Branch cuts for complex powers

  • #1
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I need to perform the following integration:

##I(s) = \frac{1}{2\pi i} \int_{\gamma}\text{d}z\ z^{-s} \frac{\text{d}\ln\mathcal{F(z)}}{\text{d}z}##,

where ##\mathcal{F(z)}## is analytic everywhere on the complex plane except at the zeroes of the function.

For the purpose of integration, the branch cut which is implied by ##z^{-s}## is chosen to be on the negative real ##z-##axis, as follows.

The following link shows the branch cut: http://s24.postimg.org/eoxg0dbo5/Stack_Exchange_Question.jpg

I understand that the function $z^{-s}$ is multivalued, but that function has $s$ complex roots over the complex plane, so should we not choose a branch that corresponds to an angle of $\frac{2\pi}{s}$ of the complex plane?
 

Answers and Replies

  • #2
jasonRF
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No one has answered for a few days so I will take a stab.
First I have a few questions: 1) is this a homework problem? 2) what is the contour ##\gamma##? 2) are there any restrictions on ##s##?
I need to perform the following integration:

##I(s) = \frac{1}{2\pi i} \int_{\gamma}\text{d}z\ z^{-s} \frac{\text{d}\ln\mathcal{F(z)}}{\text{d}z}##,

where ##\mathcal{F(z)}## is analytic everywhere on the complex plane except at the zeroes of the function.
I'm not sure what you mean by this statement about ##\mathcal{F(z)}##.

I understand that the function $z^{-s}$ is multivalued, but that function has $s$ complex roots over the complex plane, so should we not choose a branch that corresponds to an angle of $\frac{2\pi}{s}$ of the complex plane?
Almost. ##z^{-s}## is indeed a multivalued function as long as ##s## is not an integer. When ##s## is irrational it has an infinite number of values. If it is rational the denominator tells you how many values. When you dealt with more specific functions like ##\sqrt{z}##, did the power on ##z## determine the direction of the branch cut?

jason
 
  • #3
mathwonk
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a basic principle is that the branch cut be chosen to pass through all branch points and render the region simply connected. I.e. as long as the cut makes it impossible for any arc to wind around a missing branch point, it is sufficient.
 
  • #4
WWGD
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< snip>

I understand that the function $z^{-s}$ is multivalued, but that function has $s$ complex roots over the complex plane, so should we not choose a branch that corresponds to an angle of $\frac{2\pi}{s}$ of the complex plane?
Assuming ##s ## is a negative integer. Otherwise it gets much hairier. And maybe I am missing something (sorry I snipped the first part, I don't know how to paste it back in ) but if by ##ln## you mean the complex log, then ##ln f(z) ## cannot be defined at the zeros of ##f(z) ##.
 
  • #5
mathwonk
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It is true the log of a function is multiple valued in a disc containing a zero, except at the center where it is not defined. However any two of its many different choices of values, at points near the zero, differ by a constant, so the derivative of the log is well defined and single valued, except again at the zero itself. Now apparently he wants to compute an integral along a path that does not pass through a zero of f, so the integral makes sense as long as he chooses a region in which f has a holomorphic "branch", i.e. off the branch cut shown.

Apparently he also means f is analytic everywhere, since the zeroes of f are not a problem for f, they are a problem for dLn(f).

Now actually, unless s is a non integer, like 1/2, there are not necessarily branch points for z^(-s) either. I.e. z^(-3) for instance poses no problem since it is well defined near zero and even has residue zero there, so would contribute zero to the integral as long as the path avoids z=0.
 
Last edited:
  • #6
WWGD
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Thanks, Wonk, I realized it just recently, that the question makes sense only when we are working within a branch of the log. We do need the wonk, give us the wonk... :).
 

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