Singularities of a complex function

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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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