# Variable exponent causing number of singularities to change for residue?

• newmike
In summary, the function \frac{z^{-k}}{z+1} has a simple pole at z=-1 with residue (-1)^(-k). The nature of the singularities at z=0 depends on the value of k. If k is a rational number, there will be a finite number of residues corresponding to each determination of the root. If k is irrational, there will be an infinite number of residues. The origin is always a branch point in the range 0<k<1. The residue theorem can be used to evaluate the residues by choosing a suitable branch-cut.
newmike

## Homework Statement

Determine the nature of the singularities of the following function and evaluate the residues.

$$\frac{z^{-k}}{z+1}$$

for 0 < k < 1

## Homework Equations

Residue theorem, Laurent expansions, etc.

## The Attempt at a Solution

Ok this is a weird one since we've never covered anything with a non integer exponent, in fact, never with a variable exponent at all.

I realize there is a simple pole at z=-1. If I assume the numerator to be analytic and non-zero at z=-1 (which I think it true for the range of k), then I can calculate the residue at z=-1 by the simple formula: R(-1) = g(-1) / h'(-1), where g(z) is the numerator and h(z) is the denominator. If I carry through with that I get R(-1) = (-1)^(-k) which I am ok with I guess.

The problem is that I don't know what to do with the potential pole as k approaches 1. In that case we'd approach a simple pole at z=0. Alternatively, as k approached 0, we do not have a singularity at z=0 at all. So how do I handle the fact that this is a variable??

Should I try to turn this into a laurent series and go from there?

I'm no expert but would like to start the analysis. Perhaps others more qualified can comment as well. We have

$$\frac{1}{z^k(z+1)}$$

and take for example:

$$\frac{1}{z^{1/2}(z+1)}$$

We can choose the branch-cut along the positive real axis (0,infty). In that case, there are two analytic branches around the pole so two residues:

$$\mathop\text{Res}\limits_{z=-1}\left\{\frac{1}{\sqrt{z}(z+1)}\right\}=\pm \frac{1}{i}$$

Likewise for any rational root say 1/n, there will be n residues, one corresponding to each determination of the root.

I'm not sure if the root is irrational if there will be any analytic branches around the pole. Take for example:

$$\frac{1}{z^{1/\pi}(z+1)}$$

I suspect there are an infinite number of analytic determinations around the pole in this case and thus an infinite number of residues.

And since k is never one, the origin in the range 0<k<1 always remains a branch point.

Also, not too hard to numerically confirm this in Mathematica for test cases although will have to do so manually since Mathematica will always choose the negative real axis for the branch-cut.

## 1. What is a variable exponent in the context of singularities?

A variable exponent refers to the power or index of a term in a mathematical expression that determines the behavior of singularities. In the context of residue, the exponent of a variable can change the number and type of singularities present in a function.

## 2. How does a variable exponent affect the number of singularities in a function?

The variable exponent can change the number of singularities in a function by altering the power or index of a term in the expression. This can result in the appearance or disappearance of singularities, or a change in the type of singularities (such as from a pole to a removable singularity).

## 3. Can a variable exponent cause a function to have no singularities?

Yes, a variable exponent can potentially cause a function to have no singularities. For example, if the exponent of a term is a negative even number, the function may have no singularities. However, this also depends on other factors such as the overall structure of the function.

## 4. How can the behavior of singularities be predicted with a variable exponent?

Predicting the behavior of singularities with a variable exponent can be challenging and may require techniques such as contour integration and the residue theorem. Additionally, knowledge of the behavior of singularities for specific types of functions (e.g. rational functions) can also be helpful.

## 5. Can a variable exponent affect the convergence of a function?

Yes, a variable exponent can affect the convergence of a function. For example, if the exponent of a term is a negative even number, the function may converge for all values of the variable. However, if the exponent is a negative odd number, the function may only converge for certain values of the variable.

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