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{\bf Question}

Show that the points $z=1,\hspace{0.1in} z=-\frac{1}{2}$ of the
$z$-plane are inverse for the circle $C_1$ with centre $-1$ and
radius 1.

With the circle centre 1 and the radius 1 denoted by $C_2$ find a
Mobius transformation

$$w=\frac{az+b}{cz+d} \hspace{0.2in} (ad\not=bc)$$

which transforms $z=1$ to $w=-1$, $C_1$ to Re$(w)=\frac{1}{2}$ and
$C_2$ to Re$(w)=-\frac{1}{2}$.


\vspace{0.25in}

{\bf Answer}


DIAGRAM


$ABC$ are collinear $AB=\frac{1}{2}, \hspace{0.1in} AC=2
\hspace{0.1in} AO=1$, therefore $AB.AC=AO^2$

So $C$ and $B$ are inverse with respect to $C_1$

Bilinear transformations map circles and inverse points to circles
and inverses or lines and images.

So since $C_2\to L_2$ and $B'$ and $A$ are inverse with respect to
$C_2$

$A\to C \Rightarrow B'\to w=-2$

Also since $C_1\to L_1$ and $C$ and $B$ are inverse with respect
to $C_1$

$C\to A \Rightarrow B\to w=2$

So we have

$\begin{array}{rr} z &amp; w\\ 1 &amp; -1\\ -\frac{1}{2} &amp; 2\\ \frac{1}{2}
&amp; 2\end{array}$

So since $czw+dw-az-b=0$, we have

\begin{eqnarray} -c-d-a-b &amp;=&amp; 0\\ -c+2d+\frac{1}{2}a-b &amp;=&amp; 0\\
-c-2d-\frac{1}{2}a-b &amp;=&amp; 0\end{eqnarray}

$\left.\begin{array}{l} (2)-(1) \rightarrow 3d+\frac{3}{2}a=0\\
(3)-(1) \rightarrow -d+\frac{1}{2}a=0\end{array}\right\}
\Rightarrow a=d=0 \hspace{0.3in}$ so $-c-b=0$

i.e. $\ds w=-\frac{1}{z}$ - inversion in the origin, and
reflections.


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