Can Cauchy's Residue Theorem be Used for Functions with Branch Cuts?

In summary, the conversation discusses the difficulty of calculating the inverse Laplace transform of a function with branch points. The function in question has branch points at -i, 0, i, and \infty, and the discussion delves into how to correctly place branch cuts in order to evaluate the integral. The function is also shown to have a branch point at z=0, and it is noted that this is not an essential singularity. The conversation also briefly mentions using a theorem involving poles, but it is determined that this method cannot be used due to the presence of a branch cut. Finally, the method of contour deformation with Cauchy's Residue Theorem is proposed as a possible solution.
  • #1
LagrangeEuler
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Homework Statement
Find the residue of function [tex]\ln\frac{\sqrt{z^2+1}}{z}[/tex] at ##z=0##.
Relevant Equations
Residue of the function is ##c_{-1}## in Laurent series
[tex]f(z)=\sum^{\infty}_{n=-\infty}f(z)(z-z_0)^n[/tex]
First of all I am not sure which type of singularity is ##z=0##?
[tex]\ln\frac{\sqrt{z^2+1}}{z}=\ln (1+\frac{1}{z^2})^{\frac{1}{2}}=\frac{1}{2}\ln (1+\frac{1}{z^2})=\frac{1}{2}\sum^{\infty}_{n=0}(-1)^{n}\frac{(\frac{1}{z^2})^{n+1}}{n+1}[/tex]
It looks like that ##Res[f(z),z=0]=0##
 
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  • #2
You should always be extremely nervous about doing things like ##z=\sqrt{z^2}## since it might not actually be true.

I'm also not sure this question is well posed. Near ##z=0## this is going to look pretty close to ##-\ln(z)##, which is not continuous (it has a branch cut) and hence you cannot compute a residue.

I suspect the series you wrote down here only appears to work because ##z=\sqrt{z}^2## when ##0\leq arg(z) \leq \pi##, so somehow this dodges the discontinuity. I find it hard to formalize whether this is true or not.
 
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  • #3
Thank you. I am nervous. :) And I am trying to learn more through the discussion here. I did a problem, but I am not sure whether my solution is correct. Also, I am not sure is it ##z=0## essential singularity? To my mind, it is very hard to check without using ##\sqrt{z^2=z## here. Or there is a way?
 
  • #4
The formula is written as
[tex]\frac{1}{2}\ln(z+i)+\frac{1}{2}\ln(z-i)-\ln z[/tex]
z=-i,0,i are singular points. I do not think we can do Laurent series expansion around them.
 
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  • #5
Thank you. And can you tell me what type of singularities are ##-i,i## and ##0##? I am trying to calculate the inverse Laplace transform of this function by using Bromwich integral. I will get a good result using this trick but I am not sure why I can, or I cannot use it.
 
  • #6
The catch is that "singularities" are defined for meromorphic functions. So the function must be well-defined in a neighborhood of the point. That is not the case for the function and points in this example. In order to consider the function well-defined, we would have to talk about branch points on a Riemann surface on which the function is well defined.
 
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  • #7
Riemann cuts start from z=-i,0,i to infinity. We can not go around any of them to make a closed loop which is necessary for Laurent expansion, due to the cuts which leads us to another Riemann plane when we go over it.
 
  • #8
LagrangeEuler said:
Thank you. And can you tell me what type of singularities are ##-i,i## and ##0##? I am trying to calculate the inverse Laplace transform of this function by using Bromwich integral. I will get a good result using this trick but I am not sure why I can, or I cannot use it.

Don't think of inverting a laplace transform as just summing residues. That doesn't always work, as for example here.

The singularities at [itex]-i, 0, i[/itex] are branch points. They, and [itex]\infty[/itex] which is also a branch point for log, must be connected by branch cuts, positioned in such a way that the Bromwich contour does not cross them. Any decent text on complex analysis should discuss integration of functions with branch cuts.

Here I think it best to use the expansion in post #4, and place cuts as follows:
- For log(z), along the negative real axis from 0 to infinity. Use the polar representation [itex]z = r_1e^{i\theta_1[/itex] with [itex]-\pi < \theta_1 < \pi[/itex] for the argument of [itex]\log[/itex].
- For [itex]\log(z + i)[/itex], along the line [itex]z = x + i[/itex] for [itex]x < 0[/itex]. Use the polar representation [itex]z = i + r_2e^{i\theta_2[/itex] with [itex]-\pi < \theta_2 < \pi[/itex] for the argument of [itex]\log(z + i)[/itex].
- For [itex]\log(z - i)[/itex], along the line [itex]z = -i + x[/itex] for [itex]x < 0[/itex]. Use the polar representation [itex]z = -i + r_3e^{i\theta_3[/itex] with [itex]-\pi < \theta_2 < \pi[/itex] for the argument of [itex]\log(z + i)[/itex].

The resulting contour is shown in the sketch.

Image.png


Your contour integral should end up with contributions from the vertical line at [itex]z = c[/itex], which in the limit [itex]R \to \infty[/itex] is the inverse Laplace transform you want to find, as well as the six contours along the top and bottom of the cuts at [itex]\theta_i = \pm \pi[/itex]. The contributions from the three semi-circles around the ends of the cuts should vanish as the radii tend to zero and the contribution from the parts of the contour on the semi-circle of radius [itex]R[/itex] in the left half-plane should vanish as [itex]R \to \infty[/itex].
 
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  • #9
You did not latex everything. Yes for me is a bit hard to see when something is a branch point and when it is not. Thank you all for the answers.

I have one more question. In ##z=0## we have branch point. Is it also an essential singularity. Also what about the function
[tex] \frac{1}{2}\ln (1+\frac{1}{z^2})[/tex]? Is it also problem there? And is this function equivalent as the function from the problem statement.
 
  • #10
Taylor series expansion
[tex]\ln(1+z)=z-1/2\ z^2+1/3\ z^3-...[/tex]
for |z|<1 for convergence. So your formula is as OP post #1 and converges in series for |z| > 1 .
 
Last edited:
  • #11
Have you tried using this theorem?

Let ##z_0## be a ##m##-th order pole of ##f##. Then
$$\text{Res}(f(z),z_0)=\text{lim}_{z\rightarrow z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]$$
 
  • #12
docnet said:
Have you tried using this theorem?

Let ##z_0## be a ##m##-th order pole of ##f##. Then
$$\text{Res}(f(z),z_0)=\text{lim}_{z\rightarrow z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]$$

This requires the existence of an mth order pole, which we don't have because the function has a branch cut and hence isn't even continuous in a punctured neighborhood of 0.
 
  • #13
Office_Shredder said:
This requires the existence of an mth order pole, which we don't have because the function has a branch cut and hence isn't even continuous in a punctured neighborhood of 0.
My apologies. I did not see the ##\ln## next to the fraction (I need to get my eyes checked). What about using the method of contour deformation with Cauchy's Residue theorem?
 

Related to Can Cauchy's Residue Theorem be Used for Functions with Branch Cuts?

1. What is the residue of a complex function?

The residue of a complex function is the value of the coefficient of the term with a negative power in the Laurent series expansion of the function. It is denoted by Res(f, z0) and is used to evaluate the behavior of a function at a singular point.

2. How is the residue calculated?

The residue of a complex function at a point z0 can be calculated using the formula Res(f, z0) = limz→z0 [(z-z0)f(z)], where f(z) is the given function. This limit can also be expressed as the coefficient of the term with a negative power in the Laurent series expansion of the function.

3. What is the significance of the residue in complex analysis?

The residue plays a crucial role in complex analysis as it helps in evaluating the behavior of a function at a singular point. It is also used in the evaluation of contour integrals using the residue theorem, which states that the integral of a function around a closed contour is equal to 2πi times the sum of the residues of the function inside the contour.

4. Can the residue be negative?

Yes, the residue can be negative. It depends on the function and the point at which it is being calculated. The residue can be positive, negative, or zero, and it is important to consider all possible cases while evaluating the residue of a complex function.

5. In what situations is the concept of residue used?

The concept of residue is used in various situations, such as evaluating complex integrals, calculating the behavior of a function at a singular point, and solving differential equations. It is also used in the branch of mathematics known as complex analysis, which deals with functions of complex variables.

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