Poles & Residues: Find & Order of z/cosz

[nmsurobert]

Homework Statement

Show that all the singular points of the following functions are poles.
Determine the order of each pole and the corresponding residue.

z/cosz

Homework Equations

maybe cosz = 1/2 e^z+e^-z is relevant but i don't use it here

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.
lim_{z→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 don't know where to start with that mess haha

 

[Dick]

Homework Statement

Show that all the singular points of the following functions are poles.
Determine the order of each pole and the corresponding residue.

z/cosz

Homework Equations

maybe cosz = 1/2 e^z+e^-z is relevant but i don't use it here

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.
lim_{z→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 don't know where to start with that mess haha

That's pretty bad. cosz = (1/2)(e^iz+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.

 

[nmsurobert]

oops i forgot the i's. but doesn't just putting the poles into the original equation shows that theyre poles?

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

 

[Dick]

oops i forgot the i's. but doesn't just putting the poles into the original equation shows that theyre poles?

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

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.

 

[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) + 4n²π²-π²/4- 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.

 

[Dick]

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) + 4n²π²-π²/4- 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.

"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?

 

[nmsurobert]

im going to have to do more reading because i don't 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 don't know how to find any other poles.

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

 

[nmsurobert]

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

 

[Dick]

im going to have to do more reading because i don't 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 don't know how to find any other poles.

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

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?

 

[nmsurobert]

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

using the same messy method

 

[Dick]

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

using the same messy method

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 :-;

 

[nmsurobert]

i think theyre both order 1.

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

 

[Dick]

i think theyre both order 1.

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

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.