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CauchyGoursat's theorem and singularities 
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#1
Mar2713, 07:36 PM

P: 114

1. The problem statement, all variables and given/known data
Calculate the closed path integral of [tex]\frac{z+2 i}{z^3+4 z}[/tex] over a square with vertices (1i), (1,i) and so forth. 2. Relevant equations The closed line integral over an analytic function is 0 3. The attempt at a solution Alright, so first I factored some stuff, leaving me with [tex]\frac{1}{z (z2 i)}[/tex] Now, this has a simple pole at 0 and at 2i. However, 2i isn't in the square, so I'm only concerned about the one at 0. Now, I don't know how to continue from here on, formally. I know (do I?) that the answer has to be i*2pi, as that always happens to be the case when doing a closed integral of a function that has 1 singularity (and it is enclosed by the path). But how do I 'calculate' this? I tried writing z = x+iy and the same for dz, and then working it all out and parameterizing the vertices, but that gives horrible expressions to integrate and just doesn't seem right. Is there something I'm missing? Also, there is a follow up question, which concerns the same integral but now over the square with vertices (13i), (1+3i) and so forth. So this encloses both singularities, but they aren't symmetric or anything, so then I can't really go any further. (Is it even true, that when the singularities are distributed symmetrically, that they sort of cancel each others effect and your integral evaluates as 0?) Kind regards 


#2
Mar2813, 02:36 PM

P: 615

CauchyGoursat only gets you so far. You have two options:
1. Evaluate it using the typical method of line integrals. 2. Use the more general theorem. Not all contour integrals around singularities yield ##2\pi i##. They do, however, follow the relation $$\oint_{\beta}f(z) \ dz = 2\pi i \sum \mathfrak{R}_{z_i}(f(z))$$ if ##\beta## is a Jordan curve, where ##\mathfrak{R}_{z_i}## denotes the residue of the function at a point ##z=z_i## interior to ##\beta##. If you are unfamiliar with residues, I would suggest using the typical line integral method. Edit: Didn't see the bottom of your post. I see where you would get the idea that they might "cancel each other out." However, it is entirely wrong. Once you get a little further into complex analysis, you'll probably have a kind of epiphany moment where you'll start thinking "Hey! None of this makes any logical real sense in terms of what I learned from basic calculus. Better start thinking about it differently!" At that point, everything starts making sense because you'll notice that, in terms of complex numbers, it really doesn't matter what the contour is or where the poles are. As long as the contour contains the same poles and has the same winding number, you get the same answer. 


#3
Mar2813, 02:46 PM

P: 114

Thank you for your answer!
I've discussed my issue with the teacher today, and apparently he accidentally gave out exercises that were meant for later in the course. This was only one of them, and I've spent hours thinking about this one and the others, so that is a little frustrating. Either way, it is still very good to see that what I thought might have been the case is not true at all. I wasn't sure that it was, it just seemed to be so every single time. But its good to know that it is simply not the case. I look forward to getting further into the material though, I really enjoy how elegant everything is in the complex numbers. 


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