Singularities of a complex function

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Homework Help Overview

The problem involves finding and classifying singularities of the complex function (e-z) / [(z3) ((z2) + 1)]. Participants are exploring the nature of these singularities and their orders.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss identifying singularities at 0 and i, with one noting a potential oversight regarding another singularity. There is mention of using L'Hospital's rule to determine the order of the pole at i, and some participants question the application of this method.

Discussion Status

The discussion is active, with participants providing insights and corrections. One participant acknowledges a misunderstanding related to the (z^2 + 1) term, indicating that the conversation is productive and focused on clarifying concepts.

Contextual Notes

There is an indication of a basic understanding of singularities, with participants referencing their level of experience and the need for careful application of mathematical rules.

Mattbringssoda
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Homework Statement


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

Homework Equations

The Attempt at a Solution


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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!
 
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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...
 
Writing \frac{e^{-z}}{z^3(z^2+ 1)} as \frac{e^{-z}}{z^2(z- i)(z+i)} 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.
 
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!
 

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