illustration of four and two reflections in Figure [32]

Gary

I have looked at your document and it worries me a little that the distance along $L$ from $z$ to $m$ looks to be equal to the distance along $L$ from $m$ to $w$.
As I understand things, the h-distances $\mathcal{H}\{z,m\}$ and $\mathcal{H}\{m,w\}$ should be equal, not the distances along the h-line $L$, as it looks in your diagram. In the paragraph under figure 1 you do not say how you found the point $m$.

Vasco

Gary

I think the way to get over this problem is by using inversion in the circle. This results in two h-line segments which are equal in the h-plane. I would suggest another read of subsection 5, especially pages 303-305.

Vasco

Gary

Yes, I think that's where the answer to your problem lies.

Vasco

Gary

I have measured some of the distances in figure 1 and find that the length of $w$ seems to be more like $0.7$ rather than $0.75$, but that may be due to my ruler. However in figures 2 and 3 I do measure it as $0.75$. Worth a check.

As far as calculating $m$ is concerned, if you think of the right angled triangle $mc_Lc_A$ then the two sides at right angles are $R_A$ and $R_L$ and so, assuming $c_A$ is the origin, it is then possible to calculate the angle $\alpha$ of $m$ from $c_A$ and so $m=R_A e^{i(\pi-\alpha)}$.

Vasco

Gary

Did you find the reason for the discrepancy you mentioned?

I don't think it's necessary to use the cosine rule as the triangle is right-angled, but it should work the same.

Vasco

Gary

I haven't thought about this very deeply, but couldn't the fact that you are using finite lengths for infinitesimals cause some inaccuracies?

Vasco

Vasco,

I've been giving figure [32] some further study. On p. 312, in sentence 3, Needham says "Next apply the h-rotation $\mathcal{R}^{\theta}_{w}$, where $\theta$ is the angle from $d\tilde{z}$ to $dw$." To this point, we have only L, z, w, and m. If we are to apply the h-rotation, we need at least two lines in which to reflect $d\tilde{z}$, which has been calculated with reflections and the equations on p. 303. But the only line that we have is $L$. We are stuck. To make two reflections, we would need line A or B. The only way I can think of to make an h-rotation would be to begin with a set of concentric circles in the $\mathbb{C}$ plane and apply $F^{-1}(z)$ using $w$ and $\overline{w}$ as fixed points (p. 163). The pre-images of the intersecting h-lines would be rays intersecting at 0. We chose the angles of the rays and discard the lower half plane. This seems impractical, so we try another approach.

We draw lines A and B orthogonal to L. Now there are two lines (B and L) in which we can easily reflect $d\tilde{z}$. These h-lines are orthogonal, so reflection in them results in $dw = -d\tilde{z}$, which doesn't look like [32]. We need a non-orthogonal line through w to intersect with B. Needham says to "draw the h-line C through w making angle ($\theta/2$) with B. But $\theta$ is the very angle we are seeking to obtain the rotation $\mathcal{R}^{\theta}_{w}$! And if we did want to assign an arbitrary value to $\theta$, how would we draw the h-line C at that angle ($\theta/2$) through w? We can calculate the tangent of B from a ray to w and add $\theta/2$. It is not difficult to solve for the center of C on the horizon. Then we apply $\mathcal{I}_C(\mathcal{I}_B(w + d\tilde{z})$ to obtain $dw$. Thus, we have obtained $arg(dw/d\tilde{z})$ from the angle between B and C, over which we have full control.

This much enables the plotting of Figures 1-3. These seem to be correct. In figure 4, we test the idea that the motion of dz to dw can be written $\mathcal{R}^{\phi}_{a}$. That did not go well.

I have revised the figures somewhat and added a new figure.

Gary