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{\bf Question}

What sort of geometric shapes are the objects $|z-9i| \leq 15,\
|z+16i| \leq 20$? Where do the boundaries of these regions meet
the real axis? Sketch the regions. Show that the M\"obius map
$w=\left(\ds\frac{z-12}{z+12}\right)$ transforms the boundaries
into two straight lines, and find the equations of these lines.



\medskip

{\bf Answer}

$\left.\begin{array} {rcl} |z-9i| &amp; \leq &amp; 15\\ |z+16i| &amp; \leq &amp;
20 \end{array} \right\}$ are filled circles

(1) $|z-9i|&lt;15 \Rightarrow$ circle centre $9i$, radius $\leq 15$

(2) $|z+16i|\leq 20 \Rightarrow$ circle centre $-16i$, radius
$\leq 20$

(z)

PICTURE \vspace{2in}

(1) intersects $x$-axis where

\begin{eqnarray*} |x-9i| &amp; = &amp; 15\\ \Rightarrow x^2+81 &amp; = &amp; 225\\
\Rightarrow x^2 &amp; = &amp; 144\\ x &amp; = &amp; \pm 12! \end{eqnarray*}

(2) intersects $x$-axis where

\begin{eqnarray*} |x+16i| &amp; = &amp; 20\\ \Rightarrow x^2+256 &amp; = &amp; 400\\
\Rightarrow x^2 &amp; = &amp; 144\\ x &amp; = &amp; \pm 12! \end{eqnarray*}

\newpage
If we want to map circles $\rightarrow$ lines we need something to
go to $\infty$ in the M\"{o}bius mapping.

$w=\ds\frac{\alpha z+\beta}{\gamma z+\delta}$, e.g., we could map
$z=-12$ to $\infty$. Then any circle through $z=-12$ maps to a
line, so take $w=\ds\frac{z-12}{z+12}$ say.

What lines?

$w(z+12)=(z-12) \Rightarrow z=-12\ds\frac{(w+1)}{(w-1)}$

so

\begin{eqnarray*} |z-9i|=15 &amp; \rightarrow &amp;
\left|-12\ds\frac{(w+1}{(w-1)}-9i \right|=15\\ &amp; \Rightarrow &amp;
\left|w+\ds\frac{(12-9i)}{(12+9i)}\right|=|w-1|\\ |z+16i|=20 &amp;
\rightarrow &amp; \left|-12\ds\frac{(w+1}{(w-1)}+16i \right|=20\\ &amp;
\Rightarrow &amp;
\left|w+\ds\frac{(12+16i)}{(12-16i)}\right|=|w-1|\end{eqnarray*}

UGH!

Note that $\left(\ds\frac{12-9i}{12+9i}\right)$ and
$\left(\ds\frac{12+16i}{12-16i}\right)$ lie on the circle $|w|=1$.
Thus the lined bisect the chords joining two points on a circle's
circumference. Thus the two lined must pass through the centre of
$w=0$!

PICTURE \vspace{2in}

$A,\ B$ are any two points on circle centre 0. The bisector of
chord $AB$ is what we want, which by definition (equals
$\theta$'s) must pass through 0!

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