Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Apr 25, 2017 23:44:23 GMT
Vasco, Here is exercise 13: nh.ch7.ex13.pdf (118.31 KB) This should make for some good discussion. Gary |
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Post by Admin on Apr 26, 2017 5:02:12 GMT
Gary
I have a solution which I just need to write up. Yes, it certainly looks like one of those exercises.
Vasco
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Post by Admin on Apr 27, 2017 18:44:59 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 27, 2017 20:45:17 GMT
Vasco,
Your solution to ex. 13 looks very nice, particularly in the way it makes use of visualizing complex products in (i). I need to spend some time with the details of (ii)-(v).
Gary
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Post by Admin on Apr 27, 2017 22:30:10 GMT
Vasco, Your solution to ex. 13 looks very nice, particularly in the way it makes use of visualizing complex products in (i). I need to spend some time with the details of (ii)-(v). Gary Gary Thanks for the positive feedback. I have looked at your solution and I don't understand the last two sentences of part (vi). But in any case the exercise asks us to do this part directly - without using the AP. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 27, 2017 23:57:20 GMT
Vasco, Your solution to ex. 13 looks very nice, particularly in the way it makes use of visualizing complex products in (i). I need to spend some time with the details of (ii)-(v). Gary Gary Thanks for the positive feedback. I have looked at your solution and I don't understand the last two sentences of part (vi). But in any case the exercise asks us to do this part directly - without using the AP. Vasco Vasco, Yes, I need to take another look at section VI. The second to last sentence was referring back to (iii). Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 30, 2017 0:16:57 GMT
Vasco,
I've been giving Ex 13 a closer reading. What distinction are you are making in the first sentence? I don't see too much difference between saying "$f(\Gamma)$ is an origin centred circle" and "$f(z)$ is an origin centred circle". Neither one puts constraints on the shape of the loop.
(ii) I interpreted the question to mean that $\Delta$ grows until it becomes $\pi D$, because $\Delta$ is a (infinitesimal) movement of $z$ along $\Gamma$. You interpreted it as a constant infinitesimal complex number. I agree that you are correct on this. It is interesting to think of a number anchored at z rotating around the origin.
(vi) $\rightarrow$ (v)
In (v), you pointed to the information in Section VI on Maxima and Minima, but your answer was not very specific, and it doesn't seem to me that we have quite enough information to answer the question by consideration of the modular surface. That is, enough to say that there must be at least one root inside $\Gamma$. But I have made another attempt. Maybe you can see what it needs or expand your answer.
Gary
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Post by Admin on May 1, 2017 6:51:45 GMT
Vasco, I've been giving Ex 13 a closer reading. What distinction are you are making in the first sentence? I don't see too much difference between saying "$f(\Gamma)$ is an origin centred circle" and "$f(z)$ is an origin centred circle". Neither one puts constraints on the shape of the loop. (ii) I interpreted the question to mean that $\Delta$ grows until it becomes $\pi D$, because $\Delta$ is a (infinitesimal) movement of $z$ along $\Gamma$. You interpreted it as a constant infinitesimal complex number. I agree that you are correct on this. It is interesting to think of a number anchored at z rotating around the origin. (vi) $\rightarrow$ (v) In (v), you pointed to the information in Section VI on Maxima and Minima, but your answer was not very specific, and it doesn't seem to me that we have quite enough information to answer the question by consideration of the modular surface. That is, enough to say that there must be at least one root inside $\Gamma$. But I have made another attempt. Maybe you can see what it needs or expand your answer. Gary Gary The reason for my first sentence is that I wanted to make it clear that points inside $\Gamma$ do not lie on an origin-centred circle, just the points on $\Gamma$. I have re-read my solution and I think my argument in part (v) is OK. Can you explain in more detail where you think it could be more specific? If necessary I will expand my explanation to make it more specific. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 1, 2017 13:52:32 GMT
Vasco, I've been giving Ex 13 a closer reading. What distinction are you are making in the first sentence? I don't see too much difference between saying "$f(\Gamma)$ is an origin centred circle" and "$f(z)$ is an origin centred circle". Neither one puts constraints on the shape of the loop. (ii) I interpreted the question to mean that $\Delta$ grows until it becomes $\pi D$, because $\Delta$ is a (infinitesimal) movement of $z$ along $\Gamma$. You interpreted it as a constant infinitesimal complex number. I agree that you are correct on this. It is interesting to think of a number anchored at z rotating around the origin. (vi) $\rightarrow$ (v) In (v), you pointed to the information in Section VI on Maxima and Minima, but your answer was not very specific, and it doesn't seem to me that we have quite enough information to answer the question by consideration of the modular surface. That is, enough to say that there must be at least one root inside $\Gamma$. But I have made another attempt. Maybe you can see what it needs or expand your answer. Gary Gary The reason for my first sentence is that I wanted to make it clear that points inside $\Gamma$ do not lie on an origin-centred circle, just the points on $\Gamma$. I have re-read my solution and I think my argument in part (v) is OK. Can you explain in more detail where you think it could be more specific? If necessary I will expand my explanation to make it more specific. Vasco Vasco, In re-reading the relevant sections of Needham, I don't see anything that guarantees that the modular surface of an analytic function will have a pit. Why is it that it cannot look like the pizza top? There are hints that he thinks that $\Gamma$ must have an origin. I tried to provide the link in my last sentence "Since we can always translate a simple loop $\Gamma$ to include the origin, there will always be at least one root of f inside $\Gamma$." Do you not see this as a problem, or is this wrong? Gary |
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Post by Admin on May 1, 2017 17:59:08 GMT
Gary The reason for my first sentence is that I wanted to make it clear that points inside $\Gamma$ do not lie on an origin-centred circle, just the points on $\Gamma$. I have re-read my solution and I think my argument in part (v) is OK. Can you explain in more detail where you think it could be more specific? If necessary I will expand my explanation to make it more specific. Vasco Vasco, In re-reading the relevant sections of Needham, I don't see anything that guarantees that the modular surface of an analytic function will have a pit. Why is it that it cannot look like the pizza top? There are hints that he thinks that $\Gamma$ must have an origin. I tried to provide the link in my last sentence "Since we can always translate a simple loop $\Gamma$ to include the origin, there will always be at least one root of f inside $\Gamma$." Do you not see this as a problem, or is this wrong? Gary Gary In this exercise we know that the maximum value of $|f|$ is on the boundary and is the same at all points on the boundary since they all lie on a circle. So internal points must have lower values of $|f|$ than any on the boundary which means we must have at least one pit. It can't look like a pizza top if it's analytic. (see preamble to exercise 14) I don't understand what the following sentence in your document means: ...we know that all the preimages of an analytic $f(\Gamma)$ lie inside $\Gamma$. What does the phrase "the preimages of $f(\Gamma)$" mean? I don't follow this. Roots are not necessarily zero. Vasco PS you say in your post that I interpret $\Delta$ as a constant complex number, but this is not correct. I guess you mean a complex number with a constant modulus? |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 1, 2017 19:37:11 GMT
Vasco, In re-reading the relevant sections of Needham, I don't see anything that guarantees that the modular surface of an analytic function will have a pit. Why is it that it cannot look like the pizza top? There are hints that he thinks that $\Gamma$ must have an origin. I tried to provide the link in my last sentence "Since we can always translate a simple loop $\Gamma$ to include the origin, there will always be at least one root of f inside $\Gamma$." Do you not see this as a problem, or is this wrong? Gary Gary In this exercise we know that the maximum value of $|f|$ is on the boundary and is the same at all points on the boundary since they all lie on a circle. So internal points must have lower values of $|f|$ than any on the boundary which means we must have at least one pit. It can't look like a pizza top if it's analytic. (see preamble to exercise 14) I don't understand what the following sentence in your document means: ...we know that all the preimages of an analytic $f(\Gamma)$ lie inside $\Gamma$. What does the phrase "the preimages of $f(\Gamma)$" mean? I don't follow this. Roots are not necessarily zero. Vasco PS you say in your post that I interpret $\Delta$ as a constant complex number, but this is not correct. I guess you mean a complex number with a constant modulus? Vasco, Thank you. I did realize that for a circle all the points on the boundary are maximums. But if f is flat, all points are maximums. Perhaps that is the key. If they were minimums, they would have to be zeros by p. 357, paragraph 2, so every modular surface other than $\mathbb{C}$ must have a pit or a minimum on the boundary. I don't see anything in Ex. 14 relevant to the issue of whether an analytic f can be flat. A nice nonsequitor that I intended to fix. That would have to be any z on the loop. I'll take another look. True, and I think this reason is no longer needed. Yes. Gary |
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Post by Admin on May 3, 2017 5:02:55 GMT
Gary
I have looked at your latest solution to this exercise. In part (v) you write:
"Since $\Gamma$ is a circle, all points of $\Gamma$ lie in the same plane parallel to the complex plane,..."
Presumably what you meant to write is something like:
"Since $f(\Gamma)$ is a circle, all the points $|f(\Gamma)|$ lie in the same plane parallel to the complex plane,..."
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 3, 2017 5:35:17 GMT
Gary I have looked at your latest solution to this exercise. In part (v) you write: "Since $\Gamma$ is a circle, all points of $\Gamma$ lie in the same plane parallel to the complex plane,..." Presumably what you meant to write is something like: "Since $f(\Gamma)$ is a circle, all the points $|f(\Gamma)|$ lie in the same plane parallel to the complex plane,..." Vasco Vasco, Thank you. I have fixed it. Gary |
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Post by Admin on May 22, 2017 6:38:17 GMT
Gary
I have decided to remove the first sentence in my original solution, as I don't think it serves any useful purpose.
Vasco
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