\documentclass[a4paper,12pt]{article}
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\begin{document}

{\bf Question}

The region $R$ is the rectangular region in the $z=x+iy$ plane
which is bounded by $x=0,\ y=0,\ x=2,\ y=1$. Determine the region
$R'$ of the $w$ plane into which $R$ is mapped under the following
transformations.

\begin{description}
\item[(i)]
$w=z+(1-2i)$

\item[(ii)]
$w=\sqrt{2}exp\left(\ds\frac{i\pi}{4}\right)z$

\item[(iii)]
$w=\sqrt{2}exp\left(\ds\frac{i\pi}{4}\right)z+(1-2i)$

\end{description}

\medskip

{\bf Answer}

$(z)$

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\un{Notation:}

Let $0ABC \rightarrow 0'A'B'C'$ respectively

\begin{description}
\item[(i)]

$w=z+(1-2i)$ is a simple translation by $1-2i$.

(w)

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\vspace{2in}

\newpage
\noindent{Let $w=u+iv$.}

So $R'$ is a region bounded by

$\left\{\begin{array}{rcl} u & = & 1\\ v & = & -1\\ u & = & 3\\ v
& = & -2 \end{array} \right\}$


\item[(ii)]

$w=\sqrt{"}e^{\frac{i\pi}{4}}z$ is a rotation of $R$ by
$+\ds\frac{\pi}{4}$, followed by a stretching of $\sqrt{2}$:

{w}

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$w=u+iv=\sqrt{2}e^{\frac{i\pi}{4}}(x+iy)=(1+i)(x+iy)$

$\Rightarrow \left\{\begin{array}{rcl} u & = & x-y\\ v & = & x+y
\end{array} \right.$

Hence

$x=0 \rightarrow \left\{\begin{array}{rcl} u & = & -y\\ v & = & y
\end{array}\right\} \Rightarrow u=-v$

$y=0 \rightarrow \left\{\begin{array}{rcl} u & = & x\\ v & = & x
\end{array}\right\} \Rightarrow u=v$

$x=2 \rightarrow \left\{\begin{array}{rcl} u & = & 2-y\\ v & = &
2+y \end{array}\right\} \Rightarrow u+v=2$

$y=1 \rightarrow \undb{\left\{\begin{array}{rcl} u & = & x-1\\ v &
= & x+1 \end{array}\right\}} \Rightarrow \undb{v-u=2}$

\hspace{.6in} These parametrise \ \ \ \ \ eliminate parameters

\hspace{.6in}curves in the $w$ plane

In general $w=\alpha z$ accomplishes a rotation and stretching of
a region.

\newpage
\item[(iii)]
$w=\sqrt{2}e^{\frac{i\pi}{4}}z+(1-2i)$ is a rotation of $R$ by
$+\ds\frac{\pi}{4}$, followed by a stretching of $\sqrt{2}$,
followed by a translation of $1-2i$.

Thus

$u+iv=(1+i)(x+iy)+1-2i$

$\Rightarrow \left\{\begin{array}{rcl} u & = & x-y+1\\ v & = &
x+y-2 \end{array} \right.$

Therefore

$x=0 \rightarrow \left\{\begin{array}{rcl} u & = & -y+1\\ v & = &
y-2 \end{array}\right\} \Rightarrow u+v=-1$

$y=0 \rightarrow \left\{\begin{array}{rcl} u & = & x+1\\ v & = &
x-2 \end{array}\right\} \Rightarrow u-v=3$

$x=2 \rightarrow \left\{\begin{array}{rcl} u & = & 3-y\\ v & = & y
\end{array}\right\} \Rightarrow u+v=3$

$y=1 \rightarrow \left\{\begin{array}{rcl} u & = & x\\ v & = & x-1
\end{array}\right\} \Rightarrow u-v=1$

(w)

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\end{description}
\vspace{2in}

\end{document}