Residue of a complex function with essential singularity

In summary, the student is struggling with computing residues of a complex function in their mechanical engineering homework. They have received some helpful comments and have some understanding of essential singularities and series expansion, but are still unable to find a solution. The function they are working with is in the form of f(z)=exp(A*Z^N+B*Z^-N)/Z, with A, B, and N being real constants. They are attempting to compute the residue at z=0 and have written a Laurent series, but are unsure if they are on the right track. The expert responds by explaining that an essential singularity does not have a residue, and that the student's Laurent series should not have an infinite number of negative powers. They
  • #1
benygh2002
3
0

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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  • #2
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
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}}}{z-a}[/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.
 

Related to Residue of a complex function with essential singularity

What is a residue of a complex function with essential singularity?

A residue of a complex function with essential singularity is the coefficient of the term with the highest negative power in the Laurent series expansion of the function around the essential singularity.

How is the residue calculated?

The residue can be calculated using the formula Res(f,a) = (1/(n-1)!) * lim(z->a) [(d^n-1/dz^n-1)[(z-a)^nf(z)]] where n is the order of the essential singularity a and f(z) is the complex function.

Why is the residue important?

The residue helps in calculating complex integrals and finding the value of certain contour integrals. It also plays a role in determining the poles and essential singularities of a function.

Can the residue be zero?

Yes, the residue can be zero if the essential singularity is not a pole or if it is a removable singularity. In these cases, the Laurent series expansion will not have any negative powers and the residue will be zero.

How is the residue affected by the location of the essential singularity?

The value of the residue depends on the order and location of the essential singularity. As the singularity moves closer to the origin, the absolute value of the residue tends to increase. If the singularity is at the origin, the residue may be infinite.

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