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

{\bf Question}

For each of the following equations, determine whether the
equation describes a Euclidean line in ${\bf C}$ or a circle in
${\bf C}$ (or neither).  In the former case, give its slope and
$y$-intercept.  In the latter case, give its center and radius.
\begin{itemize}
\item [(a) ] $5z\overline{z} + (3+i) z + (3-i) \overline{z} +6 = 0$;
\item [(b) ] $(-2-3i) z + (-2+3i) \overline{z} +2 = 0$;
\item [(c) ] $-z\overline{z} -2i z +2i \overline{z} +1 = 0$;
\end{itemize}
\medskip

{\bf Answer}

\begin{description}
\item[(a)]
$5z\bar z+(3+i)z+(3-i)\bar z+6=0$

$z\bar z+\ds\frac{3+i}{5}z+\ds\frac{3-i}{5}\bar
z+\ds\frac{6}{5}=0$

$\left(z+\ds\frac{3-i}{5}\right)\left(\bar
z+\ds\frac{3+i}{5}\right)-\ds\frac{(3+i)(3-i)}{25}+\ds\frac{6}{5}=0$

$\left(z-\ds\frac{(-3+i)}{5}\right)\left(\bar
z-\ds\frac{(-3-i)}{5}\right) -\ds\frac{10}{25}+\ds\frac{30}{25}=0$

$\left|z-\ds\frac{(-3+i)}{5}\right|^2=-\ds\frac{20}{25}$

so \un{no solutions}.

\item[(b)]
$(-2-3i)z+(-2+3i)\bar z+2=0$ (euclidean line)

$(-2-3i)(x+iy)+(-2+3i)(x-iy)+2=0$

$-2x+3y-2iy-3ix-2x+3y+3ix+2iy+2=0$

$-4x+6y+2=0$

\hspace{1in}$\begin{array}{rcl} 6y & = & 4x-2\\ y & = &
\ds\frac{2}{3}x-\ds\frac{1}{3} \end{array}$

slope $\ds\frac{2}{3}$

y-intercept $-\ds\frac{1}{3}$

\newpage
\item[(c)]
$z\bar z-2iz+2i\bar z+1=0$

$z\bar z+2iz-2i\bar z-1=0$

$(z-2i)(\bar z+2i)-4-1=0$

$|z-2i|^2=5$

$|z-2i|=\sqrt 5$

euclidean circle center $2i$, radius $\sqrt 5$.
\end{description}

\end{document}