\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)] If $w = 2z + z^2$ show that the circle $|z|=1$
corresponds to a cardioid in the $w$-plane.
\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)]
$w = 2z + z^2$

$w+1 = (z+1)^2$

If $z$ lies on the unit circle $z+1$ lies on the circle centre 1
radius 1

\begin{center}
$\begin{array}{cc}
\epsfig{file=g77-2-1.eps, width=45mm} \ \ \ & \ \ r=2 \cos \theta
\end{array}$
\end{center}

So $r^2 = 4\cos^2 \theta$

${}$

Let $w+1 = \rho e^{i\phi} \, z+1 = re^{i\theta}$ the $\rho = r^2$
and $\phi = 2\theta$

So $\rho^2 = 16 \cos^2 \frac{\phi}{2} = 8(1+\cos\phi)$ which is a
cardioid.

\begin{center}
$\begin{array}{cccc}
\epsfig{file=g77-2-2.eps, width=30mm} & \ \longrightarrow \ &
\epsfig{file=g77-2-3.eps, width=30mm} \ \ & \ \
\epsfig{file=g77-2-4.eps, width=30mm} \\
z & \zeta = z+1 & \omega = w+1
\end{array}$

$\begin{array}{c}
\epsfig{file=g77-2-5.eps, width=30mm}\\ w
\end{array}$
\end{center}


\end{description}
\end{document}