What type of singularity is $z=0$ for $\log z$ (any branch)? What is the Laurent series for $\log z$ centered at 0, if exist? If the Laurent series has the form $\sum_{k=-\infty}^{\infty} a_kx^k$, then certainly among $a_{-1},a_{-2},...,a_{-j},...$, at least one is nonzero (or otherwise $\log z$ would be analytic at $0$). Since $\lim_{z\to 0}z\log z=0$, we must have $a_{-j}=0$ for all $j>0$, a contradiction. Hence the Laurent series centered at $0$ cannot exist.

Is the singularity at $0$ a pole? If so, what is its order? Thanks.

  • $\begingroup$ I've heard it called a, well, logarithmic singularity, but I've never actually seen the phrase defined. $\endgroup$ – user14972 Aug 3 '13 at 8:11

The singularity of $\log z$ is not an isolated singularity, so the usual classification into, pole, essential, or removable does not apply. In particular, there is no Laurent expansion about 0 and you cannot apply residue theory.

In this case the singularity is known as a branch point, and it is the typical example.

  • $\begingroup$ Then how about the integral in the link (please refer to my comments above)? Thanks. $\endgroup$ – user70520 Aug 3 '13 at 5:07
  • $\begingroup$ If you're referring to the integral of 1/(z-a) , in the link above then you use the fact that once you select a specific branch, then, within that branch, you have that d/dz(Log(z-a))=1/(z-a), i.e., within a branch, d/dzLogz=1/z . Then, integral 1/(z-a) from a to -a =Log(z-a) from a=1 to a=-1 =Log(z+1)-Log(z-1)=Log(z+1)/Log(z-1). Is that your question; did I understand you correctly? $\endgroup$ – DBFdalwayse Aug 3 '13 at 5:18
  • $\begingroup$ @user70520: The answer given there is incorrect (even if he gets the correct answer for the integral). $z=0$ is not an isolated singularity so the residue is not defined. You can still use residues to calculate the integral but you have to use a contour which can be found here, i.stack.imgur.com/ZjMpl.jpg. $\endgroup$ – Owen Sizemore Aug 3 '13 at 5:24
  • $\begingroup$ But we need the real axis... so I think the contour still need to be semi-circular, but we could exclude the boundary from the original by replacing the interval $[-r,r]$ by the arc $re^{i\theta}$, $\theta\in[0,\pi]$. Then let $r\to0$. Would this fix the solution in the link? $\endgroup$ – user70520 Aug 3 '13 at 5:27
  • $\begingroup$ Aha. I guess your new link is pretty much what I described (except you only considered half of it) after you replaced en.wikipedia.org/wiki/File:Keyhole_contour.svg :) $\endgroup$ – user70520 Aug 3 '13 at 5:30

The singularity is not a pole, since Logz (or at least its real part) does not blow up to $\infty$ as you approach 0. Maybe the best name is that $0$ is a branch point for Logz; and it is a branch point of infinite index. The branch point tells you that if you wind around the unit circle a point once, i.e., if you wind around by a value of $2\pi$, you do not ever return to your initial value , i.e., if you consider the values of {$Log(z+2n\pi)$} for $n=1,2,3,...$, these are all different values. Compare and contrast this with the case for the n-th root $z^{1/n}$ . Here $0$ is also a branch point for $z^{1/n}$ , but this time it is of index n, because $e^{i\theta/n}=e^{i\theta+2n\pi/n}$ , i.e., you return to the original value of your function after looping n times.

EDIT: As pointed out in the comments, I was wrong in my claim that |Logz| does not go to $\infty$ as $z\rightarrow 0$ ; it does, since lnx does blow up near $0$

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    $\begingroup$ But $\lim_{x\to0^+}\log x=-\infty$ $\endgroup$ – user70520 Aug 3 '13 at 4:47
  • $\begingroup$ You're right, but I think you need the function to blow up to $\infty$ in all directions, as in the case of $1/z$. Let me double-check $\endgroup$ – DBFdalwayse Aug 3 '13 at 4:49
  • $\begingroup$ Regarding math.stackexchange.com/questions/434289/… does the residue theory still apply for $z=0$ of $\log z$? $\endgroup$ – user70520 Aug 3 '13 at 4:49
  • $\begingroup$ You also need, in order for f(z) to have a pole at w, for f(z) to be analytic in $\mathbb C \{w}$, which does not happen here. $\endgroup$ – DBFdalwayse Aug 3 '13 at 4:53
  • $\begingroup$ But usually when you apply residue theory, you have a real-valued expression which restricts to a complex expression in the x-axis, and so that the integral vanishes in some contour containing part of the x-axis. In the example you linked, you want to evaluate lnx/(x^4+1), so you consider this as the restriction to the real axis of Logz/(z^4+1).Do you have some expression like that in mind, or is your question more general? $\endgroup$ – DBFdalwayse Aug 3 '13 at 5:00

$\log{z}$ is viewed as a branch point due to its multivaluedness. That is, $\log{z}$ is only determined to within an integer multiple of $i 2 \pi$. $\log{z}$ is unique within a single branch; that is, as long as a contour along which $\log{z}$ is uniquely defined does not cross a branch cut that has been defined. That $\log{z}$ blows up as $z \to 0$ is beside the point; you simply do not include branch points such as those for which the argument of the log function is zero because of the nonuniqueness of the log near there.

  • $\begingroup$ So even though $|\log z|\to\infty$ as $z\to0$, $z=0$ is not considered as a pole, but part of the branch cut. Am I right? So is $z=0$ a special kind of singularity? (and is it called "branch cut" singularity"?) $\endgroup$ – user70520 Aug 3 '13 at 19:56
  • $\begingroup$ Yes, it is a branch point singularity, similar in nature to, say $z^{\alpha}$ for $\alpha$ noninteger. $\endgroup$ – Ron Gordon Aug 3 '13 at 20:06

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