# 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 $W(\gamma,a)\in \mathbb{Z}$

## 1 Answer

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 $f(b)=2i\pi n$.