Residue of exp(1/(z+i)): How to Find and Solve for z=-i

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Homework Help Overview

The discussion revolves around finding the residue of the function exp(1/(z+i)) at the point z=-i, which is identified as an essential singularity. Participants are exploring methods to approach this problem within the context of complex analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to find the residue at an essential singularity, expressing uncertainty about the applicability of standard residue formulas. Some participants suggest expanding the function as a Laurent series and identify the residue as the coefficient of the term proportional to 1/(z+i). Others inquire about the process of expansion and seek hints on where to begin.

Discussion Status

The discussion is active, with participants providing hints and suggestions for approaching the problem. There is a focus on expanding the function and utilizing series, but no consensus on a specific method has been reached yet.

Contextual Notes

Participants note that the original poster's textbook lacks sufficient detail on Laurent series, which may be influencing their understanding and approach to the problem.

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Hi, I want to know how you find the residue of z=-i for the function exp(1/(z+i)). Clearly, the function has an essential singularity at z=-i so the good ol' formula for the residue for a pole of order m, doesn't really work here. What do I do? :)
 
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You expand the function as a Laurent series around the pole and recall that the residue at that point is the coefficient in front of term proportional to [itex]\frac{1}{z+i}[/itex]
 
okay, can you help me how that is done. My book's section os Laurent series is quite poor. The only thing I know about expanding functions as them, is that you can sometimes use geometric series. You don't need to say how I should do it, just hint me at where to start.
 
Just expand e^x as a usual Taylor series, then plug in x=1/(z+i) and it's done!
 

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