Exercise 18

Gary

I like your answer to part (ii).  The only thing that is missing, it seems to me, is a demonstration that when the shape at \(z\) is a long way from \(q\) the amplification is very small, and when the shape is near to \(q\) the amplification is very large, and so somewhere in between it must be 1.  You could use your expression for the amplification from part (i), (even though not necessary to answer part (i)), to show this.  As you have no doubt noticed your expression, when corrected, is identical to the one I derived in part (ii) using (3) on p. 124.

Vasco

Gary

In this special configuration the pre-image circles are centred at \(q=-d/c\) and so mapping (i) in (3) p. 124 transfers the circles so that they are centred at the origin, but initially the circles are described by \(|z+d/c|=\rho\).

Vasco

Gary

How do you calculate the image of the triangle? In what follows I am just guessing.
If the three vertices of the pre-image triangle are $p,q,r$, are you calculating the image vertices as $M(p),M(q),M(r)$?
If the answer is yes then it may explain what you are seeing in figure 1.

If you calculate the amplification($A$) and twist($\phi$) at a point $z$, then you should calculate the vertices of the image triangle as follows:
Let $\epsilon_1$ be a vector emanating from $z$ (i.e, one of the sides of the pre-image triangle) then the image of $z+\epsilon_1$ is
$M(z)+Ae^{i\phi}\epsilon_1$, not $M(z+\epsilon_1)$.
If you calculate the vertices of the image triangle like this, then the angle between the sides would have to be the same as the angle between the corresponding sides of the pre-image triangle and conformality would be guaranteed!.

I think that what you are seeing in your figure 1 is the effect seen in figure 11 in subsection 3 Analytic Functions on pages 197-8 - all down to the fact that we are really dealing with infinitesimal shapes. Only guessing of course.

$M(z+\epsilon_1)=M(z)+\epsilon_1M'(z)+\epsilon_1M''(z)/2+...=M(z)+Ae^{i\phi}\epsilon_1+\epsilon_1^2M''(z)/2+...$
I suspect that $M(z+\epsilon_1)$ is not a good approximation to $M(z)+Ae^{i\phi}\epsilon_1$ when $\rho<1$.

Vasco

PS I'll have a look at the source code if you think it would help.

Gary

I agree that part (ii) of question 18 on page 213 is open to more than one interpretation in terms of what it is asking you to do. When I did the problem for the first time, well over a year ago even though I only published it recently, I don't remember struggling with the meaning of (ii). Looking at it again now I can see how it could be misinterpreted. What follows is not meant to be a way to rewrite the exercise, but more of an explanation of what I thought, and still think, Needham is asking:

"An infinitesimal shape moves around on the complex plane. Imagine that it starts off by being placed at a point far away from $q$, and then moves to a point very close to $q$. Because of part (i) we can see that as it moves on this path it initially lies on a circle with a very large radius, centred at $q$, and at the end of its journey it ends up lying on a circle with a very small radius, centred at $q$. We can calculate its image under $M$ at any point on this journey and this image will lie on the corresponding image circle. We need to show that when the infinitesimal pre-image shape is at a large distance from $q$ the size of the image shape under $M$ is very small (approaching zero) and that when the pre-image shape is very close to $q$ the size of the image shape is very large, and that therefore when the pre-image shape is travelling, it will, at some point, find itself at a point where it and its image are congruent. At all other points they are similar."

Does the above correspond with your interpretation of the the exercise?

If you agree that the above is a valid interpretation of what Needham is asking, I will rewrite part (ii) to reflect this and try to make it less open to interpretation, and then if you are happy with my rewording we can post it in the errata section, but more as a clarification than a real error.

Vasco

Vasco,

I agree with your rephrasing. It is brief and clear. I also agree with the discussion paragraph.

I have corrected my calculations of twist in Figure 1. The plot now looks more like what I expected it to look like, though that was not my motive for the correction. I found the twist was a little more difficult to understand and calculate than the amplification, as it seems necessary to include z in the denominator of M'(z), which means calculation of \(\rho\) and the selection of an appropriate \(\theta\).

Gary

Vasco,

I am still tussling with 4:18. I'm finding it tricky to get a sensible application of twist. I calculated twist as arg(M'(z)) and used it in the formula for amplitwist \(f'(z) = |f'(z)|e^{i arg(f'(z))}\), but it's not working out well. The twist should then be \(arg(M'(z)) = arg((ad-bc)/(cz+d)^2)\). I noticed in your diagram that the \(\epsilon\) arrow for M(z) shows a twist of Pi, so that it points directly back to a/c. Mine don't. I am wondering how one gets that out of (3) on p. 124 or out of M'. I know that in (3), you see the minus sign on (1/z)', and that (1/z) is a factor in step (iii), so perhaps letting \(p = (ad-bc)/c^2\) we could take (p/z)' and get \(0*1/z + p(-1/z^2) = -(ad-bc)/(c^2 z^2)\). I just tried it and it straightens out the arrows, but they point away from the center rather than towards it. The effect on the small vectors is conformal.

I have reached a little better understanding of the effect of amplification on the \(\epsilon\) vectors. In the figure, I used numbers to indicate the order of amplification as \(\rho\) decreases.

dl.dropboxusercontent.com/u/35393965/nh.ch4.ex18.pdf

Gary