# Is this an appropriate proof to show that $n(\gamma,a)=-n(-\gamma,a)$

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 ?

## 2 Answers

(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.

The proof is acceptable and well written. This is not the place to post such questions.

• "This is not the place to post such questions" – I think it is, and there is even a dedicated tag [proof-verification] for such questions. (And if you consider the question off-topic then you should vote to close instead of answering.) – Martin R Oct 21 '18 at 19:54