Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Nov 23, 2015 5:53:24 GMT
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Post by Admin on Nov 23, 2015 8:26:01 GMT
Gary
Looking at your solution and at mine again, I found a typo in mine (part (i) second line of third paragraph). The expression $\sec^4\theta=1/\sec^4\theta$ should read $\sec^4\theta=1/\cos^4\theta$. I have corrected this and published the corrected version, and I am now continuing to read your solution.
I have looked at your solution and here are some comments:
Part (i) - I think that since Needham says "sketch the continuation...", he would be expecting some words to explain how you arrived at the graph in figure 1, but that's just my opinion.
Part (ii) Fine.
Part (iii) - I think it would be useful to include an explicit expression for $b_1$ and $b_2$.
Part (iv) Fine.
Part (v) - I don't understand your solution here, and your reference to conformality.
I was continuing to think about part (v) and I wondered if you were thinking of the fact that the product of the gradients of two lines which intersect at right angles is $-1$? That would work. Since the slopes of the tangent at $b_1$and at $b_2$ are $-\pi/4$ and $+5\pi/4$ then $\tan(-\pi/4)\tan(5\pi/4)=-1\cdot 1=-1$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 23, 2015 16:55:51 GMT
Gary Looking at your solution and at mine again, I found a typo in mine (part (i) second line of third paragraph). The expression $\sec^4\theta=1/\sec^4\theta$ should read $\sec^4\theta=1/\cos^4\theta$. I have corrected this and published the corrected version, and I am now continuing to read your solution. I have looked at your solution and here are some comments: Part (i) - I think that since Needham says "sketch the continuation...", he would be expecting some words to explain how you arrived at the graph in figure 1, but that's just my opinion. Part (ii) Fine. Part (iii) - I think it would be useful to include an explicit expression for $b_1$ and $b_2$. Part (iv) Fine. Part (v) - I don't understand your solution here, and your reference to conformality. I was continuing to think about part (v) and I wondered if you were thinking of the fact that the product of the gradients of two lines which intersect at right angles is $-1$? That would work. Since the slopes of the tangent at $b_1$and at $b_2$ are $-\pi/4$ and $+5\pi/4$ then $\tan(-\pi/4)\tan(5\pi/4)=-1\cdot 1=-1$. Vasco Vasco, Thank you for the good comments. Regarding the reference to conformality, this should be set off from the answers. It was just following Needham's suggestion to try out things on the computer. Regarding my answer to (v), it was a foolish mistake made in haste. A moment's reflection on tangents would have revealed that. I have redone it, and (i) as well. Now I'm going to consult the archive. After consultation of the archive, I see that I misinterpreted question (i). I see the strategy, and I will try it again later before I study your answer. Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 24, 2015 7:29:20 GMT
Vasco, I revised 4:19 (i) after having a glance at yours. It's not a copy by any means, and not as succinct but I think it raises an interesting question about the path of z^4 when Im(p) is extended to +/- infinity.
Gary |
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Post by Admin on Nov 24, 2015 7:48:57 GMT
Gary
Thanks, I'll take a look; my day is just beginning.
I agree with you that $f(p)$ goes round the origin twice. I mention this in paragraph four of part (i) in my solution, but I didn't expand on it, maybe I should have. As you write, in figure 2(b) of your solution, we see $f$ going round the origin once. The way I look at it is to note that as $|p|\rightarrow\infty,~\theta\rightarrow\pm\pi/2$, and the angle of the tangents to $f(p)\rightarrow\pm2\pi$. So I think of $f(p)$ as being the point at infinity when $\theta=-\pi/2$ and then as $p$ travels from its point at infinity ($\theta=-\pi/2)$ to $1$ and back to the point at infinity ($\theta=\pi/2),~ f(p)$ also travels from its point at infinity to $B$, then anticlockwise round the origin to $f(p)=1$ (a complete revolution), and on to $A$ and then on through $B$ again and back to the point at infinity, making a total of two anticlockwise revolutions about the origin.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 24, 2015 16:38:53 GMT
Gary Thanks, I'll take a look; my day is just beginning. I agree with you that $f(p)$ goes round the origin twice. I mention this in paragraph four of part (i) in my solution, but I didn't expand on it, maybe I should have. As you write, in figure 2(b) of your solution, we see $f$ going round the origin once. The way I look at it is to note that as $|p|\rightarrow\infty,~\theta\rightarrow\pm\pi/2$, and the angle of the tangents to $f(p)\rightarrow\pm2\pi$. So I think of $f(p)$ as being the point at infinity when $\theta=-\pi/2$ and then as $p$ travels from its point at infinity ($\theta=-\pi/2)$ to $1$ and back to the point at infinity ($\theta=\pi/2),~ f(p)$ also travels from its point at infinity to $B$, then anticlockwise round the origin to $f(p)=1$ (a complete revolution), and on to $A$ and then on through $B$ again and back to the point at infinity, making a total of two anticlockwise revolutions about the origin. Vasco I saw that you had written about two winds when I took my first brief look in the archive, so I looked no further and tried to work from there. I limited the problem to the upper half of the z plane, and that was too limiting, so I think what you have written above and in the problem is a better approach. I suspected it when I began, but I thought "I'll just try it and see if it simplifies". It didn't but it was instructional. |
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