Determine the singularities and evaluate residues

  • Thread starter brydustin
  • Start date
  • #1
brydustin
205
0

Homework Statement



f(z) = \frac{z*exp(+i*z)}{z^2+a^2}

Homework Equations



Res(f,z_0) = lim_z->z_0 (1/(m-1)!) d^{m-1}/dz^{m-1} {(z-z_o)^m f(z)}

The Attempt at a Solution



I have no clue how to do this because I don't know how to determine the order of the pole for a function of this form. For example, I could easily do this for a function like (1/(z^2+a^2))
 

Answers and Replies

  • #2
naele
202
1

Homework Statement



[tex] f(z) = \frac{z*exp(+i*z)}{z^2+a^2} [/tex]

Homework Equations



[tex]Res(f,z_0) = lim_z->z_0 (1/(m-1)!) d^{m-1}/dz^{m-1} {(z-z_o)^m f(z)}[/tex]

The Attempt at a Solution



I have no clue how to do this because I don't know how to determine the order of the pole for a function of this form. For example, I could easily do this for a function like [itex](1/(z^2+a^2))[/itex]

Hi there friend, consider expanding the numerator and the denominator separately in a Laurent series and then multiply them together to pick out the coefficient of the z^-1 term
 
  • #3
brydustin
205
0
Okay... . I'm now specifically asking how to compute it for the essential singularity, when z=∞ what is the residue?
 
  • #4
naele
202
1
I thought you needed to determine the singularities and their residues? I don't know if you've learned this yet, but if you have a function f(z)=g(z)/h(z) where h(z0)=0 and g(z0)!= 0 then the order of the pole at z0 is equal to the order of the zero at h(z0). The order of the zero can be found by looking at the order of the first non-vanishing coefficient in the series expansion of h(z) around z0. In this case, we have h(z)=z^2+a^2 so for example h(ia) = 0. If we try to expand in a series h(z)=c0 +c1(z-ia)+c2(z-ia)^2+... then c0=0 because h(ia) = 0 and c1=dh(z)/dz evaluated at z=ia which gives c1=2ia. That's non-vanishing so the order of the zero at h(ia) is 1. Hence there's a simple pole.

Then the residue at z=ia is given by g(ia)/h'(ia).

I could be wrong because it's been a while, but good luck.
 

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