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{\bf Question}

Repeat question 6A, but now finding the transformation directly
from the general form of a M\"obius map by choosing any three
points on $Re(z)+Im(z)=1$ and their images on $|w|=1$. Are the
mappings of Q6A and Q6B unique? If we demand that the region
$Re(z)=Im(z)<1$ is mapped to the interior of $|w|=1$, do your maps
satisfy this condition? If they do not, find a simple remedy in
the form of an additional transformation which is to be carried
out.


\medskip

{\bf Answer}

$w=\ds\frac{\alpha z+\beta}{\gamma z+\delta}$

Without loss of generality we divide through by

$w=\ds\frac{az+b}{cz+d}$ for $a,\ b,\ c,\ d$ to be determined.

Pick any three points on $x+y=1$, say $1,\ \ds\frac{1}{2}(1+i),\
i$.

Now pick any three image points on $|w|=1$ say $1,\ i1, -1$.

Thus we solve:

$\left.\begin{array}{rcl} \ds\frac{z}{1} & \longrightarrow &
\ds\frac{w}{1}\\ \ds\frac{1}{2}(1+i) &\longrightarrow & i\\ i &
\longrightarrow & -1 \end{array} \right\} \Rightarrow
\begin{array} {rcl} 1 & = & \ds\frac{a+b}{1+c}\\ i & = &
\ds\frac{\frac{1}{2}a(1+i)+b}{\frac{1}{2}(1+i)+c}\\ -1 & = &
\ds\frac{ai+b}{i+c} \end{array}$

Solve for $a,\ b,\ c$ to get $\left\{\begin{array} {l} b=0,\
a=\ds\frac{1}{i}\\ c=-1-i \end{array} \right.$

Thus $w=\ds\frac{z}{iz+(1-i)}$

This is clearly a different map to Q6A. This is not surprising
since we could pick any 3 points and any 3 images for the
M\"{o}bius map. Thus M\"{o}bius maps \un{cannot} be unique for
this problem: the overall shapes are correct (line $\rightarrow$
circle) but the individual maps of $z$-points may get mixed up.
You probably have a different M\"{o}bius map for this reason.

If we require $x+y<1$ to map to $|w|<1$ try a test point for

$$w=\ds\frac{\sqrt{2}[(1+i)-z]}{(1+i)z}\ \rm{and}\
w=\ds\frac{z}{iz+(1-i)}$$

say $z=0$,

$$w=\sqrt{2}\left[\ds\frac{(1+i)-0}{(1+i)0}\right] \rightarrow
\infty;\ w=0$$

Thus the solution of Q6A is \un{not} sufficient, but the solution
of Q6B \un{is}.

We can remedy the solution of Q6A by making a further transform
$w_4=\ds\frac{1}{w_3}$. Why? Well $|w|=|w_3|=1 \rightarrow
|w_4|=1$. The boundary is therefore unchanged. But $|w_3|<1$ is
mapped to $|w_4|>1$ and vice versa. Hence the new mapping
$w=\ds\frac{1}{w_3}$ is

$$w=\ds\frac{(1+i)z}{\sqrt{2}[(1+i)-z]}$$

and now the test point $z=0 \rightarrow w=0$, i.e., to the
interior of the unit circle.

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