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


{\bf Question}

\begin{description}
\item[(a)] Show that all the roots of the equation $$(1+x)^{2n+1} =
(1-x)^{2n+1}$$ are given by $$\pm i \tan\left( \frac{k\pi}{2n+1}
\right) \hspace{.2in} k = 0,1,2,\cdots , n$$By putting $n=2$ show
that
$$\tan^2\left(\frac{\pi}{5}\right)\tan^2\left(\frac{2\pi}{5}\right)=5.$$
\item[(b)] Let $z = x+iy$ and $w = u+iv$.  If $w = z^2 + 2z$ show
that the line $v=2$ is the image of a rectangular hyperbola in the
$z$-plane.  Sketch this hyperbola.
\end{description}
\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] \begin{eqnarray*} (1+x)^{2n+1} & = & (1-x)^{2n+1} \\
{\rm So \ \ \ }\frac{1+x}{1-x} & = & e^{\frac{2\pi i}{2n+1} k}
\\ x & = & \frac{e^{\frac{2\pi i}{2n+1}k} -1}{e^{\frac{2\pi i}{2n+1}k} +
1} \\ & = & \frac{e^{\frac{\pi ik}{2n+1}} -e^{-\frac{\pi i k
}{2n+1}}}{e^{\frac{\pi ik}{2n+1}} +e^{-\frac{\pi i k }{2n+1}}} \\
& =& i \tan \frac{\pi k}{2n+1} \hspace{.2in} k = -n,\cdots ,n\\ &
= & \pm i \tan \frac{\pi k}{2n+1} \hspace{.2in} k = 0,\cdots ,n
\end{eqnarray*}

Putting $n=2$.  The equation reduces to $x(x^4+10x^2 +5) =0.$

So the product of the non-zero roots is 5.

i.e.
$\ds\tan^2\left(\frac{\pi}{5}\right)\tan^2\left(\frac{2\pi}{5}\right)=5.$



\item[(b)] $z = x+iy$

$w = u+iv$

Therefore as $w = z^2 +2z, \, u +iv = x^2 -y^2 +2ixy +2(x+iy)$

So $v = 2xy +2x$

Thus $v=2$ if and only if $2xy +2x = 2$ and $y(x+1) = 1$

This is a rectangular hyperbola.

\begin{center}
\epsfig{file=g80-2-1.eps, width=60mm}
\end{center}

\end{description}
\end{document}