\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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\begin{document}

{\bf Question}

Show that the line $Re(z)+Im(z)=1$ can be written as
$|z|=|z-1-i|$. Hence, by building up the mapping in a series of
steps, find a transformation which takes this line to the unit
circle $|w|=1$.


\medskip

{\bf Answer}

Plot the region boundary:

$z=x+iy$

$Re(z)+(z)<1 \Rightarrow x+y<1$

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\put(2,-2){\vector(0,1){4}}

\put(3,0){\circle*{.125}}

\put(2,1){\circle*{.125}}

\put(5,-2){\makebox(0,0)[bl]{$x+y=1$}}

\put(2,2){\makebox(0,0)[bl]{$y$}}

\put(5,0){\makebox(0,0)[l]{$x$}}

\put(1,2){\line(1,-1){4}}

\end{picture}
\vspace{1.5in}

If we can write it as $|z|=|z-(1+i)|$, then the distance of any
$z$ (on the line $x+y=1$) from $(1+i)$ and 0 is the same.  Thus we
have to show that $x+y=1$ bisects the line from 0 to $(1+i)$.
Simple geometry comes into play:

Let $OPQ$ lie on $y=x$ with $|OP|=|PQ| \longrightarrow P$ is then
$(x,y)=(\frac{1}{2},\frac{1}{2})$

$\overline{OP}=(\frac{1}{2},\frac{1}{2})$

$\Rightarrow \overline{PQ}=(1,1)$

Thus $Q=\un{1+i}$

PICTURE \vspace{2in}

Hence

$|z-0|=|z-(1+i)|$

$\Rightarrow |z|=|z-(1+i)|$ as required.

Now to map line $\longrightarrow$ a circle.

\begin{description}
\item[(i)]
Try an inversion $w_1=\ds\frac{1}{z}$. Why?

$|z|=\infty \stackrel{w_1}{\longrightarrow} w_1=0$ finite

all other $z$ on line $x+y=1$ are finite and $\ne 0
\stackrel{w_1}{\longrightarrow} w_1=0$ finite

We know that inversions ${1}{z}$ map lines/circles to
lines/circles, but all points in $w$, which are images of $x+y=1$
points are finite. Hence image of $x+y=1$ \un{must} be a circle.

$\begin{array} {rcrcl} w_1 = \ds\frac{1}{z} & \Rightarrow &
\left|\ds\frac{1}{w_1}\right| & = &
\left|\ds\frac{1}{w_1}-(1+i)\right|\\ & \Rightarrow & 1 & = &
|1-(1+i)w_1|\\ & \rm{or} & 1 & = & |w_1(1+i)-1|\\ & \rm{or} &
\ds\frac{1}{|1+i|} & = & \left|w_1-\ds\frac{1}{1+i}\right|\\ &
\Rightarrow & \ds\frac{1}{\sqrt{2}} & = &
\left|w-1-\ds\frac{1}{1+i}\right| \end{array}$

i.e., distance from $1+i$ in $w_1$ plane is always
$\ds\frac{1}{\sqrt{2}} \Rightarrow$ a circle centre
$w_1=\ds\frac{1}{1+i}$, radius $\ds\frac{1}{\sqrt{2}}$.

(z) \setlength{\unitlength}{.5in}
\begin{picture}(4,4)
\put(0,0){\vector(1,0){4}}

\put(1,-2){\vector(0,1){4}}

\put(2,0){\circle*{.125}}

\put(1,1){\circle*{.125}}

\put(4,-2){\makebox(0,0)[bl]{$x+y=1$}}

\put(1,2){\makebox(0,0)[bl]{$y$}}

\put(4,0){\makebox(0,0)[l]{$x$}}

\put(0,2){\line(1,-1){4}}

\end{picture} \hspace{.25in} $\longrightarrow$ \hspace{.25in}
PICTURE \vspace{2in}

\item[(ii)]
Shift circle to origin $w_2=w_1-\ds\frac{1}{1+i}$

PICTURE \vspace{2in}

\item[(iii)]
Scale radius by $\sqrt{2}\ \ w_3=\sqrt{2}w_2$.

PICTURE \vspace{2in}

Set $w_3=w$ and assemble (i) $\rightarrow$ (iii).

$w=\sqrt{2}w_2=\sqrt{2}\left(w_1-\ds\frac{1}{1+i}\right)
=\sqrt{2}\left(\ds\frac{1}{z}-\ds\frac{1}{1+i}\right)$

$\Rightarrow w=\sqrt{2}\ds\frac{((1+i)-z)}{(1+i)z}$

Note that $z=0 \longrightarrow w=\infty$.
\end{description}

\end{document}