Winding number of an ellipse $\gamma$ at $0$ using $\omega(\gamma,0)=\frac{1}{2i\pi}\oint_{\gamma}\frac{1}{z}dz$? In this exercise, we are supposed to firstly find a path that parametrizes the following ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ for $a,b \in \mathbb{R}$
$\textit{I have found the following path: } \gamma:\left[0,2\pi\right] \rightarrow \mathbb{C}, t \mapsto a\cos(t)+ib\sin(t)$.
Then, we are asked to calculate the winding number of $\gamma$ at $0$.
I've found that the winding number is equal to zero.This, however, goes against my idea of what the winding number is, I'm almost sure that it should be 1.
What I've done is the following:
$\omega(\gamma,0)=\frac{1}{2i\pi}\oint_{\gamma}\frac{1}{z}dz=\frac{1}{2i\pi}\int_{0}^{2\pi}\frac{\gamma'(t)}{\gamma(t)}dt=\frac{1}{2i\pi}\ln(\left|a\cos(t)+ib\sin(t) \right| )|_{0}^{2\pi}=0$
Is my parametrization incorrect? Or is how I calculate the integral the problem? Or is the winding number really $0$ (what would mean I didn't correctly understanding what the winding number is)?
 A: Your parametrization is correct. And so is you understanding of the meaning of winding number. The problem lies in your computations: you assumed the existence of a non-existing continuous logarithm function from $\Bbb C\setminus\{0\}$ into $\Bbb C$.
You should compute\begin{align}\int_\gamma\frac{\mathrm dz}z&=\int_0^{2\pi}\frac{\gamma'(t)}{\gamma(t)}\,\mathrm dt\\&=\int_0^{2\pi}\frac{-a\sin(t)+b\cos(t)i}{a\cos(t)+b\sin(t)i}\,\mathrm dt\\&=\int_0^{2\pi}\frac{(b^2-a^2)\cos(t)\sin(t)}{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt+i\int_0^{2\pi}\frac{ab}{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt.\end{align}It turns out that$$\int_0^{2\pi}\frac{\cos(t)\sin(t)}{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt=\int_{-\pi}^\pi\frac{\cos(t)\sin(t)}{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt=0$$since the function that we are integrating is an odd function.
On the other hand, you can compute$$\int_0^{2\pi}\frac1{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt\tag1$$observing that it is equal to$$2\int_0^\pi\frac1{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt,$$since we are now dealing with an even function. Actually, it is equal to$$4\int_0^{\pi/2}\frac1{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt,\tag2$$since the function that we are integrating is symmetric with respect to $\frac\pi2$. But $(2)$ is equal to$$4\int_0^{\pi/2}\frac{\sec^2(t)}{a^2+b^2\tan^2(t)}\,\mathrm dt=\frac4{ab}\int_0^{\pi/2}\frac{\frac ba\sec^2(t)}{1+\left(\frac ba\tan(t)\right)^2}\,\mathrm dt,$$and the substitution $\frac ba\tan(t)=u$ and $\frac ba\sec^2(t)\,\mathrm dt=\mathrm du$ turns this into$$\frac4{ab}\int_0^\infty\frac1{1+u^2}\,\mathrm du=\frac{2\pi}{ab}.$$So, $(1)$ is equal to $\frac{2\pi}{ab}$, and therefore$$\omega(\gamma,0)=\frac{2\pi i}{2\pi i}=1.$$
