Closed rectifiable curve with arbitary winding number.

Question

This is an exercise from Conway's Functions of One Complex Variable(page 83, exercise 2).

Give an example of a closed rectifiable curve $\gamma$ in $\mathbb C$ s.t. for any integer $k$ there is a point $a\notin \{\gamma\}$ with $n(\gamma;a)=k$.

My solution:

$$ \gamma_1(t)=\begin{cases}\exp\{\frac{-1+i}{t}\} &t\in(0,1]\\0&t=0\end{cases}$$ is a rectifiable curve since $$\int_0^1|\gamma'(t)|\mathbb d t=\int_0^1\frac{\sqrt2}{t^2}e^{-1/t}\mathrm dt=\int_1^{\infty}{\sqrt2}{}xe^{-x}\mathrm dx=2\sqrt 2e^{-1}$$ If we connect $0$ and $\exp\{{-1+i}\}$ by a segment, then we obtain a desired closed rectifiable curve $\gamma$.

I can see the number of times $\gamma$ orbits about a given point intuitively, but how to compute the winding number rigorously? (Besides, Conway hasn't introduce homotopy there, so I don't know if there are other ways to evaluate the winding number except finding a curve homotopic to the original one?)

Answer 1

Sketch of an argument (too long for a comment): pick an $\epsilon >0$ and replace the piece of the curve inside the disc of radius $\epsilon$ centered at the origin, with one the (appropriately) oriented arcs that contains points in the first quadrant joining the points where the spiral enters the disc and where the segment from $0$ to $-1+i$ exits the disc; because the disc is simply connected the piece of the curve inside the disc is homotopic inside the disc to the arc so for any point outside of the disc and not on the curve, the winding numbers (the original curve and the modified one as above) are the same; now picking a point in the first quadrant far away st taking the segment connecting it with the origin, it hits the new curve on the chosen arc of the disc too, it is more or less obvious that the winding number wr to the modfied curve increases from $0$ far way, to some $n(\epsilon)$ just before you hit the arc, with the increase being in increment of one every time you hit the spiral and $n(\epsilon) \to \infty, \epsilon \to 0$ as you hit more and more the spiral as $\epsilon$ gets smaller and smaller