Singularities of a complex function

  • #1
Mattbringssoda
16
1

Homework Statement


[/B]
Find and classify all singularities for (e-z) / [(z3) ((z2) + 1)]

Homework Equations

The Attempt at a Solution


[/B]
This is my first attempt at these questions and have only been given very basic examples, but here's my best go:

I see we have singularities at 0 and i.

The 0 corresponds with z3, so upon inspection it's a third order pole.

To determine the order for the i singularity, I multiply the function by (z - i) and use L'Hospital's rule

Lim (z -> i ) of [ (z-i) (e-z) ] / [ (z3) ((z2) + 1) ]

= Lim (z -> i) of [ (z-i) (-e-z) + (e-z) ] / [ (5z4) + 3z2 ]

= (e^-i) / 2

Which is a finite number, and since I used the first order term (1-i), this is indeed a first order pole, according to what I've been taught.

I'm worried that I'm misunderstanding how to use L'Hospital here and was hoping I could get a second set of eyes from someone familiar with these problems.

Thanks!
 
Physics news on Phys.org
  • #2
Mattbringssoda said:
I see we have singularities at 0 and i.
You missed one. Remember that (z2+1)=(z+i)(z-i).
Mattbringssoda said:
To determine the order for the i singularity, I multiply the function by (z - i) and use L'Hospital's rule
You should slow down a bit - multiply by (z-i) is correct, but then you have (z-i) in both nominator and denominator...
 
  • #3
Writing [itex]\frac{e^{-z}}{z^3(z^2+ 1)}[/itex] as [itex]\frac{e^{-z}}{z^2(z- i)(z+i)}[/itex] it should be immediately obvious that z= 0 is pole of order three and that z= i and z= -i are poles of order one.
 
  • #4
Thank you two very much; it all stemmed down to a simple oversight that I wasn't catching from the (z^2 + 1) term. Thanks again!
 

Similar threads

Replies
2
Views
2K
Replies
5
Views
1K
Replies
2
Views
1K
Replies
14
Views
3K
Replies
7
Views
1K
Replies
9
Views
1K
Replies
17
Views
2K
Replies
1
Views
1K
Back
Top