Let $f: \mathbb{C} \setminus \{0 \} \to \mathbb{C} \setminus \{0\} $. We want to show that $f(z) = \frac{1}{z}$ maps circles into circles and lines. My professor gave the following hint: The general equation for lines and circles is
$$ \alpha(x^2 + y^2) + \beta x + \gamma y + \Delta = 0 $$
where the greek letters are obviously constants. So, given this advice, We can rewrite this in the complex plane as follows:
$$ \alpha |z|^2 + \frac{ \beta}{2}( z + \overline{z} ) + \frac{\gamma}{2i}( z – \overline{z} ) + \Delta = 0$$
So, now we apply $w = \frac{1}{z} $ and we obtain (with $|w|^2 = w \overline{w}$):
$$ \frac{ \alpha}{w \overline{w}} + \frac{\beta}{2}\bigg( \frac{1}{w} + \frac{1}{\overline{w}}\bigg) + \frac{\gamma}{2 i}\bigg( \frac{1}{w}- \frac{1}{\overline{w}}\bigg)+ \Delta = 0$$
hence,
$$ \alpha + \frac{ \beta}{2}(\overline{w} + w ) + \frac{\gamma}{2i}(\overline{w}-w) + w \overline{w} \Delta = 0 $$
Next, putting $w = u + iv$ we arrive to:
$$ \alpha + \beta u – \gamma v + (u^2 + v^2 ) \Delta = 0 $$
So, in the case when we have circles in $xy$-plane, that is when $\alpha \neq 0$, we still have circles in the $uv$-plane. So $f$ sends circles to circles if $\alpha \neq 0 $. We also have circle if $\alpha \neq 0 $ and $\Delta = 0$ which in this case $\frac{1}{z} $ sends circles to lines.
Is this a correct solution? IS there a shorter way to prove this?
Best Answer
The argument given here is correct.
An interesting alternative point of view on how to argue for essentially the same proposition is found in C. Stanley Ogilvy's charming book Excursions in Geometry.