\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

Describe the following loci in the Argand diagram:
\begin{description}
\item[(a)] $\ds \arg \left(\frac{z-z_1}{z-z_2}\right) =
\frac{\pi}{6}$
\item[(b)] $\ds |z-z_1| - |z-z_2| = 1$
\item[(c)] $\ds |z+3i|^2 - |z-3i|^2 = 12$
\item[(d)] $\ds |z+ik|^2 + |z-ik|^2 = 10k^2$
\end{description}
\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]

\begin{center}
\epsfig{file=cn-1-1.eps, width=50mm}

$\alpha_1 - \alpha_2 = \frac{\pi}{6}$
\end{center}

${}$

\item[(b)]

\begin{center}
\epsfig{file=cn-1-2.eps, width=50mm}
\end{center}


\item[(c)]

${}$

\setlength{\unitlength}{.5in}
\begin{picture}(4,4)
\put(0,2){\line(1,0){3}}

\put(1,0){\line(0,1){4}}

\multiput(1,0)(0,0.5){9}{\makebox(0,0){-}}

\multiput(0,2.5)(0.2,0){16}{\circle*{0.05}}

\put(1,0){\line(2,5){1}}

\put(1,3.5){\line(1,-1){1}}

\put(.8,0){\makebox(0,0){A}}

\put(0.8,2.7){\makebox(0,0){B}}

\put(0.8,3.5){\makebox(0,0){C}}

\put(2.1,2.2){\makebox(0,0){D}}

\put(2.1,2.7){\makebox(0,0){z}}

\end{picture}

\begin{eqnarray*} |+3i|^2 - |z-3i|^2 & = & AD^2 - CD^2 \\ & = &
AB^2 +BD^2 - CB^2 -BD^2 \\ & = & AB^2 - CB^2 \\ & = &
(AB+CB)(AB-CB) \\ & = & 6(AB-BC) \\ & = & 12 \end{eqnarray*}

So AB-BC=2, AB+BC=6.  So B is at i.

So the locus is the line $z = a+i \, \ a\epsilon{\bf R}$

${}$

${}$

\item[(d)]
${}$

\setlength{\unitlength}{.75in}
\begin{picture}(4,2)
\put(0,0.75){\line(1,0){1.5}}

\put(0.5,0){\line(0,1){1.5}}

\put(0.5,0.1){\line(2,3){0.65}}

\put(0.5,1.4){\line(2,-1){0.65}}

\put(1.15,1.075){\line(-2,-1){.65}}

\put(0.3,0.1){\makebox(0,0){-ik}}

\put(0.4,1.4){\makebox(0,0){ik}}

\put(1.25,1.1){\makebox(0,0){z}}

\end{picture}

\begin{eqnarray*} |z+ik|^2 + |z-ik|^2 & = & 2k^2 + 2|z|^2 {\rm \ \ By\ Apollonius'\ Theorem}  \\ &=&
10k^2\\  {\rm So\ \ } |z| & = & 2|k| {\rm \ \ -\ a\ circle\ }
\end{eqnarray*}
\end{description}
\end{document}