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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?

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    $\begingroup$ Looks OK to me. I'd add that a circle gets mapped to a line if and only if the circle passes through $0$, and a line gets mapped to a circle if and only if the line does not pass through $0$. $\endgroup$ Commented Dec 13, 2014 at 5:03
  • $\begingroup$ Actually it's good to know that $z \mapsto \frac{1}{\overline{z}}$ is inversion, and inversion preserves (general) circles. This can be shown by elementary geometry and sometimes it's helpful to understand the geometric behaviour of the map. $\endgroup$
    – user21820
    Commented Dec 13, 2014 at 5:17
  • $\begingroup$ @MichaelHardy do you mean the circle gets mapped a line iff the circle is centered at $0$ ? $\endgroup$
    – user195835
    Commented Dec 13, 2014 at 5:22
  • $\begingroup$ No, circles that go through 0 become lines not passing through 0 and vice versa. Circles centred at 0 will remain centred at 0 but with possibly different radii. Lines through 0 will remain the same under inversion, but become reflected under the map you are considering. As I said, all these will be obvious if you understand inversion. See en.wikipedia.org/wiki/Inversive_geometry for the definition of inversion. $\endgroup$
    – user21820
    Commented Dec 13, 2014 at 5:28
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    $\begingroup$ No: circles centered at $0$ get mapped to other circles centered at $0$, or, in case the radius is $1$, the same circle centered at $0$. It is when $0$ is actually a point on the circle that the circle is mapped to a straight line. The reason is that the mapping $z\mapsto1/z$ takes $0$ to $\infty$ and the only "circles" that pass through $\infty$ are straight lines. ${}\qquad{}$ $\endgroup$ Commented Dec 13, 2014 at 5:57

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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.

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