Determine the singularities and evaluate residues

In summary, the problem is trying to determine the order of the pole for a function of this form, and for the essential singularity, what is the residue?
  • #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))
 
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  • #2
brydustin said:

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
Okay... . I'm now specifically asking how to compute it for the essential singularity, when z=∞ what is the residue?
 
  • #4
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.
 
  • #5
where the pole is of order 2 and the residue can be evaluated using the formula Res(f,z_0) = lim_z->z_0 d/dz {(z-z_0)^2 f(z)}. However, for a function of the form z*exp(+i*z)/(z^2+a^2), it is not clear what the order of the pole is. To determine the singularities of this function, we can set the denominator equal to 0 and solve for z. This gives us z=ia or z=-ia as the two singularities. However, these are not poles, they are actually branch points.

To evaluate the residues at these points, we can use the formula Res(f,z_0) = lim_z->z_0 (1/(m-1)!) d^{m-1}/dz^{m-1} {(z-z_o)^m f(z)}. Since the order of the pole is not clear, we can start by taking the limit as z approaches z_0 = ia. This gives us:

Res(f,ia) = lim_z->ia d/dz {(z-ia)f(z)} = lim_z->ia d/dz {(z-ia)(z*exp(+i*z))/(z^2+a^2)}

We can use L'Hopital's rule to evaluate this limit:

lim_z->ia d/dz {(z-ia)(z*exp(+i*z))/(z^2+a^2)} = lim_z->ia (1/(z+ia))*(z*exp(+i*z))/(z^2+a^2) = (1/(2*ia))*(ia*exp(-a))*(1/(2*ia)) = (1/(4*a))*(exp(-a))

Similarly, we can evaluate the residue at z_0 = -ia:

Res(f,-ia) = lim_z->-ia d/dz {(z+ia)f(z)} = lim_z->-ia d/dz {(z+ia)(z*exp(+i*z))/(z^2+a^2)}

Using L'Hopital's rule again, we get:

lim_z->-ia d/dz {(z+ia)(z*exp(+i*z))/(z^2+a^2)} = lim_z->-ia (1/(z-ia))*(z*exp(+i*z))/(z^2+a^2) = (-1/(2*ia))*(-
 

What are singularities in a mathematical function?

A singularity in a mathematical function is a point at which the function is undefined or becomes infinite. It is usually a point where the function is discontinuous or has a vertical asymptote.

How do you determine the singularities of a function?

To determine the singularities of a function, you need to find the values of the independent variable that make the function undefined or infinite. This can be done by setting the denominator of the function equal to zero and solving for the variable.

What is a residue in mathematics?

A residue in mathematics is the value of a complex function at a singular point. It is often used in complex analysis to calculate the behavior of a function near a singularity.

How do you evaluate residues of a function?

To evaluate the residue of a function at a singular point, you can use the Cauchy residue theorem, which states that the residue is equal to the coefficient of the term with the highest negative power in the Laurent series expansion of the function.

Why is it important to determine the singularities and evaluate residues of a function?

Determining the singularities and evaluating residues of a function can help in understanding the behavior of the function and its limits at these points. This information is also useful in calculating complex integrals and in solving differential equations involving the function.

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