Understanding Singularities in Complex Functions

Click For Summary
SUMMARY

The discussion focuses on identifying the nature of singularities in complex functions, specifically examining three functions: f(z) = (z^3 + 3z - 2i)/(z^2 + 1) at a = i, f(z) = z/(e^z - 1) at a = 0, and e^e^(-1/z) at a = 0. The first function has a removable singularity with a limit of 0. The second function is also likely removable, as the limit approaches 1, although the power series expansion presents challenges. The third function requires analysis of limits from both negative and positive real values of z to determine its singularity type.

PREREQUISITES
  • Complex analysis fundamentals
  • Understanding of singularities: removable, poles, and essential
  • Power series expansions and convergence
  • Limit evaluation techniques in complex functions
NEXT STEPS
  • Study the classification of singularities in complex analysis
  • Learn about Laurent series and their applications
  • Explore the concept of limits in complex functions, particularly one-sided limits
  • Practice problems involving power series expansions of exponential functions
USEFUL FOR

Students and professionals in mathematics, particularly those studying complex analysis, as well as educators seeking to enhance their understanding of singularities in complex functions.

moo5003
Messages
202
Reaction score
0

Homework Statement


Determine if the following are removable, pole (with order), or essential singularities.

a) f(z) = (z^3+3z-2i)/(z^2+1) a=i

b) f(z) = z/(e^z - 1) a=0

c) e^e^(-1/z) a=0

2. The attempt at a solution

Part a is pretty straightforward, just simplify it down to (z-i)(z+2i)/(z+i) and the sing is removable with value 0.

Part b is where I'm having some trouble. I'm pretty sure its also removable since when I graphed it the limit looks like it converges to 1. Though when I expand it out into a power series I can't seem to get it to work.

z = Sigma (0 to inf over n) delta(n-1)z^n
delta = Kroniker delta function, 1 at delta(0) and 0 everywhere else.

e^z = Sigma (z^n/n!)
-1 = -Sigma (d(n)z^n)

After failing to come up with anything usefull with that method I decided to show that the actual limit was one. I couldn't seem to come up with a delta such that given an epsilon |z|<d => |f(z) - 1|<epsilon.

Overall, I was wondering if you guys could give me some hints on how to tackle the problem. :bugeye:
 
Physics news on Phys.org
For b) your idea to use series is fine. Just put in the expansion of e^z. What's the problem?
 
BTW for c), you might want to consider the limits as z->0 for z negative real and z positive real. What do you learn from considering these two limits?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
13
Views
2K
Replies
12
Views
3K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K