I am confused about the "deformation" of a closed contour that my book is doing. For reference, it is example 2.4.3 on pg. 75-76 from this free online book. The example is the integration of 1/z around a closed contour enclosing the origin. The book does not say why any special care needs to be taken here, it just goes right into doing a deformation. More specifically, this deformation essentially gets one closed contour inside of the original, except now the inside contour is traversed in such a way that the origin is no longer an issue(?). I have multiple questions about this. First, note that $\frac{1}{z} = \frac{d}{dz}(log z)$, and we know that log z has a branch point at the origin. Alright, here goes:

1.) Why does it matter how the function is behaving at the origin? We are taking a closed contour around the origin, so the function is never evaluated at this point. Our book is kind of a drag, and doesn't dwell on the finer points of integration whatsoever, so it would seem to me that if the function is defined on the path of integration then all is well. Why is this deformation necessary at all, and what is it really telling me? That as long as I go around a curve that has some orientation facing "away" from the branch point then it's fine? I seem to be missing something.

2.) I intuitively understand why the answer to this integration is $2\pi i$: log z has a branch point at the origin, so traversing this closed circuit puts us on the next branch which is known to differ by $2\pi i$. However, viewing it this way means that I was correct in saying that the above work is unnecessary, since this is just $\int_C^{}\frac{1}{z}dz = (log(r)+2\pi i)-(log(r)+0) = 2\pi i$, where I have used $z = re^{i\theta}$. Generally I can see why this reasoning would fall short, as the contour could wrap around multiple times, or do other things, but the problem specifies that it is simple and closed, so wrapping onto itself is not an issue and it can only "go up" one branch.

3.) Here's my guess at what I am incorrectly applying. In (2) I am applying the Fundamental Theorem, and perhaps the conditions for the antiderivative (log z in this case) must be analytic in some region containing the closed contour, which would include the origin and hence not satisfy my work. However, we know that log z is analytic at all points on this contour, so (again going back to (1)) why would it matter whether the integrand's antiderivative is analytic for all points inside of the contour?

Any assistance in understanding the purpose of this deformation will be greatly appreciated.

  • $\begingroup$ This thread may help too (your example is simpler but the problem is the same...). $\endgroup$ – Raymond Manzoni Feb 17 '14 at 8:36
  • $\begingroup$ Hi Raymond, great answer over on that other thread! That makes complete sense to me. However, that does not subside my confusion about the necessity of this deformation, as it seems like you view and do the integration in the same way that I see it. Yes, there is a branch point at the origin, but it seems as though moving up/down one branch is accounted for when applying the fundamental theorem. Do you see why my author is doing this? Saying that the integration of the inside circle is the same as the outside says nothing to me, because in the end he still uses the fundamental theorem. hmm. $\endgroup$ – dsm Feb 17 '14 at 16:20
  • $\begingroup$ Thanks @Daniel! In fact the first and second contour $C$ exposed don't enclose the origin (there is an 'issue' at $45$° for the second one). The point is that the integral over both contours $C$ is $0$ (the two pictures are 'topologically equivalent'). Now at the limit where $L_1$ and $L_2$ 'touch' (or rather cross) each others nothing special happens except that the contributions of $L_1$ and $L_2$ will cancel. $\endgroup$ – Raymond Manzoni Feb 17 '14 at 21:39
  • $\begingroup$ We deduce that the total (i.e. $0$) will be given by the integral over the very general contour $C_2$ minus the integral over the contour over the small perfect disk $C_1$ of radius $r$ which may easily be computed as $2\pi i$. So that the integral over any contour $C_2$ will be $2\pi i$. $\endgroup$ – Raymond Manzoni Feb 17 '14 at 21:39
  • $\begingroup$ More exactly the integral over any counterclockwise contour enclosing $0$ will be $2\pi i$. The author used $\int_C\frac {dz}z$ simply as an example and the same reasoning applies in the analytic neighbourhood of any simple pole (making the limit $r\to 0$ more interesting). $\endgroup$ – Raymond Manzoni Feb 18 '14 at 8:59

The problem is that there is actually no function $\log z$ which is analytic on $\mathbb{C} - \{0\}$, so you do not actually have an antiderivative of $1/z$ with which to apply the fundamental theorem of calculus. Only in a region with no curves encircling the origin (e.g., $\mathbb{C} - (-\infty,0]$) is it possible to define an analytic derivative of $1/z$. The basic problem is that $\log z = \log re^{i\theta} = \log r + i \theta$ and while $\log r$ is no problem (because $r$ is a positive real number and therefore has a real logarithm), $\theta$ can't be defined continuously over a curve which encircles the origin: if you start at 0, by the time you wind around once, you will get to $\pm 2\pi$ for the same point.

Edit based on comments below: The point of using a deformation is to show that the integral of $1/z$ around any simple closed curve $C$ that encloses the origin (oriented counterclockwise) is the same, regardless of the curve chosen. Any such curve can be deformed into a particular one, namely, the unit circle, for which we can easily calculate the integral of $1/z$ (by using the substitution $z=re^{i\theta}$). Thus it follows that for any such $C$ (no matter how complicated), we have $\int_C 1/z\, dz = 2\pi i$.

This does not violate what I said above about $\log z$ because neither the starting nor the ending nor any of the intermediate curves in the deformation pass through the origin. In a suitable region containing all the curves in the deformation but not the origin, there will be an analytic branch of $\log z$ for which the fundamental theorem of calculus can be used. If $C_1$ and $C_2$ are the starting and ending curves of the deformation, then we can construct a curve which we might call $C_1 - C_2$. See figure 2.4.4(b) in your link. This curve $C_1 - C_2$ goes almost all the way around $C_1$, then cuts over to $C_2$, then goes almost all the way around $C_2$ in the opposite direction, then cuts back to the starting point of $C_1$. The curve $C_1 - C_2$ is a simple closed curve that does not enclose the origin. In the region enclosed by $C_1 - C_2$ there is a branch of $\log z$; thus $\int_{C_1 - C_2} 1/z\,dz = 0$ hence $\int_{C_1} 1/z\,dz = \int_{C_2} 1/z\,dz$.

  • $\begingroup$ Hi Ted, thanks for the answer! I think I understand what you're saying, but I'm still not quite sure why the deformation is necessary. It seems like my author shows that the integration around the new inside circle is the same as on the original contour, and then proceeds to use the fundamental theorem anyways. However, this still does the exact same thing as the original integration, so I do not see why it was necessary/what it achieved; it would also seem to still violate what you said in your answer. Can you go a bit more into the details of the deformation? $\endgroup$ – dsm Feb 17 '14 at 16:27
  • $\begingroup$ @Daniel Added a few edits regarding deformations. $\endgroup$ – Ted Feb 17 '14 at 19:39
  • $\begingroup$ That's exactly what I needed to hear. I was misunderstanding why we were doing the deformation in the first place; I thought it's sole purpose was to mitigate some unwanted behavior of the original contour. I now see that it is entirely done to make the computation of any integration around a singularity point of some function a simple matter, just by cleverly using Cauchy's Theorem. Thanks for helping me out! $\endgroup$ – dsm Feb 18 '14 at 21:52

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