<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}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

Solve the equation
\begin{description}
\item[(a)] $\ds(1-z)^5 = z^5$
\item[(b)] $\ds (z+1)^6 = 56(z-1)^6$
\item[(c)] $\ds (5+z)^5 - (5-z)^5 = 0$
\item[(d)] $\ds (z - \sqrt 3 +2i)^6 + 64=0$
\item[(e)] $\ds z^6 + z^4 + z^2 +1 = 0$
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] z = 0 is not a solution.

$\ds \left(\frac{1-z}{z}\right)^5 = 1 \hspace{.2in} z =
\frac{1}{1+w} \hspace{.2in} w = e^{\frac{2k\pi i}{5}} \, \, \, h =
0, 1, ..., 4$

${}$

\item[(b)] z = 1 is not a solution

$\ds \left( \frac{z+1}{2(z-1)}\right)^6=1 \hspace{.2in} z =
\frac{2w+1}{2w-1} \hspace{.2in} w = e^{2k\pi i}{6} \hspace{.2in} k
= 0, 1, .., 5$

${}$

\item[(c)] z = 5 is not a solution

$\ds \left( \frac{5+z}{5-z}\right) \hspace{.2in} z =
\frac{5(w-1)}{w+1} \hspace{.2in} w = e^{\frac{2k\pi i}{5}} \, \,
\, k = 0, 1 .., 5$

${}$

\item[(d)]  $\ds \left( \frac{z - \sqrt 3 + 2i}{2i}\right)^6
\hspace{.2in} z = 2iw + \sqrt 3 - 2i \hspace{.2in} w =
e^{\frac{2k\pi i}{6}} \, \, \, k = 0, 1, .., 5$

${}$

\item[(e)] $\ds z^6 + z^4 + z^2 + 1 = 0$ iff $\ds
\frac{z^8-1}{z^2-1} = 0 \, \, \, z^2 \not= 1$ iff $\ds z =
e^{\frac{2k\pi i}{8}} \, \, \, \, k =1, 2, 3, 5, 6, 7$
\end{description}


\end{document}
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