Complex Analysis: Find Laurent Series for f(z) = 1/(e^-z - 1) About 0

In summary, the speaker is struggling with a math for physicists course that involves functions of complex variables, specifically Laurent series residue. They are seeking recommendations for a more beginner-friendly complex analysis book with practice problems. The specific problem they are having trouble with is finding the first 3 terms of the Laurent series for f(z)= 1/(e^-z - 1) around 0, and they have tried using residue and polar coordinates without success. They mention the book "Churchill Complex Variables and Applications" as a standard, but also recommend a free online textbook for easier understanding.
  • #1
Dathascome
55
0
Hi there, I'm taking this math for physicists course and we're doing some stuff with functions of complex variables (laurent series residue etc), and I"m having a bit of trouble.

I'm not so happy with the book we use. It's a great reference book if you know what you're doing already but terrible to learn out of (in my opinion). Does anyone know the name of a good complex analysis book, that's not to heavy from someone trying to pick it up quick but has some problems in it too?

The problem that's been driving me crazy lately is to find the first 3 terms of the laurent series for f(z)= 1/(e^-z - 1) about 0. How do I do it using and not using residue?

I've tried a few things but to no avail. I tried expanding the exponential and using the fact that 1/2*pi*int(z^(m-n-1)dz= delta function of m,n
(sorry I don't know how to write that nicer)...but that didn't really get me any where...I tried using polar coordinates again to no avail...I'm not sure what else to do.
 
Physics news on Phys.org
  • #2
A standard is Churchill Complex Variables and Applications
 
  • #3
If you are looking for an easy to understand, free textbook on complex analysis try : http://www.its.caltech.edu/~sean/book.html [Broken]

It has great examples that are worked through.
 
Last edited by a moderator:

What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of functions of complex numbers. It involves the use of concepts such as limits, derivatives, and integrals to understand the behavior of complex-valued functions.

What is a Laurent series?

A Laurent series is a representation of a function as an infinite sum of terms involving powers of a complex variable. It is a generalization of a Taylor series, which represents a function as a sum of powers of a complex variable.

How do you find the Laurent series for a given function?

To find the Laurent series for a given function, we first need to identify the singularities of the function. Then, we can use the formula for a Laurent series to express the function as an infinite sum of terms involving powers of the complex variable, centered around the singularity.

What is a singularity in complex analysis?

A singularity is a point where a function is undefined or behaves in a special way. In complex analysis, singularities can be classified as poles, essential singularities, or removable singularities.

How do you find the Laurent series for f(z) = 1/(e^-z - 1) about 0?

To find the Laurent series for f(z) = 1/(e^-z - 1) about 0, we first need to identify the singularity at z = 0. Then, we can use the formula for a Laurent series to express the function as an infinite sum of terms involving powers of the complex variable z, centered around 0. The resulting Laurent series will have a principal part and a regular part, and it can be simplified further using algebraic manipulations.

Similar threads

  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
234
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
3K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
314
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
859
Back
Top