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{\bf Question}

\begin{description}
\item[(a)] Show that all the roots of the equation $$(1+z)^{2n+1} =
(1-z)^{2n+1}$$ are given by $$\pm i \tan\left( \frac{k\pi}{2n+1}
\right) \hspace{.2in} k = 0,1,2,\cdots , n$$
\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.
\item[(c)] if $w = az+b$, where $a = 3(1-i)$ and $b = 2+3i$, then
describe what happens to any figure in the $z$-plane under this
transformation.
\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*}


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

$w = u+iv$

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

S $v = 2xy +2x$

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

This is a rectangular hyperbola.

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\item[(c)] \begin{eqnarray*} w & = & 3(1-i)z +(2 +3i) \\ & = & 3
\sqrt 2 e^{-\frac{i\pi}{4}}z + (2+3i) \end{eqnarray*} So any
figure is rotated clockwise through $45^\circ$, magnified by
$3\sqrt3$ and the translated by $(2 +3i)$

\end{description}
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