Given the closed rectifiable curve $\gamma:[0,1]\rightarrow \Bbb C$. we define $-\gamma(t)=\gamma(1-t)$.
I want to prove that $n(\gamma,a)=-n(-\gamma,a).$, where n is the winding number.
Is the following an appropriate proof ? :
I started by using the definition of the winding number
$n(\gamma,a)=\tfrac{1}{2\pi i}\int_{\gamma}\tfrac{dz}{z-a}=\tfrac{1}{2\pi i}\int_0^{2\pi}\tfrac{\gamma'(t)}{\gamma(t)-a}dt=\tfrac{1}{2 \pi i }\int_0^{2\pi}\tfrac{rine^{int}}{re^{int}}dt=n$
(this is obviously circular reasoning though. I think perhaps the following is proof enough )
$-n(-\gamma,a)=-\tfrac{1}{2\pi i}\int_{-\gamma} \tfrac{dz}{z-a}=\tfrac{-1}{2\pi i}\int_0^{2\pi}\tfrac{-\gamma '(t) dt}{-\gamma(t)-a}$
Now recall that $-\gamma(t)=\gamma(1-t)=a+re^{in}e^{-int}$. So
$-n(-\gamma, a)=\tfrac{-1}{2 \pi i}\int_0^{2\pi}\tfrac{-ne^{in}e^{-int}ri}{re^{in}e^{-int}}dt=\tfrac{n}{2\pi} \int^{2 \pi}_0dt=n$
therefore $n(\gamma,a)=-n(-\gamma,a)=n$
What do you guys think is this is a sufficient proof ?
(1) It seems that you only consider the special case $\gamma : [0, 2\pi] \to \mathbb{C},\gamma(t) = a + e^{int}$. But your proof can be adapted to work for any closed piecewise continuously differentiable $\gamma : [a,b] \to \mathbb{C}$. In fact, you have $(-\gamma) = \gamma \circ\iota$, where $\iota : [a,b] \to [a,b], \iota(t) = a + b - t $. Then the chain rule yields $(-\gamma)'(t) = \gamma'(\iota(t))\iota'(t)$. This shows $$\int_a^b\tfrac{\gamma'(t)}{\gamma(t)-a}dt = \int_{\iota(a)}^{\iota(b)}\tfrac{\gamma'(\iota(s))}{\gamma(\iota(s))-a}\iota'(s)ds = \int_{b}^{a}\tfrac{(-\gamma)'(s)}{(-\gamma)(s)-a}ds = - \int_{a}^{b}\tfrac{(-\gamma)'(s)}{(-\gamma)(s)-a}ds$$ which immediately implies $$n(\gamma,a) = - n(-\gamma,a) .$$
(2) You claim that $n(\gamma,a) = - n(-\gamma,a)$ for closed rectifiable curves (which are more general than closed piecewise continuously differentiable curves). This can easily be shown if you write down the definition of $\int_\gamma \tfrac{dz}{z -a}$ for such curves.