# Calculate the winding number of a triangle

My question is about calculating the winding number of a triangle, i.e. $$Ind_\gamma(z) = \frac{1}{2i\pi} \int_\gamma\frac{1}{w-z}dw,$$ where $\gamma$ is the path given by the border of a triangle $\Delta \subset \mathbb C$.

What I have done so far is calculating the value of the function outside of the triangle, and that is $Ind_\gamma(z) = 0 \quad \forall z \notin\Delta$ $\;$, but I don't know how to calculate it inside $\Delta$, intuitively I guess that its value is $Ind_\gamma(z) = 1 \quad \forall z \in\Delta.$

• " intuitively I guess that.." basing on what theorem? What definition of the winding number do you use? – Jack Dec 8 '17 at 20:06
• I'm not sure what kind of "calculation" you are looking for. But this follows essentially from Cauchy integral formula. – Jack Dec 8 '17 at 20:08
• The intuitive idea of the function $Ind_\gamma(z) = \int_\gamma\frac{1}{w-z}dw$ is that it tells the number of counterclockwise turns that the path makes around the point z. Link of the Wikipedia's page. @Jack – Victt Dec 8 '17 at 20:15
• The winding number in complex functions is defined by the integral you wrote times $\;\frac1{2\pi i}\;$ ... – DonAntonio Dec 8 '17 at 20:17
• You are right, I'll edit it. @DonAntonio – Victt Dec 8 '17 at 20:18

## 1 Answer

You can use Cauchy's Integral Formula, with $\;f(z)=1\;$ , and get

$$\frac1{2\pi i}\oint_\Delta\frac1{w-z}dw=f(z)=1$$

and we're done.

Another way is trying to parametrize the triangle's perimeter curve, say: take the triangle whose vertices are $\;(0,0),\,(0,1),\,(1,0)\;$, then the horizontal side is parametrized by $\;r(t)=t(1,0)+(1-t)(0,0)=(t,0),\,0\le t\;$ , and etc.