Variable exponent causing number of singularities to change for residue?

[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?

Thanks in advance.

 

[jackmell]

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.