\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Let $A$ be the circle in ${\bf C}$ with center $a = 2 + 2i$ and
radius $r = 1$.  Let $m(z) = 1 + \frac{1}{z}$.  Find the equation
of the circle $m(A)$.
\medskip

{\bf Answer}

Start by \un{putting the equation for $A$ into standard form}:

$$\begin{array}{l} |z-(2+2i)|^2=1\\ (z-(2+2i))(\bar z-(2-2i))=1\\
z \bar z-(2-2i)z-(2+2i)\bar z+|2+2i|^2=1\\ \un{z \bar
z-(2-2i)z-(2+2i)\bar z+7=0} \end{array}$$

$m(z)=1+\ds\frac{1}{z}=\omega$

$\Rightarrow \ds\frac{1}{z}=\omega-1 \Rightarrow
z=\ds\frac{1}{\omega-1}$ \un{plug into the equation for $A$}:

\medskip
$\left(\ds\frac{1}{\omega-1}\right)\left(\ds\frac{1}{\bar\omega-1}\right)
-(2-2i)\ds\frac{1}{\omega-1}-(2+2i)\ds\frac{1}{\bar\omega-1}+7=0$
\medskip
\un{clear denominators}:

$1-(2-2i)(\bar\omega-1)-(2+2i)(\omega-1)+7(\omega-1)(\bar\omega-1)=0$
\medskip
\un{and simplify}:

$1-(2-2i)\bar\omega+2-2i-(2+2i)\omega+2+2i+7\omega\bar\omega-7\bar\omega-7\omega+7=0$
\medskip
\begin{tabular} {|c|} \hline\\
$7\omega\bar\omega+(-2-2i-7)\omega+(-2+2i-7)\bar\omega+12=0$\\
\hline
\end{tabular}

which is the equation for $m(A)$.
\medskip
$7\omega\bar\omega+(-9-2i)\omega+(-9+2i)\bar\omega+12=0$
\medskip
$\left|\omega-\left(\ds\frac{9-2i}{7}\right)\right|^2=\ds\frac{1}{49}$
\end{document}