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

[LagrangeEuler]

Homework Statement

Find the residue of function $$\ln\frac{\sqrt{z^2+1}}{z}$$ at $z=0$.

Relevant Equations

Residue of the function is $c_{-1}$ in Laurent series
$$f(z)=\sum^{\infty}_{n=-\infty}f(z)(z-z_0)^n$$

First of all I am not sure which type of singularity is $z=0$?
\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}
It looks like that $Res[f(z),z=0]=0$

 

[Office_Shredder]

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.

 

[LagrangeEuler]

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?

 

[anuttarasammyak]

The formula is written as
\frac{1}{2}\ln(z+i)+\frac{1}{2}\ln(z-i)-\ln z
z=-i,0,i are singular points. I do not think we can do Laurent series expansion around them.

 

[LagrangeEuler]

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.

 

[FactChecker]

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.

 

[anuttarasammyak]

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.

 

[pasmith]

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 -i, 0, i are branch points. They, and \infty 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 z = r_1e^{i\theta_1 with -\pi < \theta_1 < \pi for the argument of \log.
- For \log(z + i), along the line z = x + i for x < 0. Use the polar representation z = i + r_2e^{i\theta_2 with -\pi < \theta_2 < \pi for the argument of \log(z + i).
- For \log(z - i), along the line z = -i + x for x < 0. Use the polar representation z = -i + r_3e^{i\theta_3 with -\pi < \theta_2 < \pi for the argument of \log(z + i).

The resulting contour is shown in the sketch.

Your contour integral should end up with contributions from the vertical line at z = c, which in the limit R \to \infty is the inverse Laplace transform you want to find, as well as the six contours along the top and bottom of the cuts at \theta_i = \pm \pi. 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 R in the left half-plane should vanish as R \to \infty.

 

[LagrangeEuler]

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
\frac{1}{2}\ln (1+\frac{1}{z^2})? Is it also problem there? And is this function equivalent as the function from the problem statement.

 

[anuttarasammyak]

Taylor series expansion
\ln(1+z)=z-1/2\ z^2+1/3\ z^3-...
for |z|<1 for convergence. So your formula is as OP post #1 and converges in series for |z| > 1 .

 

[docnet]

Have you tried using this theorem?

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

 

[Office_Shredder]

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.

 

[docnet]

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?