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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.

Sujaan Kunalan
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user70520
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    I've heard it called a, well, logarithmic singularity, but I've never actually seen the phrase defined. –  Aug 03 '13 at 08:11

3 Answers3

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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.

Rufflewind
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Owen Sizemore
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  • Then how about the integral in the link (please refer to my comments above)? Thanks. – user70520 Aug 03 '13 at 05:07
  • 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? – DBFdalwayse Aug 03 '13 at 05:18
  • @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, http://i.stack.imgur.com/ZjMpl.jpg. – Owen Sizemore Aug 03 '13 at 05:24
  • 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? – user70520 Aug 03 '13 at 05:27
  • Aha. I guess your new link is pretty much what I described (except you only considered half of it) after you replaced https://en.wikipedia.org/wiki/File:Keyhole_contour.svg :) – user70520 Aug 03 '13 at 05:30
  • I edited my above post to have the correct contour, you let the inner circle go to zero and then you note that the integral along the imaginary axis can be written in terms of the integral along the real axis (along the imaginary axis $z=ix$ and so for $z^4$ it is the same and you can break up the log by using log identities. This should work. Also you could use the contour you describe as long as you use a branch of the log that doesn't hit the contour, by for example using choosing the negative imaginary axis as a branch cut. – Owen Sizemore Aug 03 '13 at 05:33
  • Would $z=0$ considered an essential singularity according to https://en.wikipedia.org/wiki/Branch_point#Transcendental_and_logarithmic_branch_points ? – user70520 Aug 03 '13 at 20:02
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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 $0$

D. Zack Garza
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DBFdalwayse
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    But $\lim_{x\to0^+}\log x=-\infty$ – user70520 Aug 03 '13 at 04:47
  • 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 – DBFdalwayse Aug 03 '13 at 04:49
  • Regarding http://math.stackexchange.com/questions/434289/show-that-int-0-infty-frac-ln-xx41-dx-frac-pi2-sqrt216/434395?noredirect=1#comment976232_434395 does the residue theory still apply for $z=0$ of $\log z$? – user70520 Aug 03 '13 at 04:49
  • 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. – DBFdalwayse Aug 03 '13 at 04:53
  • 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? – DBFdalwayse Aug 03 '13 at 05:00
  • So for the example in the link, I'm not sure how the singularity at $z=0$ is treated...In the last solution (the one using residue theory), I think we also need a semicircular contour of the form $re^{i\theta},0\le\theta\le\pi$ that replaces the segment $[-r,r]$ on the real axis. And let $r\to0$. – user70520 Aug 03 '13 at 05:02
  • $\log{z}$ indeed blows up as $z \to 0$. You need to fix that statement, which is plain old wrong. – Ron Gordon Aug 03 '13 at 06:08
  • My bad; as Ron Gordon correctly pointed out, as z->0 lnx blows up, and therefore |logz| does also blow up. – DBFdalwayse Aug 03 '13 at 06:22
  • $\log(z)$ requires a branch cut. The standard branch cut is $\theta=\pi$. With this branch cut, $-\pi\lt\mathrm{Im}(\log(z))\le\pi$, so $\mathrm{Im}(\log(z))$ does not blow up. – robjohn Aug 03 '13 at 07:02
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$\log{z}$ 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.

Ron Gordon
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  • 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"?) – user70520 Aug 03 '13 at 19:56
  • Yes, it is a branch point singularity, similar in nature to, say $z^{\alpha}$ for $\alpha$ noninteger. – Ron Gordon Aug 03 '13 at 20:06