Points outside a circle mapped inside a circle by $1/z$ Prove or disprove the following conjecture: Let $C$ be a circle in the complex plane that passes by the points $z_1=-1$ and $z_2=a$, with $a$ real and greater that one (there are obviously infinitely many circles with that property). Then every point outside $C$ gets mapped by $1/z$ to a point inside $C$.
Note that the converse is not true. There are points inside $C$ that get mapped by $1/z$ to points also inside $C$. For example, all points in the real axis between $1$ and $a$ are inside $C$ and are mapped to points in the real axis between $0$ and $1$, which are also inside $C$, as the whole $(-1,a)$ interval in the real axis is inside the circle $C$.
For the record, I believe the conjecture to be true.

Let $C_1$ be the image of $C$ under the Möbius transformation $T(z) = \frac 1z$. Then $C_1$ is a circle, and – since $T$ maps the real axis onto itself – $C$ and $C_1$ intersect the real axis at the same angle at $z=-1$, so that the two circles are tangent to each other at that point.
It follows that $C_1$ is either “inside” $C$ or “outside” $C$. Since $\frac 1a \in C_1$ is in the interior of $C$, only the first option is possible.
Finally, $T$ maps the exterior of $C$ to the interior of $C_1$, which is contained in the interior of $C$.


Alternative solution: Let $c$ be the center of $C$. The condition $a>1$ implies that $\operatorname{Re}(c) > 0$. The radius is $r  = |c+1|$ and
$z$ is in the exterior of $C$ iff
$$
 |z-c|^2 > |c+1|^2 \\
\iff z \overline z - \overline c z - c \overline z - (c + \overline c + 1) > 0
$$
Then $w = \frac 1z$ satisfies
$$
 1 - \overline c \, \overline w - cw - (c + \overline c + 1)w \overline w > 0 \\
\iff w \overline w + \frac{1}{c + \overline c + 1} w + \frac{\overline c}{c + \overline c + 1} \overline w - \frac{1}{c + \overline c + 1} < 0 \\
\iff \left | w + \frac{\overline c}{c + \overline c + 1} \right|^2 < \frac{|c+1|^2}{(c+\overline c + 1)^2}
$$
so that $w$ is inside the circle $C_1$ with center $c_1 = -\frac{\overline c}{c + \overline c + 1}$ and radius $r_1 = \frac{|c+1|}{(c+\overline c + 1)} $ . In particular,
$$
 |w - c| \le |w - c_1| + |c_1 - c| < r_1 +  |c_1 - c| \\
 = \frac{|c+1|}{(c+\overline c + 1)}  + \frac{|c+1|(c+\overline c)}{c+\overline c + 1} = |c+1| = r
$$
so that $w$ is inside the circle $C$.
