Index of a Jordan curve

Question

Winding number theorem: If $J\subset \mathbb{C}$ is a Jordan curve and a point $z$ lies in its interior domain, then the winding number $n(J,z)=\pm 1$.

Now suppose that $J$ is smooth and we have the Jordan curve theorem. Is there any simple complex analysis proof for winding number theorem? I have found only tedious (non-analytical) proofs for the case of continuous curves.

EDIT: Possible proof should use the facts that in each component of $\mathbb{C}\setminus J$ the winding numer is constant, and winding number is zero in the unbounded component; Jordan curve theorem tells us that there are only two different components.

But I have no idea how to analytically conclude that $|n(J,z_{\text{inside}})-n(J,z_{\text{outside}})|=1.$

Answer 1

Look, for example, at pp. 85-91 of Guillemin and Pollack for a version of this valid for smooth hypersurfaces in $\Bbb R^n$.

Answer 2

Suppose $\gamma$ is a simple, closed, piecewise smooth curve in $\mathbb{C}$ defined on $[0,1]$. Choose $t_0$ where $|\gamma(t)|$ achieves its maximum. Then the curve does not intersect the halfplane $H$ which is tangent to the circle of radius $R=|\gamma(t_0)|$ at $\gamma(t_0)$, except at $\gamma(t_0)$. Then there is a ray through $\gamma(t_0)$ in the halfplane $H$ that can be extended just beyond $\gamma(t_0)$ along the straight line, in such a way that it does not intersection in other points of $\gamma$. This can be done because $\gamma$ is piecewise smooth and simple. Then you have a ray $\scr{R}$ through some $z_1$, through $z_0$ which extends to $\infty$ in the halfplane $H$.

You can then define a logarithm $\log_{\scr{R}}(z-z_1)$ with a branch cut along the extended ray. Then you can evaluate the winding integral using the antiderivative of the kernel, which is this $\log_{\scr{R}}$. It is convenient to reparameterize the curve first so that $\gamma(1)=\gamma(0)=z_0$: \begin{align} w(z_1,\gamma) & =\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z-z_1}dz \\ & =\frac{1}{2\pi i}\int_{0+}^{1-}\frac{\gamma'(t)}{\gamma(t)-z_1}dt \\ & =\frac{1}{2\pi i}\int_{0+}^{1-}\frac{d}{dt}\log_{\scr{R}}(\gamma(t)-z_1) dt\\ & =\left.\log_{\scr{R}}(z-z_1)\right|_{\mbox{one side of branch cut}}^{\mbox{other side of branch cut}} \\ & = \frac{\pm 2\pi i}{2\pi i} = \pm 1. \end{align}

Answer 3

Here's a quick-and-dirty proof of the inside-vs-outside thing (with a little bit of hidden topology).

1. Pick $t_0$ with $|\gamma(t_0)|$ maximal. Let $u = \gamma(t_0)$. By smoothness, $s = \gamma'(t_0)$ is perpendicular to the ray from the origin to $u$.

2. Define $\mu(t) = 2 -s^{-1}(\gamma(t) - A)$. Then $\mu$ is just $\gamma$, translated to the origin, rotated so the tangent to the origin is vertical and the entire curve lies to the right of the origin, and then translated so the entire curve is to the right of the line $x = 2$, with vertical tangent there. I'll prove the result for $\mu$ instead of $\gamma$.

3. Let's look at $z_0 = 1$ and consider the winding integral for $\mu$ around $z_0$. Because $z_0$ is exterior to $\mu$, this integral is zero.

4. Now draw a "keyhole" path consisting of an almost complete counterclockwise circle around $z_0 = 1$ (starting at $z = 1.1 + \epsilon i$, finishing at $1.1 - \epsilon i$), travelling right to $z = 2$, traversing $\gamma$ counterclockwise, and then returning to $1.1 + \epsilon i$. The winding integral $I$ for this is (by a homotopy argument) the same as the winding integral for the point $z_1 = 2 + \delta$, for small enough $\delta$, which is what we'd like to evaluate, as it's the winding number for the "interior" component (or its negative, depending on the original orientation of $\gamma$).

5. But this winding integral is the winding integral fror $\gamma$ about $z0$ (which we know is zero) plut the winding integral for a ccw circle around $z_0$, plus two terms that cancel as $\epsilon \to 0$. The winding integral for a tiny circle around $z_0$ is obviously $1$, so we're done.

This differs only slightly from @COVID_20's approach, in the sense that you never need to think about log branches except for a very nicely-defined integral over a circle, which you've presumably already seen back when looking at the Cauchy integral formula or something.