# 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 $$C$$.