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Post by mondo on Oct 20, 2022 23:13:02 GMT
You can find my solution proposal in the attached document. Vesco, do you agree with my solution? Attachments:exerices_5.pdf (103.63 KB)
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Post by Admin on Oct 21, 2022 15:06:06 GMT
Mondo
But so does $\frac{e^{i\phi}}{\overline{z}}$ where $\phi$ is any angle.
Vasco
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Post by mondo on Oct 21, 2022 18:05:09 GMT
Mondo But so does $\frac{e^{i\phi}}{\overline{z}}$ where $\phi$ is any angle. Vasco Vasco, I don't think I can agree with it. $\frac{e^{i\phi}}{\overline{z}}$ will additionally rotate every point by the angle $\phi$ and this is not what this mapping does? |
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Post by Admin on Oct 21, 2022 18:24:11 GMT
Mondo But so does $\frac{e^{i\phi}}{\overline{z}}$ where $\phi$ is any angle. Vasco Vasco, I don't think I can agree with it. $\frac{e^{i\phi}}{\overline{z}}$ will additionally rotate every point by the angle $\phi$ and this is not what this mapping does? Mondo My point is that you say that the 2 transformations exchange the points inside and outside the unit circle and then you say that $1/\overline{z}$ does this and therefore it's the answer. But so does my function, and so you haven't found the unique solution. Rotating the points still leaves them either inside or outside which is what you claim the transformations do. Put simply the transformations do more than just exchange the points. You say that the transformation doesn't rotate the points but you haven't proved that. All you say is that the points are exchanged but this is true of rotated points. Vasco |
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Post by Admin on Oct 21, 2022 18:47:11 GMT
Mondo
1. You say the transformation in the exercise exchanges points inside and outside the unit circle.
2. We know that the transformation $1/\overline{z}$ also exchanges points inside and outside the unit circle
3. It is not logical to say that these two facts mean that the 2 transformations are the same.
Vasco
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Post by Admin on Oct 21, 2022 18:55:38 GMT
Mondo
Your answer does not show that the transformation is $1/\overline{z}$
Vasco
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Post by Admin on Oct 21, 2022 19:02:35 GMT
mondo
Your argument is similar to an argument such as:
1. This man speaks English
2. English men speak English
3. So the man is an Englishman. Fallacious.
Vasco
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Post by Admin on Oct 22, 2022 9:48:16 GMT
Mondo
In your answer you say "... the net effect is that every point inside the unit circle $C$ is now its exterior and vice versa,..."
But you need to say where precisely the initial point ends up. What you have written is not precise enough. The words inside and outside are too vague. If I gave you an initial point outside the unit circle, how would I know where to put the mapped point? Somewhere inside? But where?
Vasco
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Post by mondo on Oct 22, 2022 18:39:29 GMT
Mondo In your answer you say "... the net effect is that every point inside the unit circle $C$ is now its exterior and vice versa,..." But you need to say where precisely the initial point ends up. What you have written is not precise enough. The words inside and outside are too vague. If I gave you an initial point outside the unit circle, how would I know where to put the mapped point? Somewhere inside? But where? Vasco Vasco, I agree with your comments. My "solution" as it is now is more a "guess". I try to create a geometric proof for it . It is easy to show that characteristic points like $1$ or $0$ will remain on their positions however I need to show that some point $p$ outside the unit circle, will be mapped to a point $\bar{p}$ inside it, according to a given function $f$. Something similar to figure [21a] p.142 however how are we sure that $\bar{a} = I_k(a)$ and $a = I_K(\bar{a})$ as author claims there? |
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Post by Admin on Oct 23, 2022 7:13:22 GMT
Mondo In your answer you say "... the net effect is that every point inside the unit circle $C$ is now its exterior and vice versa,..." But you need to say where precisely the initial point ends up. What you have written is not precise enough. The words inside and outside are too vague. If I gave you an initial point outside the unit circle, how would I know where to put the mapped point? Somewhere inside? But where? Vasco Vasco, I agree with your comments. My "solution" as it is now is more a "guess". I try to create a geometric proof for it . It is easy to show that characteristic points like $1$ or $0$ will remain on their positions however I need to show that some point $p$ outside the unit circle, will be mapped to a point $\bar{p}$ inside it, according to a given function $f$. Something similar to figure [21a] p.142 however how are we sure that $\bar{a} = I_k(a)$ and $a = I_K(\bar{a})$ as author claims there? Mondo Best not to use $\bar{p}$, $\bar{a}$ etc unless you mean the conjugate. You will cause confusion. Vasco |
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Post by mondo on Oct 24, 2022 6:29:59 GMT
Vasco, I agree with your comments. My "solution" as it is now is more a "guess". I try to create a geometric proof for it . It is easy to show that characteristic points like $1$ or $0$ will remain on their positions however I need to show that some point $p$ outside the unit circle, will be mapped to a point $\bar{p}$ inside it, according to a given function $f$. Something similar to figure [21a] p.142 however how are we sure that $\bar{a} = I_k(a)$ and $a = I_K(\bar{a})$ as author claims there? Mondo Best not to use $\bar{p}$, $\bar{a}$ etc unless you mean the conjugate. You will cause confusion. Vasco Yes, actually I wanted to make it $\hat{p}$. Anyway, I still can't get figure [21a] - how does it shows that $I_K(a) = \hat{a}$? Why the $\sqrt{2}$ is involved? |
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Post by Admin on Oct 24, 2022 10:06:49 GMT
Mondo
As it says on page 142 just above [21], you need to look at [12] on page 134 and [3] on page 127 from which [12] is derived.
On page 127 read the paragraphs below [3] which consider the case when $L$ intersects $K$. In [12] the line $K$ from [3] becomes the plane $\Pi$ which in [21] becomes the complex plane and $K$ becomes the sphere centred at $N$ that intersects $\Sigma$ along its equator.
Then try again to understand [21] and the paragraphs below it.
Vasco
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Post by mondo on Oct 24, 2022 21:07:48 GMT
Ok I think I got it - Figure [21a] actually shows us a case where a circle $\Sigma$ passing through the center of inversion $K $got mapped to a line (a real line of $C$ in this case). Hence we can conclude that a circle (line) got mapped to a unit circle $\Sigma$ and hence circle was preserved. The $\sqrt(2)$ comes from the fact that we want to map point $a$ at a distance $1$ from $N$ to the point $S$ at a distance $2$. For this we can refer to the formula (4) from p.125 to get $\frac{R^2}{\bar{z} - \bar{q}} = \frac{\sqrt{2}^2}{1} = 2$. Is my understanding correct?
It is difficult to see because author advertised this example as a mapping of a "generic circle" while fig [21a] shows us a special circle - a line (is considered a circle that passes through a point at infinity).
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Post by Admin on Oct 25, 2022 14:44:06 GMT
Ok I think I got it - Figure [21a] actually shows us a case where a circle $\Sigma$ passing through the center of inversion $K $got mapped to a line (a real line of $C$ in this case). Hence we can conclude that a circle (line) got mapped to a unit circle $\Sigma$ and hence circle was preserved. The $\sqrt(2)$ comes from the fact that we want to map point $a$ at a distance $1$ from $N$ to the point $S$ at a distance $2$. For this we can refer to the formula (4) from p.125 to get $\frac{R^2}{\bar{z} - \bar{q}} = \frac{\sqrt{2}^2}{1} = 2$. Is my understanding correct? It is difficult to see because author advertised this example as a mapping of a "generic circle" while fig [21a] shows us a special circle - a line (is considered a circle that passes through a point at infinity). Mondo From what you have written above I don't think you have understood it properly. I will write a short one-page document explaining the way I see it and publish a link to it. I remember when I first read this about 7 years ago I had to study it for a long time. Everything in the book is difficult and chapter 3 is especially difficult. The only way to really understand each chapter is to do the exercises independently. Notice that $K$ and $\Sigma$ in figure 21 are not circles but spheres. The figure shows a cross section through $N$ and the real axis. Vasco |
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Post by mondo on Oct 25, 2022 16:18:36 GMT
From what you have written above I don't think you have understood it properly. I will write a short one-page document explaining the way I see it and publish a link to it. I remember when I first read this about 7 years ago I had to study it for a long time. Everything in the book is difficult and chapter 3 is especially difficult. The only way to really understand each chapter is to do the exercises independently. Notice that $K$ and $\Sigma$ in figure 21 are not circles but spheres. The figure shows a cross section through $N$ and the real axis. Vasco Thank you Vasco. Yes, it is a demanding chapter, I am impressed you remember so much after 7 years, forum and users like me definitely help with that  Yes I am aware $K$ and $\Sigma$ are spheres and what we see on [21a] is a vertical cross section through $N$ and real axis. Hence this figure must depicts a circle to real line mapping as this is the only thing we can represent on a real line. |
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