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Need some help with computing index number (winding number). I need to compute the index number: $n(\gamma,0)$ where $\gamma(t)=\cos t + 3 i \sin t$ for $0 \leq t \leq 4\pi$

I tried to use parametrization, but the integral seem quite ugly. By the use of computer the answer is 0, but it can't be because the index counts the number of circuit on this ellipse, it should be 2 as i see it.

Maybe you have a way to compute it?

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    $\begingroup$ @Matt: You deleted the factor of $4$; I think that was included by intention. $\endgroup$ – joriki Nov 22 '11 at 14:27
  • $\begingroup$ @joriki: Unintended. Thanks, joriki. $\endgroup$ – Rudy the Reindeer Nov 22 '11 at 14:38
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Clearly the winding number is $2$, because $2\pi$ worth of $t$ is one circuit around the origin, and there are two of them. This argument ought to be sufficient everywhere except in specialized classroom situations where you're required to use some particular method to determine the winding number. Since you're not specifying a method I assume this is not the case for you.

If you do want some kind of "more rigorous" argument, I recommend something like:

Let's construct a continuous argument function $f(t)$ for $\gamma$. Since $\gamma(0)=1$, let's set $f(0)=0$.

For $0\le t\le \pi$ the imaginary part of $\gamma(t)$ is nonnegative, so let's make $f(t)\in[0,\pi]$ in this interval. This leads, in some way we don't need to care about, to $f(\pi)=\pi$.

For $\pi\le t\le 2\pi$ the imaginary part of $\gamma(t)$ is nonpositive, so let's make $f(t)\in[\pi,2\pi]$ in this interval. This leads, in some way we don't need to care about, to $f(2\pi)=2\pi$.

For $2\pi\le t\le 3\pi$ the imaginary part of $\gamma(t)$ is nonnegative, so let's make $f(t)\in[2\pi,3\pi]$ in this interval. This leads, in some way we don't need to care about, to $f(3\pi)=3\pi$.

... and so forth ...

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  • $\begingroup$ At a given $t$ the argument is $\tan^{-1}(3\tan t)$, which has derivative ${1 \over 1 + 9\tan^2 (t)}(3\sec^2(t)) \geq 0$. So the argument is increasing. It is zero at multiples of $2\pi$ so the curve wraps around twice. $\endgroup$ – Zarrax Nov 22 '11 at 15:13
  • $\begingroup$ I'll be very glad if you can explain it again. I didn't so much understand...Thanks. $\endgroup$ – bond Nov 22 '11 at 19:04
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The winding number $n(\gamma, a)$ is defined by $$n(\gamma, a) = {1 \over 2\pi i} \int_{\gamma} {1 \over z - a}\,dz$$ So you are looking for $${1 \over 2\pi i} \int_{\gamma} {1 \over z}\,dz$$ The Cauchy integral formula will give that the integral over each circuit of this is $1$, so the overall answer is $2$.

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    $\begingroup$ Doesn't the Cauchy integral formula suppose that we already know that $\gamma$ winds around the origin the appropriate number of times? $\endgroup$ – Henning Makholm Nov 22 '11 at 14:35
  • $\begingroup$ I think you can assume this for the ellipse. If not, scale the ellipse by 3 in the vertical direction. Surely you can assume it for a circle centered at the origin. $\endgroup$ – Zarrax Nov 22 '11 at 14:38
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    $\begingroup$ The point if that once you know how many times $\gamma$ winds around the origin, that's the winding number right there. The detour via the integral formula adds no value to the computation. $\endgroup$ – Henning Makholm Nov 22 '11 at 14:40
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    $\begingroup$ The definition of $n(\gamma,0)$ is the integral, not "how many times $\gamma$ wraps around the origin". So I am interpreting the question as asking the student to interpret it in terms of the Cauchy integral formula and correctly figuring out what the curve is, not asking in detail why an ellipse wraps around the origin once. The latter follows by scaling the circle anyhow. $\endgroup$ – Zarrax Nov 22 '11 at 14:46
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    $\begingroup$ Before you're allowed to use Cauchy's integral formula, you need to prove that its conditions are satisfied, among which is that the curve must have winding number 1. Once you have done this, you have already solved the original problem. There's nothing left to use the integral formula for. $\endgroup$ – Henning Makholm Nov 22 '11 at 14:55

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