason is but I'm not entirely sure.

As we didn't specify what type of region G was lets say that its an annulus. lets also say that w is in the centre of this annulus. Then the winding number of both curves around the point w is no longer zero infact now they can be different. but then we dont neccarily have it that $\int_{\gamma_0}f(z)dz=\int_{\gamma_1}f(z)dz$ even though f is holomorphic in G. and so we can assume that the two curves are not homotopic.

I'm not sure if this reasoning is correct ,if it is the correct notion I think it's not explained very formally as it were .

So I have three questions:

i)Is the reasoning in my attempt section correct ?

ii)Is my reasoning as to why homotopy is important here correct ?

iii) Is there any suggestions ( If I am correct in my reasoning behind the importance of homotopy ) as to how I can present it more formally ?

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  • $\begingroup$ (i) Not exactly. As your example in (ii) with $G$ an annulus shows, we don't necessarily have a disk containing $G$ but excluding $w$. $\ $ At (ii) you correctly write that $\int_{\gamma_1}\frac{dz}{z-w}$ is not necessarily zero. Do you already know that if $\gamma$ is nullhomotopic within $G$, then $\int_\gamma f(z)dz=0$ for every holomorphic $f$? $\endgroup$ – Berci Oct 22 '18 at 22:34
  • $\begingroup$ @Berci ah okay I see where I went wrong with part i) I gave an example that contradicted my own method...silly mistake. as for (ii) by nullhomotopic do you mean gamma homotopic to zero ? (i.e. gamma is homotopic to a constant curve . This is how it was defined in class although I'm not sure what a "constant curve" is.) If this is what you mean I know that $\int_{\gamma}f=0$ for gamma homotopic to zero and f holomorphic. Would I be right in saying that if we use an annulus for G that gamma is not homotopic to zero? $\endgroup$ – Voltron Oct 22 '18 at 22:57
  • $\begingroup$ Yes, neither $\gamma_1$ nor $\gamma_2$ need not be homotopic to zero. (That zero in general can be any other point: the only value of the constant path -- should be within $G$.) But we can connect them by a path and reverse one of them, so that the composition becomes nullhomotopic. $\endgroup$ – Berci Oct 22 '18 at 23:00
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(i) $\ $ Not exactly. As your example in (ii) with $G$ an annulus shows, we don't necessarily have a disk containing $G$ but excluding $w$.
$\ $ At (ii) you correctly write that $\int_{\gamma_1}\frac{dz}{z-w}$ is not necessarily zero.

(ii) $\ $ Well, if $\gamma_1$ and $\gamma_2$ are not homotopic, then, yes, their winding number might be different.
$\ $ So, being homotopic is an important condition.

(iii) $\ $ Cut both curves e.g. at their starting points, and connect these two points by a path $p$ within $G$.
$\ $ Consider then the closed curve $\gamma:=\gamma_1p\gamma_2^\smallsmile p^\smallsmile$, where $u^\smallsmile$ denotes the reversed path of the path $u$.
$\ $ Show that $\gamma$ is null homotopic within $G$.
$\ $ Therefore, as $w\notin G$, we have $\int_\gamma\frac{dz}{z-w}=0$.
$\ $ Deduce then $\int_{\gamma_1}\frac{dz}{z-w}=\int_{\gamma_2}\frac{dz}{z-w}$.

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  • $\begingroup$ thank you for your answer. I can't fully digest it right now because I have to go to sleep soon but tomorrow I'll give it a proper read and try to fix my answer..... I may have more questions if you'd be kind enough to answer them $\endgroup$ – Voltron Oct 22 '18 at 23:20
  • $\begingroup$ I was a unsure of how to use your gamma definition to show that $\gamma$ is null homotopic, But I think I fixed the errors in logic I had been using in my proof in the answer below . Would you mind checking it for me ? $\endgroup$ – Voltron Oct 24 '18 at 16:37
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Let $\gamma_1,\gamma_2$ be two closed rectifiable curves in a region G with $\gamma_0$ homotopic to $\gamma_1$ in G.

$\Rightarrow \int_{\gamma_0}f(z)dz=\int_{\gamma_1}f(z)dz$. this is from cauchy's theorem for open sets. (It didn't explicitly say G was open but I used it anyway .This may be a mistake )

If f is holomorphic on .

Consider now

$$n(\gamma_0,w)=\tfrac{1}{2\pi i}\int_{\gamma_0}\tfrac{1}{z-w}dz$$

Well $w\in \Bbb C-G, z\in G$ so therefore $f(z)=\tfrac{1}{z-w}$ is holomorphic in G.

As such $\int_{\gamma_0}f(z)=\int_{\gamma_1}f(z)$

Which means $\int_{\gamma_0}\tfrac{1}{z-w}dz=\int_{\gamma_1}\tfrac{1}{z-w}dz$

In other words

$$\tfrac{1}{2\pi i}\int_{\gamma_0}\tfrac{1}{z-w}=\tfrac{1}{2 \pi i}\int_{\gamma_1}\tfrac{1}{z-w}$$

So therefore

$n(\gamma_0,w)=n(\gamma_1,w)$

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  • $\begingroup$ How is that theorem of Cauchy proved? I guess it's just the same technique as in my answer. Assuming the homotopy is $(\gamma_t)$, it's probably convenient to take $p(t)=\gamma_t(0)$.. $\endgroup$ – Berci Oct 24 '18 at 22:43

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