2
$\begingroup$

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 $\operatorname{Ind}(γ,z) = 0$. This case is clear to me.

But what if the curve is such $z$ is outside the curve but inside its convex envelope? The winding number should still be 0, but it seems to me that I cannot find a disk containing the curve and not the “singularity”.

Edit: pick the function $f:= \frac{1}{z-z_0}$ and, looking at the picture, if $z_0$ is enough far away we see that $\int_{\gamma}\frac{1}{z-z_0}=0$ since we can inscribe the curve $\gamma$ in a disk and, since $z_0$ is outside the disk we have that $f$ is holomorphic on the disk and by Cauchy null intergal theorem we can conclude that the index of gamma at $z_0$ is 0. But suppose now I want to apply a similar reasoning for the point $z_0$ as in the picture. It is still outside the curve, but is it possible to adapt a similar reasoning to conlcude the index of the curve at $z_0$ is still 0 or should I follow the continuity/discrete-valued argument described by Martin? Thanks!

enter image description here

$\endgroup$
2
  • $\begingroup$ How do you define “outside the curve”? $\endgroup$
    – Martin R
    May 23 at 8:20
  • $\begingroup$ The intuitive notion of outside a closed curve $\endgroup$
    – Crash Bandicoot
    May 23 at 8:32

1 Answer 1

Reset to default
3
$\begingroup$

Let $\operatorname{Im}(\gamma)$ denote the image of $\gamma$. $$\Bbb C \setminus \operatorname{Im}(\gamma)$$ is an open subset, and exactly one of its connected components is unbounded.

The function $$ z \mapsto \operatorname{Ind}(\gamma,z) $$ is continuous and integer-valued on each component of $\Bbb C \setminus \operatorname{Im}(\gamma)$. It follows that $\operatorname{Ind}(\gamma,z)$ is constant in each component of $\Bbb C \setminus \operatorname{Im}(\gamma)$.

In particular, since $$ \lim_{z \to \infty} \operatorname{Ind}(\gamma,z) = 0 \, , $$ $\operatorname{Ind}(\gamma,z) = 0$ on the unbounded component of $\Bbb C \setminus \operatorname{Im}(\gamma)$.

$\endgroup$
5
  • $\begingroup$ I see! Thanks… so in general the idea is to use Cauchy null integra on disk to prove that $ \lim_{z \to \infty} \operatorname{Ind}(\gamma,z) = 0 \, , $ and then use continuity of Index wrt $z$ to show that it is zero also on the “hidden” points, correct? Where for “hidden” points I mean all this points that are outside the closed curve but inside its convex hull. $\endgroup$
    – Crash Bandicoot
    May 23 at 8:40
  • 1
    $\begingroup$ $\lim_{z \to \infty} \operatorname{Ind}(\gamma,z) = 0 $ is a simple calculation, you don't need Cauchy for that. Then continuity and integer-valuedness implies that it is zero on the whole unbounded component. The concept of “convex hull” is not used here. $\endgroup$
    – Martin R
    May 23 at 8:42
  • $\begingroup$ I understand your proof and it is fine for me: I edited my question, so that I can explain better my doubt. Thank you for your kind help (I'll accept your answer then) $\endgroup$
    – Crash Bandicoot
    May 23 at 8:58
  • 2
    $\begingroup$ @CrashBandicoot: The Cauchy integral theorem applies to functions which are holomorphic in a simply connected domain. In your example, $\gamma$ is a Jordan curve, and that encloses a simply connected domain by the Jordan curve theorem. So you can use the Cauchy integral theorem in your example to conclude that the winding number is zero. But that is a lot of machinery, compared to the above argument. $\endgroup$
    – Martin R
    May 23 at 9:31
  • $\begingroup$ Alright, that’s the feedback I was looking for! Thank you very much! $\endgroup$
    – Crash Bandicoot
    May 23 at 10:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .