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Residue of a complex function with essential singularity 
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#1
Jul911, 04:23 AM

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1. The problem statement, all variables and given/known data
Hello friends, I'm a student of mechanical engineering and I have a problem with computing residues of a complex function. I've read some useful comments. Now I ve got some ideas about essential singularity and series expansion in computing the residue. However, I still can't find the solution to my problem. I arrived at a complex function in the process of finding a solution to a mechanical problem. Then I have to obtain the residues to proceed to the next steps. 2. Relevant equations The function has the following form: f(z)=exp(A*Z^N+B*Z^N)/Z where A, B and N are real constants (N>=3). 3. The attempt at a solution I want to compute the resiude at z=0. I wrote the Laurent serie of f but got an infinite sum. I do not even know if I am at the right direction. I would be really thankful if someone could give me a hint on this and put me back in the right direction. 


#2
Jul911, 07:34 AM

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Thanks
PF Gold
P: 39,339

An essential singularity does not have a residue.
However, your Laurent series should not have an infinite number of negative powers and so this is NOT an essential singularity. [tex]e^{Az^n+ Bz^{n}}= e^{Az^n}e^{Bz^{n}}[/tex] is analytic and so has a Taylor's series with no negative power terms. Dividing by z gives a series have "[itex]z^{1}[/itex]" as its only negative power. The residue is the coefficient of [itex]z^{1}[/itex]. 


#3
Jul911, 10:49 AM

P: 3

Thanks for your reply.
1) You mentioned that "an essential singularity does not have a residue". I'm a bit confused about this. For example: [itex]f\left(z\right)=e^{z+\frac{1}{z}}[/itex] has an essential singularity at z=0. However. after writing Laurent expansion of f(z) the residue at z=0 can be obtained as a serie: [itex]Res\left(f,0\right)=1+\frac{1}{2!}+\frac{1}{2!3!}+\frac{1}{3!4!}+...[/itex] 2) Laurent expnasion of my desired function f(z) is: [itex]f\left(z\right)=\frac{e^{Az^n+Bz^{n}}}{z}[/itex] [itex]=\frac{\sum{(Az^n+Bz^{n})}^k}{z k!}[/itex] (k=0 to infinity) in which the coefficients of 1/z terms give the residue as a serie (but the serie does not converge). 3) I would be thankful if you could tell me what I should do to find the residue of the following function which is similar to the previous function f(z): [itex]g\left(z\right)=\frac{e^{Az^n+Bz^{n}}}{za}[/itex] where a is a constant. What should I do to find the poles? At what point I should write the Taylor expansion? Thanks for your help. 


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