Expanding f(z) in a Laurent Series

In summary, the conversation discusses expanding the function f(z)=1/z(z-2) in a Laurent series that is valid for the annular region 0<|z-3|<1. The equations for this expansion are provided, including the Taylor series for 0.5(1/(z-2)) and the question of why the answer for 0.5(1/z) is not 1/2(z-3) *∑((-1)k*(3/(z-3))k). The significance of the annular region is also mentioned as being important in understanding the answer to the original question.
  • #1
Pouyan
103
8

Homework Statement


Expand the function f(z)=1/z(z-2) in a Laurent series valid for the annual region 0<|z-3|<1

Homework Equations


I know 1/z(z+1) = 0.5(1/(z-2)) - 0.5(1/z)

Taylor for 0.5(1/(z-2)) is : ∑(((-1)k/2) * (z-3)k) (k is from 0 to ∞)For the second 0.5(1/z) the answer is a Taylor : ∑((1/6)*(-1/3)k * (z-3)k)

But why the answer for 0.5(1/z) is not 1/2(z-3) *∑((-1)k*(3/(z-3))k) (k is from 0 to ∞) ?

The correct answers for both are from Taylor but I thought for the second is a Laurent
 
Physics news on Phys.org
  • #2
Did you consider the annular region you were given?
 
  • Like
Likes Pouyan
  • #3
vela said:
Did you consider the annular region you were given?
Yes but I can't draw it in this page !
 
  • #4
You don't need to draw it.

What's the significance of the annular region? Why did the problem bother giving it to you? If you understand that, it's the answer to your original question.
 

FAQ: Expanding f(z) in a Laurent Series

What is a Laurent Series?

A Laurent Series is a representation of a complex function, f(z), as an infinite sum of terms. It is used to express a function in terms of its power series expansion around a point, including both positive and negative powers of (z-a), where a is the center of the series.

Why is it important to expand f(z) in a Laurent Series?

Expanding f(z) in a Laurent Series allows for a better understanding of the behavior of a function in a given region. It can also help with calculating integrals and derivatives of the function, as well as finding the zeros and poles of the function.

What is the difference between a Taylor Series and a Laurent Series?

A Taylor Series only includes positive powers of (z-a), whereas a Laurent Series includes both positive and negative powers. This allows for a Laurent Series to be used for functions with singularities or poles, while a Taylor Series is only valid for functions that are analytic in a given region.

How do you find the coefficients in a Laurent Series?

To find the coefficients in a Laurent Series, one can use the formula for the nth derivative of f(z) at the center, a, divided by n!, multiplied by (z-a)^n. Another method is to use the Cauchy Integral Formula to calculate the coefficients.

What is the region of convergence for a Laurent Series?

The region of convergence for a Laurent Series is the set of all points in the complex plane where the series converges. This can be determined by finding the radius of convergence, which is the distance from the center, a, to the nearest singularity or pole of the function.

Similar threads

Calculating Laurent series of complex functions
Replies
3
Views
740
Finding residues with Laurent series.
Replies
1
Views
1K
Expanding f(z) in a Laurent Series for |z|>3
Replies
3
Views
1K
Find the Laurent Series of a function
Replies
9
Views
1K
Prove if ##a\cdot c = b \cdot c## then ##a = b## using Peano postulates
Replies
2
Views
799
Link between Z-transform and Taylor series expansion
Replies
2
Views
2K
Laurent expansion of ##ze^{1/z}##
Replies
12
Views
4K
On pointwise convergence of Fourier series
Replies
3
Views
1K
Solution verification of ODE using Frobenius' method
Replies
3
Views
873
What is the residue of f(z) = e^(-2/z^2) using a laurent series?
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top