# winding number example

Let $$\gamma:[a,b]\rightarrow \mathbb C$$ be a closed curve and$$Int(y)=\{z\in\mathbb C-tra(\gamma): ind(z)\neq0\}, \\ Ext(y)=\{z\in\mathbb C-tra(\gamma): ind(z)=0\},$$ where $ind(z)\;$is the winding number.

Are there any examples where $Int(\gamma)$ and $Ext(\gamma)$ are not connected ?

• What is $y$ and what is $tra(\gamma$ (one parenthesis?)
– zhw.
Commented Jun 28, 2016 at 18:06
• I think that "tra" denotes the "trace" of the curve (i.e., the image of the function $\gamma$). Commented Jun 28, 2016 at 18:11

Let $C_1,C_2,C_3$ be the circles $|z| = 1, |z-2|=1, |z-4|=1.$ Starting at $1,$ let $\gamma$ be the closed contour that traces out the bottom half of $C_2,$ then traces once around all of $C_3,$ then traces the top half of $C_2.$ That brings us back to $1.$ We're not done: Continue $\gamma$ to go once around $C_1$ with positive orientation, then once around $C_1$ with negative orentation. We're back at $1,$ so $\gamma$ is closed. Then the winding number of $\gamma$ inside $C_1$ and $C_2$ are both $1,$ the winding number inside $C_1$ is $0,$ and the winding number in the exterior of all three circles is $0.$ Thus both $\text { Int }\gamma,\text { Ext }\gamma$ are not connected.

If you define, on $[a, b] = [0, 1]$

$$\gamma(t) = \begin{cases} \frac{1}{2t-1} - 1 & t > \frac{1}{2} \\ \frac{1}{2t-1} + 1 & t < \frac{1}{2} \\ 0 & t = \frac{1}{2} \end{cases}$$

then one of those two regions consists of all of $\mathbb C$ except for the $x$-axis, which is disconnected. The other is empty. Of course, the function I've defined isn't even continuous, so you have to expect this sort of thing.

You probably meant to write "suppose that $\gamma: [a, b] \to \mathbb C$ is a continuous (or even differentiable?) function with $\gamma(a) = \gamma(b)$. Then ... "

This example shows that those missing hypotheses are important.

• ...and now you've edited the problem to include the idea that $\gamma$ is a loop. Commented Jun 28, 2016 at 18:13

If $\gamma$ is a continuous and non-self-intersecting loop, then the interior and exterior are each connected; that's the Jordan curve theorem. However, that doesn't hold if $\gamma$ intersects itself. Consider a lemniscate: it partitions the plane into one unbounded region with winding number zero, and two bounded regions with winding number one. WolframAlpha gives a parametrization of a lemniscate.

• I was actually thinking of a lemniscate. Just couldn't come up with a parametrization. Thanks for helping me out. Commented Jun 28, 2016 at 18:02
• google.com/… Commented Jun 28, 2016 at 18:12
• I added a link to a parametrization for you. Commented Jun 28, 2016 at 18:15