exercise p. 307, (39)

Gary

Here is my version. I will now look at yours.

The formula for inversion in a circle $K$ centred at the point $q$ on the real axis, and of radius R, is given by

$\mathcal{I}_K(z)=\frac{q\bar{z}+(R^2-q^2)}{\bar{z}-q}$.

If we define the Möbius transformation $M_K$ as

$M_K(z)=\frac{qz+(R^2-q^2)}{z-q}$,

it follows that $\mathcal{I}_K(z)=M_K(\bar{z})$, and $\overline{M_K(\bar{z})}=M_K(z)$.
If we define a second circle $J$ with centre on the real axis then we can write
$\mathcal{I}_J(z)=M_J(\bar{z})$ and $\overline{M_J(\bar{z})}=M_J(z)$.
If we now compose these two inversions then we find that
$\mathcal{I}_J[\mathcal{I}_K(z)]=\mathcal{I}_J\mathcal[M_K(\bar{z})]=M_J[\overline{M_K(\bar{z})}]=M_J[M_K(z)].$

So the composition of the two inversions is equal to the composition of the two corresponding Möbius transformations. Since the two Möbius transformations both have real constants $a,b,c,d$, then the composition must also have real constants.

So if we write the composition Möbius transformation as

$M(z)=\frac{az+b}{cz+d}$ with $a,b,c,d$ real,

and then write $z=x+iy$ where $y>0$, and substitute, we find that

$M(z)=\frac{ax+b+iay}{cx+d+icy}=\frac{(ax+b+iay)(cx+d-icy)}{(cx+d+icy)(cx+d-icy)}$

Since $M$ transforms the upper half plane to itself we must have

Im$[M(z)]>0$ or $\frac{ay(cx+d)-cy(ax+b)}{(cx+d)^2+c^2y^2}>0$

Since $y>0$ this simplifies to $ad-bc>0$.

So finally, a motion of type $\mathcal{M}\equiv\mathfrak{R}_{L_2}\circ\mathfrak{R}_{L_1}$, where $L_1$ and $L_2$ are h-lines, corresponds to a Möbius transformation

$M(z)=\frac{az+b}{cz+d}$, where $a,b,c,d$ are real and $(ad-bc)>0$.

Vasco

Vasco,

I don't understand the need to create and compose two Möbius transformations. On p. 307, paragraph 3, ln 5, Needham says "Composing two such functions, we find [exercise] ...", where he was referring to the inversion $\mathcal{I}_{K}$. So it appears that he is only calling for the composition of two inversions to create one Möbius transformation.

Also, I am finding it impossible to distinguish tildes from overstrike lines, or are they all overstrike lines? I see that they probably are.

Gary

Vasco,

I wouldn't say the text prohibits your solution but I don't think it invites it. It seems a harder way to conceptualize it.

Gary

Vasco,

It's certainly worthwhile reading that subsection and the following ones, but the quote in paragraph 2, p. 172 again calls for only two reflections. The algebra with $M(z) = \mathcal{I}_{S}(\mathcal{I}_{L}(z))$ was not too bad and the results with the more direct approach were easier to interpret.

I am finding subsection 7 slow going, but I'm making some progress. It would be a help if I could reproduce the diagrams, but I'm finding it difficult to plot the orthogonal circles under the different conditions presented by h-rotations, limit rotations, and h-translations (figures [27], [29], [30]).

Gary