Winding Number iff statement

[Poopsilon]

Homework Statement

Let D ⊂ C be open, connected, and bounded. Suppose the boundary of D consists of a finite number of piecewise differentiable simple closed curves: α0,...,αN, with α1,...,αN contained in the interior of α0. Suppose α0 is oriented in the positive direction and α1, . . . , αN oriented in the negative direction. Let α = α0 + · · · + αN . Show that for z ∈ ℂ\α, χ(α, z) = 0 if and only if z ∈ D.

Homework Equations

Well this formula is probably important: $χ(α, z) = \frac{1}{2\pi i}\int_\alpha \frac{dζ}{ζ-z}$.

I'm not sure what others might be important, maybe Cauchy's Integral Formula & Theorem, possibly some residue theorems..

The Attempt at a Solution

I mean χ(α, z) measures the winding number of a point z, so at first it seems that χ(α, z) = 0 when z is not in D, yet that's the opposite of what the theorem is saying for left to right.

So let's say I start with the right to left direction by assuming z ∈ D, then that means I need to prove that $\int_\alpha \frac{dζ}{ζ-z} = 0$.

Well by Cauchy's Integral Theorem, if I can prove that $f(ζ) = \frac{1}{ζ-z}$ is analytic in some simply connected region containing our curve α, then that would suffice.

Now I'm picturing D as some sort of disk like blob with some holes in it, where a0 is its outer boundary and a1 through aN are the boundaries of its holes. Thus since ℂ\α is open I will be able to sneak a disk around each of the holes which excludes z, wherever it might be in D, and thus $f(ζ) = \frac{1}{ζ-z}$ will be analytic on each one of those disks (since they are simply connected) and thus the integral $\int \frac{dζ}{ζ-z} = 0$ for the curves α1 through αN.

As for α0, I'm thinking that we could look at the region outside some disk that is just inside a0, and close enough to it to exclude z. And then from here I'm thinking that this region actually is simply connected through the point at infinity ( gosh I sure wish I knew more topology ). And then from that I can show that $\int_{\alpha_0} \frac{dζ}{ζ-z} = 0$.

But then this entire argument worries me, because for the other direction I assume z is not in D, then I can just stick a disk around all of D which excludes z and then again by Cauchy's Integral Theorem $f(ζ) = \frac{1}{ζ-z}$ is analytic in that disk and so its integral is zero for all the curves α0 through αN. So I think maybe this all spiel is wrong, maybe some of you could help me out, thanks.

 

[sunjin09]

If you imaging a simple ring shaped open region, where α0 is the outer circle, α1 is the inner circle, what you try to prove is certainly not true ...

 

[Poopsilon]

I'm sorry but I can't do anything with that comment.