# Can the winding number be infinite?

Let $z$ be a point in the complex plane, and $\gamma$ be a closed curve. Is it possible that $$n(\gamma,z) = \frac{1}{2\pi i}\int_\gamma \frac{dw}{w-z}$$ becomes unbounded? In other words, is it possible to find a curve $\gamma$ such that it winds around a fixed point infinitely many times?

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Assuming $\gamma$ has domain $[0,1]$ and $z$ doesn't lie on the curve, then $z$ must be bounded away from the curve. Then the curve would have infinite length, which I think is not possible if it is a closed curve. – MPW Aug 20 '14 at 1:54
I believe that the curve must be rectifiable (i.e. have finite length) by hipothesis. At least Wikipedia makes this hipothesis when defining line integral (in particular winding number). – Luiz Cordeiro Aug 20 '14 at 2:23
The locus of the curve is compact and bounded away from $z$. The curve is uniformly continuous. By choosing sufficiently small balls, we can cover the curve by a finite number of these small balls in such a way that shows that the index of the curve around $z$ is finite. – copper.hat Aug 20 '14 at 2:57

Without loss of generality, we show that the winding number around $0$ is finite (the general proof follows by translating the complex plane).
Remember the alternative definition of the winding number: if $\gamma$ is a closed path, we can choose a continuous choice of argument $\theta:[0,1] \to \mathbb R$ such that $$\frac{\gamma(t)}{|\gamma(t)|}=\mathrm e^{i\theta(t)}$$
The winding number can then be defined as$$w(\gamma, 0)=\frac{1}{2\pi}\big(\theta(1) - \theta(0)\big)$$
But $\theta$ is a continuous function on a compact set so it is bounded. In particular the winding number is finite.