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Post by mondo on Jul 22, 2023 22:47:02 GMT
Yes and then in respect to $r$ and $\theta$. These two calculations give different twist. When calculating with respect to $z$ we get $2z$ hence amplification is $2|z|$ and twist $e^{i\theta}$. However when calculated with respect to $\theta$ (and only the part responsible for the twist, as if we are on the unit circle) I get $d/d\theta [e^{2i\theta}] = 2ie^{2i\theta}]$. Have you read the last paragraph on page 189? Angles in this context are angles between intersecting lines such as those in figure 1 Do you mean angles between these orthogonal lines that form a square? If so, then as we approach $0$ these squares are smaller and smaller but are still squares. I think the idea with duplication of angles is shown on figure [1] by angle $\theta$ on the LHS and $2\theta$ on the RHS. We see that some arbitrary point $z$ has an angle $\theta$ and after the $z^2$ mapping it is $2\theta$ now. |
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Post by Admin on Jul 23, 2023 8:18:18 GMT
Mondo
The amplitwist concept only applies to analytic functions of a complex variable $z$, not just any function.
For the second part of your question, I mean any two intersecting curves at any point are mapped to different curves that intersect, but the angle between the curves is preserved. [1] is a special case where the curves intersect at right angles.
Vasco
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Post by Admin on Jul 23, 2023 8:33:43 GMT
Mondo
These questions indicate to me that you do not understand fully the ideas introduced in chapter 4. I would suggest you read it again, all of it and slowly, and then do the exercises at the end of the chapter.
If you don't do this you will waste a lot of time thinking about things which are not relevant.
Vasco
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Post by Admin on Jul 23, 2023 12:42:16 GMT
Mondo
If you don't fancy reading the chapter again, try doing the exercises, and if you get stuck read the relevant parts of chapter 4 to help you solve the problem
Vasco
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Post by Admin on Jul 23, 2023 14:17:59 GMT
Mondo
Another thing I wanted to say is that in general the mapping is from squares to squares only in the limit when the squares become infinitesimal.
Vasco
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Post by Admin on Jul 23, 2023 17:44:58 GMT
Mondo
Do you understand what it means to say that a complex mapping is conformal?
Vasco
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Post by mondo on Jul 23, 2023 18:31:58 GMT
Thank you Vasco for all the hints. Yes I am going to reread entire chapter 4.
Yes, a conformal mapping is one where angles (sign and magnitude) are preserved for the mapping in the entire region.
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Post by Admin on Jul 23, 2023 18:43:39 GMT
Mondo
OK but you need to be clear what you mean by angle here. it is not the angle of a complex number, but the angle between two intersecting curves in the complex plane.
Vasco
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Post by mondo on Jul 23, 2023 18:54:17 GMT
Yes Vasco, this is the angle I had in mind writing reply #66. However on p.190, when author talks about a critical point $z=0$, it sounds like he claims the conformality breaks there but I can't visualize it. Figure [1] shows that as we approach 0 the conformality is maintained still - as I said earlier, squares from the LHS are squares still on the RHS (only smaller). When we actually hit $0$ who cares about conformality if every complex number is gone?
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Post by Admin on Jul 23, 2023 19:08:25 GMT
Mondo
No. At the origin on the left take the angle between any two rays passing through the origin. These rays are mapped to two rays again through the origin, but the angle between them at the origin is now double what is is on the left. So conformality breaks down at the origin.
Vasco
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Post by mondo on Jul 23, 2023 19:12:24 GMT
Yes, that makes a perfect sense! However why is it also doubled at $\infty$?
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Post by Admin on Jul 23, 2023 19:18:07 GMT
Mondo
Read subsection 4 on pages 144-146.
Vasco
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Post by mondo on Jul 24, 2023 6:51:10 GMT
Vasco, thank you. I think the double angle issue at critical points $z=0$ and $z=\infty$ is clear now. However how about this challenge $\frac{d}{dz}z^2 = (\partial_x + i\partial_y)(x+iy)^2 = \partial_x{(x^2-y^2 + 2ixy)} + i\partial_y{(x^2-y^2 + 2ixy)} = 2x + 2iy - 2iy -2x = 0$
This method utilizes the "Powerful computational tool" introduced on page 517.. these derivatives in respect to different variables are really confusing - I would expect to get the same result but what I get is in direct conflict with the result obtained from a different "method".
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Post by Admin on Jul 24, 2023 7:07:33 GMT
Mondo
It's your mistake! $\frac{d}{dz}$ is not equal to $(\partial_x+i\partial_y)$
Vasco
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Post by mondo on Jul 24, 2023 7:10:44 GMT
Also, Mondo It's shown algebraically on page 198!!!! Vasco No it's not what I wanted. It just shows how to represent a derivative as the "amplitwist". While I want to get an algebraic feel of the fact that angles are not doubled beside the origin. This is how I would show this: let's take two complex numbers $A=e^{i\theta}$ and $B=e^{i(\theta+h})$ so $B$ is very close to $A$ on the unit circle. Now, let's apply the complex mapping $z^2$, our point A is mapped to $e^{2i\theta}$ while $B = e^{2i(\theta + h)}$ and now a crucial observation in respect to the origin $z=0$ angles of both are doubled, but in respect to each other they are only $2h$ apart! Hence the angles between them is definitely not doubled. So to summarize the doubling of angle that author is talking about in the first paragraph of page 190 is in regards to the origin. The phrase he uses "except at z=0" is confusing a bit. and what you said today Mondo No. At the origin on the left take the angle between any two rays passing through the origin. These rays are mapped to two rays again through the origin, but the angle between them at the origin is now double what is is on the left. So conformality breaks down at the origin. Vasco is more or less my idea in the reply I gave the day before, just rephrased and simpler. |
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