# winding number of paths

Let $$c:[0,1]\to\mathbb{R}^2\backslash\{\mathbf{0}\}$$ be a closed path with winding number $$k$$. Let $$\tilde{c}=\rho(t)c(t)$$, where $$\rho:[0,1]\to(0,\infty)$$ is function satisfying $$\rho(0)=\rho(1)$$. Determine the winding number of $$\tilde{c}$$.

What I did is that: Let $$\alpha_0 = \frac{-ydx+xdy}{x^2+y^2} = f_1dx+f_2dy$$, and $$\int_{\tilde{c}}\alpha_0 = \int_0^1 f_1(\rho(t)c_1(t))(\rho'(t)c_1(t)+\rho(t)c_1'(t))dt$$. But I don't know how to keep going from here. Any advice?

## 1 Answer

Does not

$$k_c = \dfrac{1}{2\pi i}\displaystyle \int_0^1 \dfrac{c'(t)}{c(t)} \; dt? \tag 1$$

indeed it does; then

$$k_{\bar c} = \dfrac{1}{2\pi i}\displaystyle \int_0^1 \dfrac{(\rho(t)c(t))'}{\rho(t) c(t)} \; dt = \dfrac{1}{2\pi i}\int_0^1 \dfrac{\rho'(t) c(t) + \rho(t)c'(t)}{\rho(t) c(t)} \; dt$$ $$= \dfrac{1}{2\pi i} \displaystyle \int_0^1 \dfrac{\rho'(t) c(t)}{\rho(t) c(t)} \; dt + \dfrac{1}{2\pi i} \int_0^1 \dfrac{\rho(t)c'(t)}{\rho(t)c(t)} \; dt$$ $$= \dfrac{1}{2\pi i} \displaystyle \int_0^1 \dfrac{\rho'(t)}{\rho(t)} \; dt + \dfrac{1}{2\pi i}\int_0^1 \dfrac{c'(t)}{c(t)} \; dt = \int_0^1 \dfrac{\rho'(t)}{\rho(t)} \; dt + k_c; \tag 2$$

now since

$$\rho:[0, 1] \to (0, \infty), \tag 3$$

we have

$$(\ln \rho)(t))' = \dfrac{\rho'(t)}{\rho(t)}, \tag 4$$

whence

$$\displaystyle \int_0^1 \dfrac{\rho'(t)}{\rho(t)} \; dt = \int_0^1 (\ln \rho(t))' \; dt = \ln (\rho(1)) - \ln (\rho(0)) = 0 \tag 5$$

since

$$\rho(0) = \rho(1); \tag 6$$

thus we are left with

$$k_{\bar c} = k_c, \tag 7$$

$$OE\Delta$$.

The point is, of course, that the winding number $$k_c$$ does not really depend on the radius $$r(t) = \vert c(t) \vert$$ of $$c(t)$$, as long as we have $$r(t) \ne 0$$ on $$c(t)$$.

Nota Bene: In fact, the above argument is just a re-write of the usual demonstration that the winding number of $$c(t)$$ makes any sense at all, since in polars we have

$$c(t) = r(t)e^{i\theta(t)}, \tag 8$$

whence

$$c'(t) = r'(t) e^{i\theta(t)} + r(t) i \theta'(t) e^{i\theta(t)}, \tag 9$$

$$\dfrac{c'(t)}{c(t)} = \dfrac{r'(t) e^{i\theta(t)} + r(t) i \theta'(t) e^{i\theta(t)}}{r(t)e^{i\theta(t)}} = \dfrac{r'(t)}{r(t)} + i\theta'(t), \tag{10}$$

which yields

$$\displaystyle \int_0^1 \dfrac{c'(t)}{c(t)} \; dt = \int_0^1 \dfrac{r'(t)}{r(t)} \; dt + i \int_0^1 \theta'(t) \; dt = \ln r(1) - \ln r(0) + i \Delta \theta = i \Delta \theta, \tag{11}$$

where $$\Delta \theta$$ is the increment incurred by $$\theta$$ as we traverse the closed path $$c(t)$$; as such,

$$\Delta \theta = 2\pi n, \; n \in \Bbb Z, \tag{12}$$

so the winding number is indeed a sensible construct. End of Note.

• Thanks. But I’m not in the complex plane. So is it the same in R^2? – QD666 Feb 17 at 20:25
• @QD666: whoops! my bad; that's what I get for reading/answering questions before coffee! Hang tight, will edit. Cheers! – Robert Lewis Feb 17 at 20:38
• @QD666: sorry this is taking so damn long; I actually had to break from MSE for awhile and do my real life; still at it, however . . . – Robert Lewis Feb 18 at 3:30
• No worries, it’s fine – QD666 Feb 18 at 3:39