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Post by Amep on Sept 27, 2017 0:36:44 GMT
Hello,
I am curious why in part i figure 1 there are singularities at z=in, where n is an integer, when the solution clearly states z=2ni.
I am also curious why part ii states the problem as the 3rd root of (z^4 - 1), when the problem in the book is the 5th root. However, I don't think this should change any branch points. Figure 2 also denotes all these branch points as singularities.
I am not sure if these are errors, or if I am missing something, so any insight on this would be greatly appreciated.
Thank you.
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Post by Admin on Sept 27, 2017 7:00:35 GMT
Hi
Welcome to the forum!
You are right about the errors/typos. I have corrected them all plus a couple more which I noticed when checking.
Thanks for bringing these to my attention. I have replaced the solution with a corrected version.
Vasco/Admin
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Sept 27, 2017 7:24:56 GMT
Hello, I am curious why in part i figure 1 there are singularities at z=in, where n is an integer, when the solution clearly states z=2ni. I am also curious why part ii states the problem as the 3rd root of (z^4 - 1), when the problem in the book is the 5th root. However, I don't think this should change any branch points. Figure 2 also denotes all these branch points as singularities. I am not sure if these are errors, or if I am missing something, so any insight on this would be greatly appreciated. Thank you. Amep, Vasco can give you better answers, but in case you are in a hurry, I will try. In (i), I will refer to the information on pp. 90-92. In (i), there are no singularities at $z = in$, where n is odd. For odd integers n, $f(z) = \frac{1}{e^{i n \pi} - 1} = -\frac{1}{2}$ with denominator > 0, so it is not a singularity. It appears to me that all the small white circles have crosses. Judging from the first paragraph of Vasco’s answer, the small white circles may be intended to be singularities, but as I explained above, I don’t think the odd values of in qualify. So I think your question is valid. In part (ii), I think the 3rd root of (z^4 - 1) must be a typographical error. My book also shows a 5th root. It does appear that Figure 2 denotes all the branch points as singularities using small circles with crosses, but they are clearly not, as he himself states. I don’t think it affects the rest of the answer. Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 27, 2017 15:16:28 GMT
Vasco and Amep,
Excuse my syncopated reply. Vasco must have posted the revision while I was working on this, and I failed to check for replies. But perhaps no harm done. Perhaps Vasco could say something about how one identifies branch points of a function such as that in (i).
Gary
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Post by Admin on Sept 27, 2017 16:19:40 GMT
Gary
Yes we were obviously typing our replies to Amep at the same time! As you say, no harm done.
The function in (i) does not have any branch points as it is a single-valued function.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 27, 2017 20:36:58 GMT
Gary Yes we were obviously typing our replies to Amep at the same time! As you say, no harm done. The function in (i) does not have any branch points as it is a single-valued function. Vasco Vasco, How do we know it is single-valued? As z traverses the unit circle centered at K some number of times, it cycles through the same values with each revolution. Then $f(z) = \frac{1}{e^{\pi z}-1}$ must also return to an initial value after each full cycle of z. Taking the definition on p. 92, if q is a branch point of some multifunction f(z), and f(z) first returns to its original value after N revolutions round q, then q is called an algebraic branch point of order (N-1). We know there will be revolutions around some q, because f(z) is analytic by the Cauchy-Riemann equations, so it is conformal and will send a circle to a circle. The data appear to indicate a very tiny circle with repeating values after one revolution interspersed with the singularities, which is what one would expect, but it doesn't make a nice plot. In fact, it plots as wings in quadrants II and III rather than a circle. Gary |
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Post by Admin on Sept 27, 2017 20:54:58 GMT
Gary
First way of seeing that $f$ is single-valued is to say: $e^{\pi z}$ is single-valued, and so $f$ is single-valued.
Second way: as $z$ goes round a point, $f$ returns to its initial value after 1 revolution of $z$. That's the definition of a single-valued function at the top of page 92.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 27, 2017 21:04:27 GMT
Gary First way of seeing that $f$ is single-valued is to say: $e^{\pi z}$ is single-valued, and so f is single-valued. Second way: as $z$ goes round a point $f$ returns to its initial value after 1 revolution of $z$. That's the definition of a single-valued function at the top of page 92. Vasco Vasco, Thank you. It's a point on which I am easily confused, even after looking it up. Gary |
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