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{\bf Question}

Write down an explicit M\"obius transformation taking the circle
$\{ z\in {\bf C}\: |\: |z-2| =3\}$ to the circle $\{ z\in {\bf
C}\: |\: {\rm Re}(z) + {\rm Im}(z) = 1\}$.
\medskip

{\bf Answer}

Take 3 points on $A=\{|z-2|=3\}$, say $z_1=2+3i, z_2=-1, z_3=5$,
and 3 points on $B=\{\rm{Re}(z)+\rm{Im}(z)=1\}$, say
$w_1=1,w_2=i$, and $w_3=-2+3i$. Now find $m \in$ M\"ob$^+$ with
$m(z_k)=w_k,\ \ k=1,2,3$. Start by finding $n(z)$ with $n(2+3i)=0,
n(-1)=\infty$ and $n(5)=1$, and $p(z)$ with $P(1)=0, p(i)=\infty$
and $p(-2+3i)=1$ and then set $m=p^{-1} \cdot n$.

\begin{eqnarray*} n(z) & = & \ds\frac{z-(2+3i)}{z+1} \cdot
\ds\frac{6}{5-(2+3i)}\\ & = & \ds\frac{6}{3-3i}\cdot
\ds\frac{z-(2+3i)}{z+1}\\ & = &
\ds\frac{6z-6(2+3i)}{(3-3i)z+(3-3i)}
\end{eqnarray*}

\begin{eqnarray*} p(z) & = & \ds\frac{z-1}{z-i} \cdot
\ds\frac{-2+3i-i}{-2+3i-1}\\ & = & \ds\frac{z-1}{z-i} \cdot
\ds\frac{-2+2i}{-3+3i}\\ & = &
\ds\frac{2(z-1)}{3(z-i)}=\ds\frac{2z-2}{3z-3i} \end{eqnarray*}

$$p^{-1}(z)=\ds\frac{-3iz+2}{-3z+2}$$

\begin{eqnarray*}
m(z) & = & p^{-1}(n(z))\\ & = & \ds\frac{-3i n(z)+2}{-3n(z)+2}\\ &
= &
\ds\frac{-3i(6z-6(2+3i))+2((3-3i)z+(3-3i))}{-3(6z-6(2+3i))+2((3-3i)z+(3-3i))}\\
& = & \ds\frac{(6-24i)z-48+30i}{(-12-6i)z+42+48i} \end{eqnarray*}

(There are many others.)

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