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)?

  • 4
    $\begingroup$ Don't use logarithms willy-nilly in complex analysis. That's pretty much the source of all beginner mistakes! By your logic, the winding number of the unit circle is also $0$, but again it is not: $\frac{1}{2\pi i}\int_{|z|=1}\frac{1}{z}\,dz=1$, as you can easily verify by direct parametrization of the unit circle. For the ellipse, are you allowed to use the fact that winding number is preserved by homotopy? $\endgroup$
    – peek-a-boo
    May 19 at 9:12
  • $\begingroup$ Directly identify real and imaginary parts, I think $\endgroup$
    – FShrike
    May 19 at 9:14
  • $\begingroup$ If $f(t)$ is a complex function, the derivative of $\ln |f|$ is not $\frac{f'}{f}$. Instead it is $\frac{\Re(f'\overline{f})}{f\overline{f}}$. $\endgroup$ May 19 at 9:15
  • $\begingroup$ @Gribouillis Since $\bar{f}$ is not differentiable (in general) how are you finding the chain rule expression there? $\endgroup$
    – FShrike
    May 19 at 9:21
  • $\begingroup$ @Gribouillis $\ln|f|$ is not analytic is any domain, unless $|f|$ is constant. See here. $\endgroup$
    – Evan Aad
    May 19 at 9:22

1 Answer 1


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$$\omega(\gamma,0)=\frac{2\pi i}{2\pi i}=1.$$

  • $\begingroup$ Actually, the OP's computation was correct for the real part because $\int_0^{2\pi}\Re\left(\frac{\gamma'}{\gamma}\right) d t = \int_0^{2\pi}(\ln |\gamma(t)|)' d t = 0$ $\endgroup$ May 19 at 9:54

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