Winding number (demonstration)

How could I explain mathematically, that the winding number of a closed curve $\gamma$ around $a$ ($a \notin \gamma$) gives always an integer value. $$W(\gamma,a)=\frac{1}{2\pi i} \int_{\gamma} \frac{dw}{w-a}$$ where $W(\gamma,a)\in \mathbb{Z}$

Let $\gamma=\{\gamma_1,\gamma_2,\ldots,\gamma_n\}$ where $\gamma_i$ corresponds to a curve defined on $[a_i,b_i]$. Let $b_i = a_{i+1}$ for $i=1,\ldots,n-1$. Now, $\gamma$ is defined on the continuous interval $[a,b]$ with $a=a_1$ and $b=b_n$. $\gamma$ is differentiable on $(a_i,b_i)$ where at the end points $\gamma$ is left and right derivatives. Define $$f(t) = \int_a^t\frac{\gamma'(t')}{\gamma(t')-\alpha}dt'$$ Then $f$ is continuous on $[a,b]$ and differentiable for $t\neq a_i,b_i$. By the Fundamental Theorem of Calculus, $$f'(t) = \frac{\gamma'(t)}{\gamma(t)-\alpha}.$$ Consider $\frac{d}{dt}e^{-f(t)}(\gamma(t)-\alpha)$. Then we have $$e^{-f(t)}\gamma'(t)-f'(t)e^{-f(t)}(\gamma(t)-\alpha) = 0$$ There is a constant $K$ such that $K=e^{-f(t)}(\gamma(t)-\alpha)\iff\gamma(t)-\alpha=Ke^{f(t)}.$ Since $\gamma$ is a closed curve, $\gamma(a) = \gamma(b)$. $$Ke^{f(a)}=\gamma(a)-\alpha=\gamma(b)-\alpha=Ke^{f(b)}$$ Therefore, $K\neq 0$ so $e^{f(a)}=e^{f(b)}$. There exist an $n\in\mathbb{Z}$ such that $f(b)=f(a)+2i\pi n$, but $f(a)=0$ so $f(b)=2i\pi n$.
• I have a question: why $K\neq 0$?
• @RFZ : $K \neq 0$ because $\gamma(a) - \alpha \neq 0$. Commented Mar 8 at 5:30
• Am I missing something? Why is $\frac{d}{dt}e^{-f(t)}(\gamma(t)-\alpha) = 0$ ?
Let z(t) is the parametric equation of the curve. $$Z(t)=x(t)+iy(t)$$ and $$|z(t)|=r(t)=\sqrt{x(t)^{2}+y(t)^{2}}$$. by using this transformation: $$w(t)=r(t)^{-1}z(t)$$ we can tie down the $$\Gamma$$ to a circle($$\Gamma^{*}$$)
let $$z=re^{i\theta}$$ be any point on the curve($$\Gamma^{*}$$). \begin{align} \implies dz=e^{i\theta}dr+ire^{i\theta}\\ \frac{dz}{z} = \frac{dr}{r} + id\theta \hspace{1cm}\because{z\ne0} \\ \frac{dz}{z} = d[\ln{r}]+id[\theta]\\ \int_{\Gamma}\frac{dz}{z}=[i\theta]^{a}_{b}=2\pi iN\\ \end{align} $$\because \text{curve is closed a=b so N should be an integer.}$$ \begin{align} N=\frac{1}{2\pi i}\int_{\Gamma^{*}}\frac{dz}{z}\\ \implies N=\frac{1}{2\pi i}\int_{\Gamma}\frac{dz}{z}\\ \end{align} $$\because \text{Cauchy integral theorem, integral is same around any piecewise smooth closed curves.}$$ We can extend this to any other point similarly so your integral gives always an integer.