# Winding number of external point

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!

• How do you define “outside the curve”? May 23 at 8:20
• The intuitive notion of outside a closed curve May 23 at 8:32

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 $$\Bbb C \setminus \operatorname{Im}(\gamma)$$.
• 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. May 23 at 8:40
• $\lim_{z \to \infty} \operatorname{Ind}(\gamma,z) = 0$ 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. May 23 at 8:42
• @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. May 23 at 9:31