@Kari $|z - a| = r$ is the right way to go about it, yes. If $z' = 1/z$, then $r = |z - a| = |1/z' - a| = |1 - az'|/|az'| = |1/a - z'|/|z'|$. Aka, distance between $z'$ and $0$ is a constant multiple of the distance between $1/a$ and $z'$. Can you tell what those points are?
Note also that $a \neq 0$ here is required for that little manipulation to work, but after you have figured this one out, you should think a bit about why it's not really scandalous. Plugging in $1/a = \infty$ for $a = 0$, naively, you can interpret the result.
About the last thing, a complex number at infinity can be seen as extending a circle's radius to infinity so $z$ would follow the path of a circle to be equidistant from this point at infinity.
@Kari Be careful! $|z'-1/a| = r/|a| \cdot |z'|$. Nobody said $r = |a|$, in which case they would indeed by equidistant from both $1/a$ and $0$ hence $z'$ would trace out a line.
What if $r/|a| \neq 1$? Would they still trace out a line? Check a few examples, maybe.
@Kari Yup, if $a = 0$, $|z| = r$ maps to $|z'| = 1/a$. A small circle centered $0$ maps to a large circle centered at $0$ aka "a small circle in a neighborhood of $\infty$".
That seems strange in reality. I remember trying to fold a paper into a sphere but there were always 4 holes between the quadrants. I'm guessing the infinity stuff fills them in?
Sort of kind of. You're pinching the boundary circle of the disk.
More to the point, under such an identification, $1/z$ becomes a function from the Riemann sphere to itself. It turns the Riemann sphere "upside down", in a sense, sending infinity to 1/infty = 0 and 0 to 1/0 = infinity.
You still need to figure out what happens to the rest of the circles :) Up till now, you proved that under the image of the inversion map, (1) circles centered at $0$ goes to circles centered at $0$ and (2) circles centered at $a$ with $r = |a|$, that is to say, circles passing through the origin, map to straightlines.
What about circles not passing through the origin? That is to say, $r \neq |a|$.
I don't have a general result on where they go and what their radii are but I feel that would just come from going through the algebra with a general $a = b+ci$.
I've read that for any set $S$ of odd natural numbers there exist sequences $a_n$ so that $\sum\limits_{n=1}^\infty a_n^s$ converges iff $s\in S$, has anyone read something like that and has a reference?
@Kari That's not quite pictorially obvious, no. Note that the point maximum distance from $0$ in the circle is precisely the point on the line with minimum distance from $0$. If you go back and forth, you'll see there are points on the image of the circle by the inversion map which has same distance from $0$. That's more or less line-like.
@Kari Hyperbola has two disjoint components though. A circle can never map to a hyperbola, because it's connected.
I guess I don't have an easy way to "see" that the image is distinct from parabola. Whatever I said thus far works equally well for a parabola.
Oh well, looks like writing formulas win over visualization this time :)
But modulo details that you can figure out, you have proved that it maps circles to circles when considered as a map from the Riemann sphere to itself. That's nice, isn't it?
@Kari It's sort of like a rotation, more than reflection (indeed, there's no right notion of "reflection" for complex plane or Riemann sphere which is moreover holomorphic).
Namely, take $1$ and $-1$ and the corresponding points in the Riemann sphere (the "east" and "west" pole). Take an axis joining them and half-rotate around that axis.
Nothing :) When you pass to the Riemann sphere, image of that just becomes a circle crossing the point at infinity (north pole) in the Riemann sphere. You have adjoined $\infty$: there's no problem at all. It's a circle.
On $\Bbb C$ it's a line because $\infty$ isn't there: we cannot see it "crossing $\infty$" like any honest to god circle would do passing from any other point in $\Bbb C$. So we see it as a line.
