A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
Compute the Laurent expansion of z/((z1)(2z)) for each of the following regions: abs(z) <1 and abs(z2)<1.
 2 years ago
Compute the Laurent expansion of z/((z1)(2z)) for each of the following regions: abs(z) <1 and abs(z2)<1.

This Question is Closed

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Partial fraction decomposition would probably be a good start.

Hennie
 2 years ago
Best ResponseYou've already chosen the best response.0In the expression (A/(z1)) and B/(z2) is the value for A=1 and B=2 ?

Hennie
 2 years ago
Best ResponseYou've already chosen the best response.0For the laurent's expansion should I then use (1/(1z)) and ((1/(1z/2)) and expand each term seperately ?

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1It should be \[ \frac{1}{z1}  \frac{2}{z  2} \] so yes you're right... and now yes, you want to expand these two terms around those two points. For example, for z<1, we can rewrite that as \[ \frac{1}{1  z} + \frac{1}{1  z/2} = \sum_{n=0}^\infty z^n+ \sum_{k=0}^\infty (\frac{z}{2})^k \] you can simplify that if you'd like, bringing them both under the same summation sign and all that stuff, but that's the idea. Since z<1, these series converge. For z2<1, they don't necessarily converge, so you have to expand them about the point z = 2 for the second part, but the same deal applies.
Ask your own question
Ask a QuestionFind 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.