<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}


{\bf Question}

\begin{description}
\item[(i)] Verify the identity $$z^5 -1 = (z-1)z^2 \left\{ \left( z
+ \frac{1}{z}\right)^2 + \left( z + \frac{1}{z}\right) -1
\right\}$$ and putting $\ds z = \cos \left( \frac{2\pi}{5}\right)
+ i \sin \left(\frac{2\pi}{5}\right)$ and evaluate $\ds\cos \left(
\frac{2\pi}{5}\right)$
\item[(ii)] If $w = \cosh z$ show that the lines $x=x_0, \, y=y_0$
correspond the ellipse and hyperbolas $$ \frac{u^2}{\cosh^2x_0} +
\frac{v^2}{\sinh^2x_0} = 1, \, \frac{u^2}{\cos^2y_0} -
\frac{v^2}{\sin^2y_0} = 1$$ Also show that the whole of the
$w$-plane corresponds to any strip of the $z$-plane of the $\pi$
bounded by lines parallel to the $x$-axis.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(i)] $\ds  (z-1)z^2\times \left\{ \left(z+\frac{1}{z} \right)^2
\left(z + \frac{1}{z}\right) - 1 \right\}$

$\ds = (z-1)z^2\left\{z^2 + 2+\frac{1}{z^2} + z+\frac{1}{z} -
1\right\} $

$= (z-1)(z^4+z^3+z^2+z+1) = z^5-1$

Putting $z = \cos \frac{2\pi}{5} + i\sin\frac{2\pi}{5}
\hspace{.2in} z^5 - 1=0$ and $z + \frac{1}{z} = 2\cos
\frac{2\pi}{5}$

So $4c^2 + 2c -1=0$ and $c &gt;0$ which gives $\ds c = \frac{\sqrt
5-1}{4} = 0.309016994372$

${}$
\item[(ii)] $w = \cosh z$ so $u +iv = \cosh (x+iy) = \cosh x \cos
y + i \sinh x \sin y$

If $\ds x = x_0 \hspace{.2in} \frac{u^2}{\cosh^2x_0} +
\frac{v^2}{\sinh^2 x_0} = \cos^2y + \sin^2 y = 1$

If $\ds y = y_0 \hspace{.2in} \frac{u^2}{\cos^2y_0} +
\frac{v^2}{\sin^2 y_0} = \cosh^2x + \sinh^2 x = 1$

The system of hyperbolas covers the w-plane and $\cos^2y_0$ and
$\sin^2y_0$ are periodic with period $\pi$.  So any strip parallel
to the $x$-axis of width $\pi$ will map onto the whole of the
$w$-plane.




\end{description}
\end{document}
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