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{\bf Question}

\begin{description}
\item[(a)] Left $\ds x=e^{i\theta} \, \, (\theta \not= 0).$  Prove
that $$\frac{z+1}{z-1} = -i\cot \frac{1}{2} \theta.$$ Find all the
values of $z$ satisfying $z^7 = -1$, and hence solve the equation
$(w+1)^7 = -(w-1)^7$.  Deduce the roots of $w^6 + 21x^4 + 35w^2+7
= 0$
\item[(b)] Sketch the region of the complex plane satisfying the
two conditions $1 < |z| <2$ , $0 \leq \arg \, z \leq
\frac{\pi}{4}$
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] $\ds \frac{e^{i\theta}+1}{e^{i\theta}-1} =
\frac{e^{\frac{1}{2}i\theta}+e^{-\frac{1}{2}i\theta}}{e^{\frac{1}{2}i\theta}
- e^{-\frac{1}{2}i\theta}} = \frac{\cos \frac{1}{2} \theta}{i\sin
\frac{1}{2}\theta} = - i \cot \frac{1}{2}\theta$

${}$

Solutions of $z^7 = -1$ are $$z = \exp\left(\frac{2k+1}{7} \pi
i\right) \hspace{.2in} k = 0, 1, \cdots, 6$$

So $\ds \frac{w+1}{w-1} = z$ $$w = -i\cot \left( \frac{k +
\frac{1}{2}}{7} \pi i\right) \hspace{.2in} k = 0, 1, \cdots , 6$$

${}$ $\ds (w+1)^7 + (w-1)^7 = 0$ reduces to

$\ds 2w(w^6 + 21w^4 + 35w^2 +7) = 0$

So w = 0 or $\ds x = -i\cot \frac{k + \frac{1}{2}}{7} \pi i
\hspace{.2in} k = 0, 1, 2, 4, 5, 6$
\item[(b)]

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