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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?

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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.

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    $\begingroup$ Thanks. But I’m not in the complex plane. So is it the same in R^2? $\endgroup$ – QD666 Feb 17 at 20:25
  • $\begingroup$ @QD666: whoops! my bad; that's what I get for reading/answering questions before coffee! Hang tight, will edit. Cheers! $\endgroup$ – Robert Lewis Feb 17 at 20:38
  • $\begingroup$ @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 . . . $\endgroup$ – Robert Lewis Feb 18 at 3:30
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    $\begingroup$ No worries, it’s fine $\endgroup$ – QD666 Feb 18 at 3:39

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