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


2 Answers 2


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

  • $\begingroup$ I have a question: why $K\neq 0$? $\endgroup$
    – RFZ
    Commented Mar 6 at 3:54
  • 2
    $\begingroup$ @RFZ : $K \neq 0$ because $\gamma(a) - \alpha \neq 0$. $\endgroup$
    – Nebzat
    Commented Mar 8 at 5:30
  • $\begingroup$ Am I missing something? Why is $\frac{d}{dt}e^{-f(t)}(\gamma(t)-\alpha) = 0$ ? $\endgroup$
    – MPW
    Commented yesterday

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.


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