Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Nov 9, 2015 22:38:09 GMT
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Vasco
GaryVasco
Posts: 14
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Post by Vasco on Nov 10, 2015 10:46:18 GMT
Gary
I have one or two issues with your solution:
1. \(|(1/z)'|\) is the absolute value of the amplification...
In the amplitwist concept the amplitwist at a point is just a complex constant such as \(pe^{i\phi}\) with \(p\) being the amplification and \(\phi\) being the twist. So there is no need to talk of the absolute value of the amplification since the amplification is by definition a real number \(\geq 0\). Perhaps you were intending to write: \(|(1/z)'|\) is the absolute value of the amplitwist.
I assume that when you write absolute value this means the same as modulus.
2. \(f(z)=1/z=ae^{i\alpha}z\) (p. 199 and by Figure 1, below)
On page 199 the second example function is \(f(z)=Az\) where \(A=ae^{i\alpha}\) is a complex constant. It turns out that in this case the amplitwist \(f'(z)=A=ae^{i\alpha}\). It does not follow that any function, \(f(z)=(1/z)\) in this case, can be written as amplitwist\(\times z=ae^{i\alpha}z\). It is just fortuitous that in this case \(1/z\) can be written as amplitwist\(\times (-z)=(-1/z^2)\cdot -z\) and so 3 below works. By considering other cases such as \(f(z)=z^2\) and \(f(z)=e^z\) we can see that \(f(z)\) cannot generally be expressed as \(ae^{i\alpha}z\).}
In Figure 1 of your solution the solid vector at \(1/z\) is not equal to \(1/\epsilon\) as shown on your diagram. Since \(z\) is mapped to \(1/z\) and \(z+\epsilon\) is mapped to \(1/(z+\epsilon)\) then the length of the solid vector at \(1/z\) is
\(|1/r-1/(r+|\epsilon|)|=\frac{|\epsilon|}{r(r+|\epsilon|)}=|\epsilon|/r^2+\)terms in \(|\epsilon|^2\) and higher.
So neglecting higher order terms we can see that the amplification is \((1/r^2)=1/|z|^2\). I think that that this would be a legitimate way of arriving at the required result using your alternative approach to the one suggested in the exercise.
3. The contraction in this case is the ratio of \(|1/\overline{z}|\) to \(|z|\).
Presumably you are using 2 above to lead you to this conclusion. However, logically 2 is not correct and so 3 is not correct. Also, having written 2 above and assuming it to be correct, all you need to do is re-arrange and you will obtain \(ae^{i\alpha}=1/z^2\), without any more work.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 10, 2015 20:29:03 GMT
Vasco,
You have put your finger on what appears to be a misconception I have had, that amplitwist always takes the form aei Alpha in analytic functions.
Regardless of my difficulty there, my approach was actually to inspect the figure and to notice that the length of 1/z changed from that of z by the ratio of |1/zBar| to |z|. One can reason from prior knowledge of 1/z and it’s conjugate and inversion within a circle that this is the amplification, and that 1/z is a twist of -2 Theta followed by multiplication of ze-i2Theta by the amplification, so z -> Az seemed appropriate. Where I run into serious trouble is when I compare the amplitwist obtained this way with (1/z)’—the signs are different. Yet with a conformal and analytic function, should they not be the same? Or is f’ = amplitwist just a lucky coincidence with some functions?
Regarding your point 2, I don’t quite follow the calculations of the length of the epsilon vector. I’ll work on it. The labels in my figure are sloppy, but the f(epsilon) vectors were calculated as you suggest, as 1/(z+e) and 1/(z+ie) and drawn from point to point. Lengths were not calculated. Other than illustrating equal twist and amplification, I don’t see what is gained by doing the calculations of amplitwist on these vectors instead of on a generic z, since under I(z) they must behave like all other points on the complex plane other than 0.
I will keep current revisions at the same Dropbox link for a time.
Gary
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Post by Admin on Nov 10, 2015 20:49:36 GMT
Gary
You are right that the amplitwist at a point is of the form \(ae^{i\alpha}\) and yes \(f'=\)amplitwist, but the amplitwist is the effect of the function on infinitesimal vectors not on \(z\).
Hope this makes sense. More later regarding your other points.
Vasco
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Post by Admin on Nov 10, 2015 21:37:48 GMT
Gary
Looking at your diagram, you can see that the twist for the infinitesimal vector at \(z\) is \(-2\theta-\pi\) not \(-2\theta\), and so if if \(a\) is the amplification then the amplitwist is \(ae^{i(-2\theta-\pi)}=ae^{-i\pi}\cdot e^{-2\theta}=-ae^{-2\theta}\) and lo and behold a minus sign appears! . That comes from thinking about the effect of the function on infinitesimal vectors and not on \(z\).
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 10, 2015 23:10:37 GMT
Vasco,
Yes! That clears up the mystery. Now I'm very happy to have a reliable procedure for deriving amplitwist from figures. I think I went astray with Figure [12b], where the twist of (a epsilon) equals the rotation of z. I mistakenly concluded this must always be the case so I was just inspecting the change in z.
Also, consider p. 198, where f'(z) = |f'(z)|e^{i arg[f'(z)]. This shows that a = |f'(z)| and everything else is twist. So in 16, |f'(z)| = |(1/z)'| and e^{i arg[f'(z)] = -e^{-i 2Theta}.
Regarding the case of z^2, where (z^2)' = 2z, how do we apply this to eps vectors if we wanted to plot them? Do we write (2eps eps) = 2eps^2 = (|2eps^2| e^{i 2Arg(eps)} eps) (assuming they emanate from the origin)? I'll try a plot.
Typos: e^{-2Theta} --> e^{-i2Theta} twice.
Gary
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Post by Admin on Nov 11, 2015 9:17:40 GMT
Vasco, Regarding the case of z^2, where (z^2)' = 2z, how do we apply this to eps vectors if we wanted to plot them? Do we write (2eps eps) = 2eps^2 = (|2eps^2| e^{i 2Arg(eps)} eps) (assuming they emanate from the origin)? I'll try a plot. Gary
Since the amplitwist of \(z^2\) is \(2z=2re^{i\theta}\), amplification \(2r\) and twist \(2\theta\), we can write \(2re^{i\theta}\cdot \epsilon\) for the amplitwisted vector. So \(\epsilon\) at \(z=re^{i\theta}\) is multiplied by \(2r\) and rotated through \(\theta\) to obtain the infinitesimal vector at \(z^2\).
See also page 194.
Vasco
Watch out for the origin - it's a critical point for \(z^2\). |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 11, 2015 16:11:09 GMT
Vasco, Regarding the case of z^2, where (z^2)' = 2z, how do we apply this to eps vectors if we wanted to plot them? Do we write (2eps eps) = 2eps^2 = (|2eps^2| e^{i 2Arg(eps)} eps) (assuming they emanate from the origin)? I'll try a plot. Gary
Since the amplitwist of \(z^2\) is \(2z=2re^{i\theta}\), amplification \(2r\) and twist \(2\theta\), we can write \(2re^{i\theta}\cdot \epsilon\) for the amplitwisted vector. So \(\epsilon\) at \(z=re^{i\theta}\) is multiplied by \(2r\) and rotated through \(\theta\) to obtain the infinitesimal vector at \(z^2\).
See also page 194.
Vasco
Watch out for the origin - it's a critical point for \(z^2\).Vasco, I am able to partially confirm your statement via trial and error with a plot. With \(2\epsilon \epsilon\) the result was huge \(\epsilon\) arrows, a bit of a contradiction, due to the squaring of the absolute values, but with \(2re^{i\theta}\cdot \epsilon\) the plot displays what appears to be the appropriate amplitwist: dl.dropboxusercontent.com/u/35393965/test_amplitwist_zSq.pdfYes, Figure [5] clearly gives the answer. I don't know how I missed it. Do you know why the epsilons are not rendering properly in my first sentence? Gary |
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Post by Admin on Nov 11, 2015 17:25:12 GMT
Gary
You need to enclose the maths between backslash( and backslash). I have edited it so you can see what I have done if you quote it.
What do you mean when you type \(2\epsilon \epsilon\) ?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 11, 2015 20:48:33 GMT
Vasco
Thank you.
>What do you mean when you type 2ϵϵ ?
At that point, I thought that one would apply \((z^2)' = f'(z)\) as \(f'(\epsilon)\), so \(f'(\epsilon)\cdot\epsilon = 2\epsilon\cdot\epsilon\).
Gary
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