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

{\bf Question}

\begin{itemize}
\item[a)]
Let $x,y$ be the real and imaginary parts of the complex number
$z$ and let $w=\tan z$.  By finding the modulus and argument of
$\ds\frac{w-i}{w+i}$ in terms of $x,y$ show that a line
$y$=constant of the $z$-plane transforms to a circle of the
$w$-plane, and that a line $x$=constant transforms to an arc of a
circle through $i$ and $-i$.


\item[b)]
Find all solutions of $\tan z=2-i$.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds w=\tan z=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}=
-i\frac{e^{2iz}-1}{e^{2iz}+1}$

so $\ds e^{2iz}w+w+ie^{2iz}-i=0 \hspace{0.3in}
e^{2iz}=-\frac{w-i}{w+i}$

So $\ds\frac{w-i}{w+i}=-e^{-2iz}=-e^{-2ix}e^{2y}=
e^{-2ix+i\pi}e^{2y}$

So $\ds\left|\frac{w-i}{w+i}\right|=e^{2y} \hspace{0.3in}
\arg\left(\frac{w-i}{w+i}\right)=\pi-2x$

So if $y$=constant $\ds\hspace{0.3in}
\left|\frac{w-i}{w+i}\right|$=constant - circle of Apallonius.

If $x$=constant $\ds\hspace{0.3in}
\arg\left(\frac{w-i}{w+i}\right)$=constant - circular arc.


DIAGRAM



\item[b)]
$\tan z=2-i$, so $\ds e^{2iz}=-\frac{2-i-i}{2-i+i}=-1+i=
\sqrt2e^{\frac{3\pi i}{4}}=\alpha$

$2iz=\ln|\alpha|+i(\arg\alpha+2n\pi)$

$\ds\hspace{0.3in}
=\frac{1}{2}\ln2+i\left(\frac{3\pi}{4}+2n\pi\right)$

$\ds z=\frac{1}{4i}\ln2+\frac{3\pi}{8}+n\pi \hspace{0.3in}
n\in{\bf Z}$

\end{itemize}

\end{document}