Residue of a complex function with essential singularity

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The discussion revolves around computing the residue of a complex function with an essential singularity, specifically the function f(z) = exp(A*Z^N + B*Z^-N)/Z. The student is confused about the nature of essential singularities, noting that they do not have residues, yet examples show that residues can be computed from Laurent series expansions. Clarifications indicate that while the function has an essential singularity at z=0, the Laurent series should not yield an infinite number of negative powers, implying it is not an essential singularity in this context. The conversation also touches on how to approach similar functions, such as g(z) = e^(Az^n + Bz^-n)/(z-a), and the need to identify poles for residue calculations. Understanding these concepts is crucial for progressing in their mechanical engineering problem.
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Homework Statement



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.



Homework 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).

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.
 
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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.

e^{Az^n+ Bz^{-n}}= e^{Az^n}e^{Bz^{-n}}
is analytic and so has a Taylor's series- with no negative power terms. Dividing by z gives a series have "z^{-1}" as its only negative power. The residue is the coefficient of z^{-1}.
 
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:

f\left(z\right)=e^{z+\frac{1}{z}}

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:

Res\left(f,0\right)=1+\frac{1}{2!}+\frac{1}{2!3!}+\frac{1}{3!4!}+...2) Laurent expnasion of my desired function f(z) is:

f\left(z\right)=\frac{e^{Az^n+Bz^{-n}}}{z}

=\frac{\sum{(Az^n+Bz^{-n})}^k}{z k!}

(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):

g\left(z\right)=\frac{e^{Az^n+Bz^{-n}}}{z-a}

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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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