Singularities of a complex function

[Mattbringssoda]

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!

 

[Svein]

I see we have singularities at 0 and i.

You missed one. Remember that (z²+1)=(z+i)(z-i).

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...

 

[HallsofIvy]

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.

 

[Mattbringssoda]

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!