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To calculate an integral I need to find the winding number of $$\gamma(t)=\sin(2t)+i\cos(t)$$ around $\frac{i}{2}$. Graphically it looks like it is $1$. How can I use calculus to show this rigorously? I don't know how to calculate $$\int_{-\pi/2}^{3\pi/2}\frac{2\cos(2t)-i\sin(t)}{\sin(2t)+i\cos(t)-\frac{i}{2}}dt$$Do I even have to? (Directly)

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    just use a u-substitution, and set u=denominator – Saketh Malyala Dec 01 '19 at 20:51
  • I don't see how you compute a winding number by doing a substitution. The whole point is that $\log$ is multivalued and you need to know how many times you wind around the branch point. – Ted Shifrin Dec 01 '19 at 20:57
  • @TedShifrin well, let's see. a wikipedia formula for winding number is $I = \displaystyle \frac{1}{2\pi i}\int_{\alpha}^{\beta}\frac{\gamma'(t)}{\gamma(t)-z},dt$ – Saketh Malyala Dec 01 '19 at 20:59
  • @SakethMalyala: Trust me. I know this. I've taught it dozens of times. But if you evaluate the integral by substitution, your antiderivative will merely be the multivalued logarithm. Now what? Try it going around the unit circle one time, five times. Can you tell the difference with your approach? – Ted Shifrin Dec 01 '19 at 21:02
  • i was afraid that'd happen. curious, though, does this formula serve any good? @TedShifrin – Saketh Malyala Dec 01 '19 at 21:04
  • @SakethMalyala that formula is exactly what the asker is using to produce that integral – Ben Grossmann Dec 01 '19 at 21:05
  • The asker is open to sugestions :) – Ruben Kruepper Dec 01 '19 at 21:05
  • wait, i remember it better now. it was something about $\sum i\Delta \arg$ and how over the entire loop, the magnitude cancels out but the argument is cumulative, and that's what results in the integral looking like this. – Saketh Malyala Dec 01 '19 at 21:06
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    @RubenKruepper: You could pick a generic direction and see how many times $\gamma(t)-i/2$ hits that direction. In general, this is not so easy to do exactly, but maybe here you can with a little trigonometry. Try the positive real direction. When is $\gamma(t)-i/2$ a positive real number? – Ted Shifrin Dec 01 '19 at 21:09
  • @TedShifrin for $t=\pi/3$, but I'm not exactly sure what argument you're trying to make with "pick a generic direction"? – Ruben Kruepper Dec 01 '19 at 21:21
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    Right: Precisely one value of $t$. I'm saying that you can count the winding number about $P$ by seeing how many times the vector (from the point $P$) turns around the point, and it is a theorem that in order to do this it must cross a "generic" direction exactly that many times. I'm using a little bit of differential topology here, and I'm not giving you a proof. But it's about choosing what's called a regular value of the "direction map" $\dfrac{\gamma(t)-P}{|\gamma(t)-P|}$ to the unit circle. – Ted Shifrin Dec 01 '19 at 21:25
  • Do you know that the winding number is homotopy invariant? – Paul Frost Dec 01 '19 at 23:58

2 Answers2

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It suffices to define a consistent arg function for $\ln(\gamma(t)-\frac i 2)$ on $[\frac \pi 2,\frac{3\pi}2]$. I suggest the following: $$f(t)=\left\{\begin{array}{cc}-\frac \pi 2,&t =\frac \pi 2,\pi,\frac{3\pi}2\\ \tan^{-1}\left(\frac {2\cos(t)-1}{2\sin(2t)}\right)-\pi,&\frac \pi 2<t<\pi\\ \tan^{-1}\left(\frac{2\cos(t)-1}{2\sin(2t)}\right),&\pi<t<\frac{3\pi}2\end{array}\right.$$ This is well defined and continuous, so the winding number is zero.

[Edit] More details: paraphrasing Wikipedia, if $\gamma$ is a closed curve parameterized by $t\in [\alpha,\beta],$ the winding number of $\gamma$ about $z$ is $${\rm Ind}_{\gamma}(z)=\frac 1{2\pi i}\int_{\alpha}^{\beta}\frac{\gamma'(t)}{\gamma(t)-z}~dt .$$ In the special case here $z=\frac i2$ and we are considering the behavior about how the curve $\gamma(t)$ is winding around $\frac i 2$. The argument (angle) of $\gamma(t)$ with respect to $\frac i 2$ can be seen from the complex number $$\gamma(t)-\frac i 2=\sin(2t)+\frac{2\cos(t)-1}2i.$$ Taking hint from polar coordinates, the angle of this complex number is related to $$\tan^{-1}\left(\frac {2\cos(t)-1}{2\sin(2t)}\right).$$ As in the definition of $f(t)$, this angle is wavering between the values $$-\frac \pi 2\sim \delta-\pi\sim- \frac \pi 2\sim -\delta\sim -\frac \pi 2,$$ where $\frac \pi 2>\delta>0$ and $-\delta$ is the maximum of $f$ on $[\frac \pi 2,\frac{3\pi}2]$. Note that here the arctan takes values in $(-\frac\pi 2,\frac \pi 2), $but the function $f$ takes values in $[\delta-\pi,-\delta]$, a subinterval of $(-\pi,0)$. Nevertheless $f(\frac\pi 2)=f(\frac{3\pi}2)=-\frac\pi 2$, so the total change of angle equals zero.

Pythagoras
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  • If I'm not mistaken (I could easily be, as it's past my bedtime, and I find it rather hard to think of the problem in these terms), your $f(t)$ is not a value of $\arg(\gamma(t) - i/2)$; rather it is the value, for $z = \gamma(t)$, of a continuous argument of $i/2 - z$ that takes values in the interval $(0, \pi)$. I'm sorry if that's not quite an accurate statement; what I'm pretty sure of is that you have mistakenly replaced $z - i/2$ with $i/2 - z$, but the exact details are confusing, to me at least! – Calum Gilhooley Dec 02 '19 at 02:28
  • @CalumGilhooley I think we can agree on the definition of argument (angle) but your use of $z$ may cause confusion, because in the formula quoted from Wikipedia by Saketh Malyala, it would refer to the constant $z=\frac i 2$. I will try to write a paragraph in my original answer to explain. – Pythagoras Dec 02 '19 at 05:13
  • @CalumGilhooley Thanks! Now I see what you meant. I fixed the problem. Please let me know if it works now. I have to say that my initial intention was to define a consistent arg function which doesn't need to be the angle (namely I allowed a shift by a constant). But I do apologize for causing some confusion. – Pythagoras Dec 02 '19 at 23:02
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    Your revised expression for $f(t)$ is correct. It agrees with this: $$ \operatorname{atan2}\left(\cos(t)-\frac12, \ \sin(2t)\right) \quad \left(\frac\pi2 \leqslant t \leqslant \frac{3\pi}2\right). $$ See atan2 - Wikipedia. – Calum Gilhooley Dec 07 '19 at 02:09
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It looks as if you mean $\gamma$ to be defined on the interval $[0, 2\pi]$, although the limits of the integral expression give a different and misleading impression - and as the other answer correctly states, the winding number is zero if the parameter interval is $[\pi/2, 3\pi/2]$.

Let $\gamma_1$, $\gamma_2$, $\gamma_3$ be the restrictions of $\gamma$ to the intervals $[0, \pi/2]$, $[\pi/2, 3\pi/2]$, $[3\pi/2, 2\pi]$, respectively.

I'm afraid this hastily-produced figure lacks a caption, and the point with parameter value $3\pi/2$ has overwritten the point with parameter value $\pi/2$, because both are at the origin: Sorry, no time to do this properly.

$\gamma_2$ is a closed curve, contained in the open half-plane $\{z : \Im z < 1/2\}$, on which there is a continuous choice of argument for $z - i/2$, for example one with values in $(-3\pi/2, \pi/2)$, so the winding number of $\gamma_2$ about $i/2$ is $0$.

Each of $\gamma_1$, $\gamma_3$ separately also has a continuous choice of argument for $z - i/2$, although their concatenation, the closed curve $\gamma_1 + \gamma_3$, does not.

(Correction: that was very badly put! Every curve has a continuous choice of $\arg(z - w)$, for any point $w$ not on the curve; and indeed the very concept of winding number I'm appealing to here depends on that proposition. I should have said something to the effect that the construction of such a function for $\gamma_1 + \gamma_3$ has to be done piecewise, most naturally for $\gamma_1$ and $\gamma_3$ separately; and this is because $\gamma_1 + \gamma_3$ isn't contained in any domain on which there is a continuous choice of argument for $z - i/2$. I hope this hasty correction will suffice.)

For instance, the principal value of $\arg(z - i/2)$ could be used for $\gamma_1$, and the winding number of $\gamma_1$ about $i/2$ is then computed as: $$ \frac{\arg(0 - i/2) - \arg(i - i/2)}{2\pi} = \frac{-\pi/2 - \pi/2}{2\pi} = -\frac12. $$

For $\gamma_3$, you could choose a continuous argument function, call it $\arg_3$, whose value varies in the range $(0, 2\pi)$ (undefined on the non-negative real axis), and the winding number of $\gamma_3$ about $i/2$ is then computed as: $$ \frac{\arg_3(i - i/2) - \arg_3(0 - i/2)}{2\pi} = \frac{\pi/2 - 3\pi/2}{2\pi} = -\frac12. $$

The winding number of $\gamma$ about $i/2$ is the sum of the winding numbers of $\gamma_1$, $\gamma_2$, $\gamma_3$ about $i/2$, which evaluates to: $$ -\frac12 + 0 - \frac12 = -1. $$