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\newcommand{\ds}{\displaystyle}
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{\bf Question}

Given $a\in {\bf C} -\{ 0\}$, set $g(z) = az$.  Prove that $g$
takes circles in $\overline{\bf C}$ to circles in $\overline{\bf
C}$.
\medskip

{\bf Answer}

Let $\alpha z \bar z+\beta z + \bar \beta \bar z + \gamma=0$ be
the equation of a circle in $\bf C$, call the circle $A$.
$(\alpha,\gamma \in \bf{R},\ \beta \in \bf{C}).$

$g(z)=az=\omega$ so $z=\ds\frac{1}{a}\omega$: plug into the
equation for $A$:

$$\alpha\left(\ds\frac{1}{a}\omega\right) \left(\ds\frac{1}{\bar
a}\bar\omega\right)+\beta\ds\frac{1}{a}\omega+\bar\beta\ds\frac{1}{\bar
a}\bar\omega+\gamma=0$$

\un{$\ds\frac{\alpha}{|a|^2}\omega\bar\omega+\ds\frac{\beta}{a}\omega+
\bar{\left(\ds\frac{\beta}{a}\right)}\bar\omega+\gamma=0$}.


(note that since $\alpha =0$ if and only if
$\ds\frac{\alpha}{|a|^2}=0$ $g$ takes circles in $\bf{C}$ to
circles in $\bf{C}$ and lines in $\bf{C}$ to lines in $\bf{C}$).

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