Homework Help: Poles and residues

1. Feb 18, 2015

nmsurobert

1. The problem statement, all variables and given/known data
Show that all the singular points of the following functions are poles.
Determine the order of each pole and the corresponding residue.

z/cosz

2. Relevant equations
maybe cosz = 1/2 ez+e-z is relevant but i dont use it here

3. The attempt at a solution
the pole i found was z = π/2 + 2nπ and -π/2 +2nπ

i attempted to find the residues using limits.
limz→π/2 + 2nπ (z-(π/2 + 2nπ)) z/cosz

doing some annoying algebra and just evaluating at π/2 + 2nπ i get 0/0. so i can use L'hopitals rule, right?
doing so i get my limit to equal π/2 + 6nπ

is that right?

i know i can use lorent series but i honestly dont know where to start with that mess haha

2. Feb 18, 2015

Dick

That's pretty bad. cosz = (1/2)(eiz+e-iz) so your relevant equation is wrong and you do want to use it to show the only poles are on the real axis. Given that you do have the right poles on the real axis. And yes, you can use l'hopital. But show how you got that limit.

3. Feb 18, 2015

nmsurobert

oops i forgot the i's. but doesnt just putting the poles into the original equation shows that theyre poles?

i didn't realize it was "pretty bad" LOL

4. Feb 18, 2015

Dick

It shows they are poles, it doesn't show they are the only poles. That's not my only complaint. The limit to compute the residue is bad and I can't tell why because you didn't show your work.

5. Feb 18, 2015

nmsurobert

lim z → (z-(π/2 + 2nπ)) z/cosz

z2-π/2z - 2nπz

then evaluate at z = π/2 + 2nπ

π2/4 + π/2(2nπ) +π/2(2nπ) + 4n2π22/4- π/2(2nπ) π/2(2nπ) - 4n2π2 = 0

cos(π/2 + 2nπ) = 0

0/0

l'hopital rule
2z - π/2 - 2nπ/ - sinz evaluate at π/2 + 2nπ

2(π/2) + 4nπ -π/2 -2nπ / -sin(π/2 + 6nπ)

= -π/2 - 2nπ

so my first calculation was different but its still the same method.

6. Feb 18, 2015

Dick

"Different" is what I wanted since the first one was bad. That one looks ok (with some typos). Now what about the -π/2+2nπ series and what about showing that the ONLY poles are on the real axis?

7. Feb 18, 2015

nmsurobert

im gonna have to do more reading because i dont know what youre talking about with "-π/2+2nπ series" i thought i take the limit and the limit is the residue.

i also need to do some more reading because i dont know how to find any other poles.

im happy ive made it this far though, i had no idea what the hell i was looking at 2 hours ago.

8. Feb 18, 2015

nmsurobert

thank you though, even though youre kind of a dick. ;-)

9. Feb 19, 2015

Dick

You said "the pole i found was z = π/2 + 2nπ and -π/2 +2nπ". So far you've only found the residue at π/2 + 2nπ. What about -π/2 +2nπ?

10. Feb 19, 2015

nmsurobert

sorry, i went to sleep then to class lol.
the next one should be -π/2 +2nπ.

using the same messy method

11. Feb 19, 2015

Dick

Right again. It's not really all that bad is it? Now can you show they are the only poles using your relevant equation? And what's the order of the poles? Yes, I'm kind of a dick :-;

12. Feb 19, 2015

nmsurobert

i think theyre both order 1.

im not sure how to use the equation to show that. im guessing it has something to do with using the ln|z| + iarg though. not sure if thats right.

13. Feb 19, 2015

Dick

You are correct about the order, can you say why? As to why they are the only poles, if $cos(z)=0$ then $e^{iz}=-e^{-iz}$, right? Put $z=a+ib$ and show $b$ must be 0.