1

Hi guys in this integral

$$\int_{+\partial D}\dfrac{\sin\left(\dfrac{1}{z}\right)\cos\left(\dfrac{1}{z-2}\right)}{z-5}\mathrm{dz}$$

where $D=\{z\in\mathbb{C}:|z|<3\}$, is $z=5$ a pole, and are $z=2$ and $z=0$ essential singularities ?

If the domain is $|z|<3$, is the integral zero?

Dylan
  • 16,575
Marià
  • 241
  • 1
  • 2
  • 8
  • in which sense ? – Marià Jul 08 '18 at 13:21
  • Yes essential singularities for the transcendental functions because their Taylor/Laurent expansions have non-zero coefficients for infinitely many negative exponents. By the way, is this Rome or where? – mathreadler Jul 08 '18 at 13:21
  • So 5 is a pole ? but it doesn't fit in the domain so the integral i zero ? , no this is naples southern italy :D – Marià Jul 08 '18 at 13:25
  • Okay, yes at $z=5$ is a pole, the Cauchy theorem of residues states that any closed contour along a "simple" path (in the right orientation - importante!) with be determined by sum of residues inside the loop. But try to make a habit to read more carefully the theorems when you realize you miss something to be able to calculate. And if it is not stated explicitly in the theorem often there can be examples you can read and learn. – mathreadler Jul 08 '18 at 13:29
  • sorry but i haven't understand how i can procede to get the result of the integral... – Marià Jul 08 '18 at 14:45
  • is the integral = 0 ? – Marià Jul 08 '18 at 14:59

1 Answers1

0

Hint You have two options,

  1. To rotate one way and consider residues on inside of circle, then you will get two essential singularities which in general can be nasty to evaluate (but not always!).
  2. Or you can flip direction of contour and instead consider outside of circle. Then you will get the "nice" pole $z=5$ inside of your area, and potentially a pole at complex infinity will also be "inside" since you turned the circle "inside out" so to speak.

It was a while ago I did this, so I don't remember all the details you will need to refer to to be allowed to do nr 2.

mathreadler
  • 25,824
  • so if a take the poles that are outside the circle how are they related to the poles at infinty? – Marià Jul 08 '18 at 15:24
  • @Maria If it works the way i remember they would be complementary in the sense that sum of all outside + pole at infinity = sum of all inside but opposite sign. Take formula at wikipedia $-Res\left(\frac 1 {z^2}f(\frac 1 z ),0\right)$ for $f(z) = \frac 1 z$ you see it will evaluate to precisely - (the pole at $z=0$). – mathreadler Jul 08 '18 at 15:29
  • This is by necessity of continuity, imagine a loop shrinking to a point (which is not a pole of the function). It's integral would need to go to 0 for a function nice enough to be analytic at the point you squeeze it around. Then all poles will be outside and pole at infinity would need to match sum of all others but negative sign in order to sum up to the obvious integral which in that case would be 0. – mathreadler Jul 08 '18 at 15:33
  • but in zero we have an essential singularity not a pole... i'm not undersanding this... – Marià Jul 08 '18 at 15:52
  • Yes, that is why approach 2 would probably be preferrable here as I suspect residue at infinity is easier to calculate than residues of essential singularities at 0 and 2. Let it take some time, read the theorems, read calculation examples from lectures if you have them, take small pauses in between. – mathreadler Jul 08 '18 at 15:56
  • but i havent'really understood the 2 method.... i can't see the relationship between the the pole in 2 and the infinite residues – Marià Jul 08 '18 at 15:59
  • The residue theorem ties values of a complex integral along a closed loop to the values of the residues inside the loop. What counts as "inside" the loop depends on which orientation you traverse the loop. Clockwise gives one value, counter-clockwise another. They should be - each other. But what you can gain on is if the residues outside are easier to calculate than the ones inside. Then you can select the opposite orientation of the contour and calculate the outside residues if they look less complicated to evaluate. – mathreadler Jul 08 '18 at 16:03
  • Ok i get it , but if the loop doesn't have any residues inside but only 5 that is outside we have this situation: 5 is outsiode the loop and no one is inside the loop. how can the residues be the same ? – Marià Jul 08 '18 at 16:17
  • @Marià If no residues inside then the integral will be $0$ since it is a sum of 0 terms and the sum of the residues outside should also be $0$. So the residue at $\infty$ "balances out" the other ones outside so to speak. – mathreadler Jul 08 '18 at 16:21
  • Ahhh ok so in that case, beacuse z=5 is outside the loop (|z|<3) the integral is = 0? – Marià Jul 08 '18 at 16:25
  • You have two inside the circle $|z|\lt 3$: the essential singularities. Those may be tricky to calculate especially if it was long time ago you calculated series. You have two outside, $z=5$ and $z=\infty$ you can choose to calculate which sum you think will be easiest. – mathreadler Jul 08 '18 at 16:30
  • but i remember that you can calcoulate residues only for poles and not for essentials singualritier – Marià Jul 08 '18 at 16:38
  • @Marià It is possible to calculate residues of essential singularities, but it becomes more complicated than normal poles. You need to find the coefficient for the $z^{-1}$ term in the Laurent power series expansion. If you are skilled in series calculation you can do it that way also. – mathreadler Jul 08 '18 at 16:43
  • 1
    ahhh ok now i got it , cause calculate the residues of the essential singularities is hard , i can calcolate the residues at infinite add the residue at 5 ( because infinite and 5 are both external) and after change sign and i get the sum of the residues of the singularities inside the loop , is that correct? – Marià Jul 08 '18 at 16:43
  • @Marià Yes that sounds right. If you want an example you can check this question https://math.stackexchange.com/questions/434606/calculate-residue-at-essential-singularity/845625 and the answer by Simon who shows how to calculate an essential singularity and the comment by Daniel Fischer of how to avoid to do it by calculating residue at infinity instead. – mathreadler Jul 08 '18 at 17:47
  • so to see if i have understand everything.... if i have essential singualirties inside the loop instead outside i have poles , i sum all the resiodues otutsides , so the sum of the internal residues will be -(outsides residues) , right ? – Marià Jul 08 '18 at 17:53
  • @Marià yes, just be careful so you don't forget pole at infinity. It's so far away so it will always be outside. – mathreadler Jul 08 '18 at 17:56
  • yes , so for instance if i have 2 essentnial singualrrities inside the loop and one pole outside. the sum of the residues of the essential singualrities ( so the residues inside) = - (res(poles)+res(infinite)) that's right ? – Marià Jul 08 '18 at 18:06
  • Yes. Have you tried it? – mathreadler Jul 08 '18 at 18:09
  • no i will try tonight now i got to rest a bit... thlk so much you made my day – Marià Jul 08 '18 at 18:13