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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)?

user21820
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cnymfais
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    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? – peek-a-boo May 19 '22 at 09:12
  • Directly identify real and imaginary parts, I think – FShrike May 19 '22 at 09:14
  • If $f(t)$ is a complex function, the derivative of $\ln |f|$ is not $\frac{f'}{f}$. Instead it is $\frac{\Re(f'\overline{f})}{f\overline{f}}$. – Gribouillis May 19 '22 at 09:15
  • @Gribouillis Since $\bar{f}$ is not differentiable (in general) how are you finding the chain rule expression there? – FShrike May 19 '22 at 09:21
  • @Gribouillis $\ln|f|$ is not analytic is any domain, unless $|f|$ is constant. See here. – Evan Aad May 19 '22 at 09:22
  • How am I then supposed to calculate the integral? – cnymfais May 19 '22 at 09:26
  • @FShrike $t$ is a real parameter, not a complex parameter. My comment explains why using $\ln |\gamma|$ to compute the integral is wrong. – Gribouillis May 19 '22 at 09:28
  • @cnymfais Substitute the explicit expression for $\gamma$ and the explicit expression for $\gamma'$ in the integral, and then calculate separately the real and the imaginary parts. – Evan Aad May 19 '22 at 09:30
  • @EvanAad I never said $\ln |f|$ was analytic. – Gribouillis May 19 '22 at 09:46

1 Answers1

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

José Carlos Santos
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  • Actually, the OP's computation was correct for the real part because $\int_0^{2\pi}\Re\left(\frac{\gamma'}{\gamma}\right) d t = \int_0^{2\pi}(\ln |\gamma(t)|)' d t = 0$ – Gribouillis May 19 '22 at 09:54