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{\bf Question}

 For the circle $A$ given in Problem 1, determine the general
form of an element of ${\rm M\ddot{o}b}(A) =\{ m\in {\rm
M\ddot{o}b}\: |\: m(A) = A\}$.

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\noindent Further, determine which elements of ${\rm
M\ddot{o}b}(A)$ {\bf do not} interchange the two discs determined
by $A$, and which {\bf do}.
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{\bf Answer} %note ref. to qu 1%

$m \in $M\"ob$(\bf{R})$ has one of the following forms:

\hspace{0.5in} $p(z)=\ds\frac{az+b}{cz+d}\ \ a,b,c,d \in \bf{R},\
ad-bc=\pm 1.$

\hspace{0.5in} $q(z)=\ds\frac{a\bar{z}+b}{c\bar{z}+d}\ \ a,b,c,d
\in \bf{R},\ ad-bc=\pm 1.$

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Calculate $mpm^{-1}$ and $mqm^{-1}$ with $m$ as in problem 1.

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$mpm^{-1}$

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$=\left(\begin{array} {cc} -4-8i & 4+6i\\ -2 & 1+i
\end{array}\right)\left(\begin{array}{cc} a & b\\ c & d \end{array}
\right)\left(\begin{array} {cc} 1+i & -4-6i\\ 2 & -4-8i
\end{array}\right)$

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$=\left(\begin{array}{cc} (4-12i)a-(8+16i)b &
(-32+56i)a+(-48+16i)b\\+(-2+10i)c+(8+12i)d &
+(-20+48i)c+(32+-56i)d\\ \\-2(1+i)a-4b &
(8+12i)a+(8+16i)b\\+2ic+2(1+i)d & +(2-10i)c+(4-12i)d
\end{array} \right)$

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where $a,b,c,d \in \bf{R}$ and $ad-bc=\pm 1$ ($mpm^{-1}$ does
\un{not} interchange the two discs determined by $A$ if and only
if $ad-bc=1$).

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$mqm^{-1}$: \un{careful with $\bar z$s in the compositions}

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$mqm^{-1}$

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$=\left(\begin{array} {cc} -4-8i & 4+6i\\ -2 & 1+i
\end{array}\right)\left(\begin{array}{cc} a & b\\ c & d \end{array}
\right)\left(\begin{array} {cc} 1-i & -4+6i\\ 2 & -4+8i
\end{array}\right)$

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since $m^{-1}(z)$ gets conjugated in $q(z)$

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\newpage
$\left(\begin{array}{cc} (-4-8i)a+(4+6i)c & (-4-8i)b+(4+6i)d\\
-2a+(1+i)c & -2b+(1+i)d
\end{array} \right) \left(\begin{array}{cc}
1-i & -4+6i\\ 2 & -4+8i
\end{array} \right)$

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$= \left[\begin{array} {cc} (-12-4i)a-(8+16i)b &
(64+16i)a+80b\\+(10+2i)c+(8+12i)d & -52c+(-64+8i)d\\
\\-2(1-i)a-4b & (8-12i)a+(8-16i)b\\+2c+2(1+i)d
& +(-10+2i)c+(-12+2i)d
\end{array} \right]$

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where $a,b,c,d \in \bf{R}$ and $ad-bc=\pm 1$ ($mqm^{-1}$ does
\un{not} interchange the two discs determined by $A$ if and only
if $ad-bc=-1$).
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