Residue of a complex function with essential singularity

benygh2002
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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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