Does f(z) have a singularity at infinity and how can its residue be obtained?

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The function f(z) = ze^(iz) / (z^2 + a^2) exhibits a singularity at infinity. To analyze this, one must transform the variable using w = 1/z and evaluate the limit as w approaches 0. The limit of exp(i/w) as w approaches 0 diverges, confirming the presence of a singularity. The residue at infinity can be determined using L'Hôpital's rule, and the implications of this singularity are relevant in various physical theories.

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does f(z)=\frac{ze^{iz}}{z^2+a^2} have a singularity at infinity?

if so, how do i get the residue there?
 
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Does f(w):=f(1/z) has a singularity at w=0 is what you must ask yourself.

Btw - you're really not in the right forum.
 
Last edited:
okay so i transform z -> 1/w then take lim w-> 0... if it blows up then i do have a singularity... how do i get lim w->0 of exp(i/w) ?

well first, i think i need l'hospital's (for the whole function). then, can i use the fact that when taking a limit it can be approached along any line on the Z-plane? i.e. use the path along i-axis ?

i think the conclusion will be that it blows up. three follow up questions. 1. how do i get the residue at infinity? 2. what is the conclusion in a case wherein the limit does not exist? 3. Is the singularity at infinity and/or its residue useful? (i mean i know the finite singularities are useful in integration, does this arise in some physical theory?)

i am very sorry for posting in the wrong forum.. thanks for all the help
 

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