Is z=0 a pole in the contour integral ∫sinθ/(a-sinθ)dθ from -\pi to \pi?

[zezima1]

Trying to teach myself contour integration, but I'm not so good at it. I want help with evaluating the closed integral:

∫sinθ/(a-sinθ)dθ from -\pi to \pi

So I substitute z= e^iθ, and sinθ = -i/2(z-z^-1) and dθ = -ie^-iθdz

So our integral becomes:

∫-i/2(z-z^-1)/(a+i/2(z-z^-1) dz

Is this correct so far? If so, what do I do from this point? I suppose I want to find the order of the pole for this function?

 

[clamtrox]

Where is the -i/z from dθ=-i/z dz?

So next you should convince yourself that the integration region is bounded by a simple curve, and then find all the poles inside this curve.

 

[aaaa202]

Where is the -i/z from dθ=-i/z dz?

oops, I forgot that on here :)

Okay, so I thought that's what I should do, but how do you determine the order of the pole for such a messy expression? i.e. how can you in general be sure that the pole you get in the denominator is not canceled off by something in the numerator? :)

 

[scurty]

oops, I forgot that on here :)

Okay, so I thought that's what I should do, but how do you determine the order of the pole for such a messy expression? i.e. how can you in general be sure that the pole you get in the denominator is not canceled off by something in the numerator? :)

You simply plug the number into the numerator (remember that numerator here is everything besides the $(z-a)^m$ term where a is the pole) and if the result is not 0 then m is your order.

Your integral is correct, but in the case of finding poles, it is best to clean it up a bit.

$\displaystyle \int_{\partial \mathbb{D}} \frac{-\frac{i}{2} (z-z^{-1})}{a+ \frac{i}{2} (z-z^{-1})} dz$

(I rewrote it so the integral is easier to view by people)

You converted it correctly, since you substituted for $e^{it}$, you are integrating around the unit disk and since $t \in [-\pi,\pi]$ this becomes contour integral now. The easiest way to find the poles would be to multiply through by (2z)/(2z) to get rid of all the fractions and then consider what make the integrand undefined. Like I mentioned above, simply write the integrand as $\frac{g(z)}{(z-a)^m}$ and verify $g(a) \neq 0$. Then $m$ is the order of the pole $a$. Then use the Residue Theorem.

The residues eventually come out to clean numbers, so don't be deterred by the messy algebra.

Edit: Some of your residues might be outside of $\mathbb{D}$ depending on your choice of $a$. Remember to take that into account.

 

[clamtrox]

Edit: Some of your residues might be outside of $\mathbb{D}$ depending on your choice of $a$. Remember to take that into account.

There are three cases to consider here (assuming a is real): a = 0, 0<a≤1 and a>1. Only the latter one is of any interest, since a=0 is trivial, and the integral does not converge if 0<a≤1.

 

[HallsofIvy]

You say you are doing this as a contour integral in the complex plane but you haven't said what contour!

 

[scurty]

There are three cases to consider here (assuming a is real): a = 0, 0<a≤1 and a>1. Only the latter one is of any interest, since a=0 is trivial, and the integral does not converge if 0<a≤1.

a<-1 is also a valid case.

 

[zezima1]

Has been a while, but going back into solving this integral.

So I end up with the integral:

∫ (z^-1-z)/(2az + iz²-i) dz

Now the roots of the equations in the denominator are clearly poles. But how do I know, that you can't get higher order poles by reducing the fraction further? And secondly: The z^-1 in the numerator is undefined for z=0 - will that make a problem?

 

[clamtrox]

Has been a while, but going back into solving this integral.

So I end up with the integral:

∫ (z^-1-z)/(2az + iz²-i) dz

Now the roots of the equations in the denominator are clearly poles. But how do I know, that you can't get higher order poles by reducing the fraction further? And secondly: The z^-1 in the numerator is undefined for z=0 - will that make a problem?

Clearly z=0 is a pole. In this case it's easy to find all the poles. Just write the integrand as \frac{P(z)}{Q(z)} where P and Q are polynomials. Then find all the roots of Q, and those are the poles.

 

[zezima1]

So my expression is not correct for evaluating the integral? Please show me how you get that ratio of polynomials you want? :(

 

[scurty]

So my expression is not correct for evaluating the integral? Please show me how you get that ratio of polynomials you want? :(

You just multiplied the denominator by $2z$ and then multiplied the top by 2/i for some reason? The whole reason for multiplying by $\frac{2z}{2z}$ is to get rid of the zeroes in the numerator (and "transfer" them down to the denominator, so to speak).

So, go back and multiply $\displaystyle \int_{\partial \mathbb{D}} \frac{-\frac{i}{2} (z-z^{-1})}{a+ \frac{i}{2} (z-z^{-1})} dz$ by 2z/2z and you will hopefully see what to do from there!

 

[zezima1]

Well, in your version, where did the 1/iz term go?
The original integral is:

∫(1/2i(z^-1-z)/(a-1/2i(z-z^-1) dz/iz

How do you come to your version from that? Seems to me like you left out the iz term under dz, which I then put back in? Maybe I'm just a bit tired, but I should think something is wrong.

 

[zezima1]

Hi all, I think I managed to find the poles inside the unit circle, which as ai - i√(a²-1) in agreement with my solutions notes. However, I also found that z=0 is a pole, which my book does agree on. Who is right, and why? :)