A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Show that the singular point of the following function is a pole. Determine the
order m of that pole and the corresponding residue B.
f(z)=(1coshz)/(z^3)
anonymous
 3 years ago
Show that the singular point of the following function is a pole. Determine the order m of that pole and the corresponding residue B. f(z)=(1coshz)/(z^3)

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I really wish my university had an applied complex variables class :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think we can at least start this if i recall \(\cosh(z)=\cos(iz)\) if so, we can expand in a taylor series

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we can try \[\cosh(z)=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this should give \[1\cosh(z)=\sum_{k=1}^{\infty}\frac{z^{2k}}{(2k)!}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Which would then give: \[\sum_{k=1}^{\infty}\frac{z^{2k3}}{(2k)!}\] right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0divide by \(z^3\) first term is \(\frac{1}{2z}\) and all the others have \(z\) in the numerator

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@malevolence19 yes, that is what i get too. or shift the index either way.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So the the residue is just 1/2?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so b1 comes out to be 1/2.yeah?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes if my memory serves me in this context the residue is the coefficient of the \(z^{1}\) term

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, 1/2 i forgot about that in front.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i would not place much money on this, because it has been a while, but i am pretty sure it is correct

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and pole is??i don't have any idea what the heck is it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It would just be zero right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It it was like: \[\frac{B}{(z\lambda)}\] then the pole would be lambda? I think, but I'm going off very little knowledge.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lambda is a singular point..not sure about pole

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, the pole is at 0, because that is where the denominator is undefined. it is a simple pole because it is a zero of order one

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0^the denominator becomes undefined at singular points...so every singular point is a pole??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think its pole is 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a "pole" is a kind of singular point think of a simple example from pre calc class \[\frac{x^24}{x2}\] vs \[\frac{x^2}{x2}\] in the first example, 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this thing implies m=1 at b1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in both examples, 2 is a singular point because the function is not defined at 2 but in the first example, 2 is a removable singularity, because you can remove it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in the second example, 2 is a pole, in pre calc a "vertical asypmtote"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0however it is spelled

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isolated singular point.what is it??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what you posted was about the order of the pole, not the pole itself

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in your case above, we had \(\frac{1}{2z}\) plus stuff with positive exponents, means you have a pole of ORDER 1, not a pole at \(z=1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here is a nice succinct explanation (very short) with example http://www.solitaryroad.com/c612.html

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0implies order of pole=k if bk is the coefficient of z^1 term.right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the order of the pole is the multplicty of the zero on the denominator, to put it in pre calculus terms, example \[\frac{1}{(z5)^3}\] has a pole of order 3 at \(z=5\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{z2}\] has a pole of order 1 at \(z=2\) a pole of order one is called a simple pole
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.