Analyzing Singularities at z=2 & -1/3

In summary, the question is about stating the nature of singularities at z=2 and z=-1/3. The person is unsure of how to do this and is seeking help. Another person suggests reviewing the definition of a pole of order n instead of expanding the series. The person who asked the question also wants to understand the technique of using a laurent expansion to determine the nature of a singularity.
  • #1
AlBell
11
0

Homework Statement



I have been asked to state the precise nature of the singularities at z=2 and z=-1/3 in
t6rtb6.png


Homework Equations



I know the laurent series is given by
vymfz4.png


The Attempt at a Solution



I think I need to expand the series out into a laurent series around z=2 and z=-1/3 but I am really stuck on how to do this and would really appreciate a bit of help! From some examples I have seen I need to manipulate the denominator into a known series and then expand this but I am unsure of how to do that. Thanks
 
Physics news on Phys.org
  • #2
You don't need to expand anything. You just need to review the definition of a pole of order n.
 
  • #3
I know I don't have to but I thought that a laurent expansion was another way to find out the nature of a singularity, like what the order of the pole was, and I just wanted to try and get a grasp of this technique as well and get the same answer for both techniques
 

FAQ: Analyzing Singularities at z=2 & -1/3

What is the significance of analyzing singularities at z=2 and -1/3?

The values of z=2 and -1/3 represent singularities in a mathematical function. Analyzing these singularities can provide insights into the behavior of the function and its limitations.

How do you determine if a singularity exists at a specific value of z?

A singularity exists at a specific value of z if the function becomes undefined or infinite at that point. This can be determined by looking at the function's equation and identifying any values of z that cause the denominator to equal zero.

Can a singularity at z=2 or -1/3 be removed?

In some cases, it is possible to remove a singularity by using mathematical techniques such as factoring or substitution. However, in other cases, the singularity is inherent in the function and cannot be removed.

How do singularities at z=2 and -1/3 affect the overall behavior of the function?

Singularities can cause discontinuities or abrupt changes in the behavior of a function at the point where they occur. This can result in unexpected or undefined outputs for certain inputs.

Are there any real-world applications for analyzing singularities at z=2 and -1/3?

Yes, analyzing singularities is an important aspect of studying mathematical functions and can be applied in various fields such as physics, engineering, and finance. By understanding the behavior of a function at singularities, we can make more accurate predictions and solve real-world problems.

Similar threads

Calculating Laurent series of complex functions
Replies
3
Views
748
Singularities and Laurent series
Replies
1
Views
1K
What is the singularity of e^(-1/z^2)?
Replies
25
Views
10K
Elliptic functions, removable singularity, limits,
Replies
1
Views
2K
On weakly singular equations and Frobenius' method
Replies
7
Views
968
Expanding f(z) in a Laurent Series for |z|>3
Replies
3
Views
1K
Expanding Laurent Series around Singularities
Replies
26
Views
2K
Classifying singularities of a complex function
Replies
2
Views
1K
Laurent series of z^2sin(1/(z-1))
Replies
2
Views
3K
Find and classify the isolated singularity
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top