$\bar{\Bbb C}$ is the extended complex plane.
Let $f\colon \bar{\Bbb C}\to \bar{\Bbb C} $ be a univalent function that preserves circles(if $K$ is a circle in $\bar{\Bbb C}$, then so is $f(K)$). Show that $f$ is either
1) $f(z)=\dfrac{az+b}{cz+d}$
or
2) $f(z)=\dfrac{a\bar{z}+b}{c\bar{z}+d}$
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we don't assume that $f$ is continuous.
Source: An introduction to complex function theory, Bruce P. Palka P469
Hint: Consider $f$ of the form that fixes $0$, $1$, $\infty$. take a look at the images of the real axis , the imaginary axis, the line $x=1$, and the circle $\{z\colon |z-\frac12|=\frac12\}$. How to show that $f$ is either the identity or the reflection in the real axis? $f$ is univalent, so $f$ map parallel lines to parallel lines
This is a good question, please reopen it. Sincere thanks!
Sketch of the proof ($f$ fixes $0,1,\infty$ implies $f(z)=z$ or $f(z)=\bar z$ as we can clearly get there with an extended plane automorphism) with geometric details upon request though they should be easily seen. We use Euclidean notions below, so lines are lines and circles are circles so to speak.
$f$ fixes $\infty$ hence it sends lines to lines and circles to circles and by univalency, it preserves the cardinality of intersections hence, for example, parallel lines go to parallel lines and tangent circles to tangent circles. Also horizontal lines are sent to horizontal lines as $f$ fixes the real axis
So if $f(iy)=L_0, f(1+iy)=L_1, f(\{z\colon |z-\frac12|=\frac12\}= C$, $L_0||L_1$ passing through $0,1$ respectively and $C$ is a circle passing through both $0,1$ and $L_0 \cap C=0$, $L_1 \cap C =1$, hence $L_0, L_1$ are tangent to $C$ and since they are parallel it follows that the center of $C, 0,1$ are collinear hence $C=\{z\colon |z-\frac12|=\frac12\}$ and then clearly $L_0=iy, L_1=1+iy$, so $f$ fixes all those, so in particular $f$ preserves verticality too.
The horizontal tangents to $C$ are then either fixed or switched as they still must be horizontal tangents under $f$. In both cases, the line $x=\frac{1}{2}$ is fixed since the tangency points are either fixed or switched, hence doing intersections, the point $x=\frac{1}{2}, y=0$ is fixed.
Similarly, if a circle is fixed, its horizontal and vertical tangents are either fixed or switch (because they stay horizontal or vertical tangents respectively), but in any case, the horizontal and vertical lines joining the respective tangency points are fixed, hence the center is fixed!
Repeating the argument above with the circles that contain the fixed points $0,\frac{1}{2}$, and $\frac{1}{2}, 1$ respectively and are centered on the real axis, we get that they are fixed hence their centers are fixed, so $\frac{1}{4}, \frac{3}{4}$ are fixed and so are the vertical lines through them, which immediately implies that the circle centered at $\frac{1}{2}$ and of radius $\frac{1}{4}$ is fixed too (by uniqueness with vertical tangent lines at fixed $\frac{1}{4}, \frac{3}{4}$
By induction, all points $\frac{c}{2^n}, 0 \le c \le 2^n, c$ integral are fixed as are all circles centered at them and of radius $\frac{1}{2^{n+1}}$ as long as $c \ne 0, c \ne 2^n$, so in other words as long as they stay inside the strip $x=0, x=1$ and we can use the same uniqueness as above.
Noting that for a fixed circle like above, the segment on the real axis inside the circle is preserved under $f$ (as such are defined by the fact that vertical lines through any point on it intersect the circle in two points!) we get that $f$ sends all segments $[\frac{c}{2^n}, \frac{c+1}{2^n}]$ (inside the strip) to themselves, hence $f$ is the identity there by usual limiting arguments, so all vertical lines inside the strip are fixed.
Coming back to the original circle $C$ and its horizontal tangents, let's remember that they are either fixed or switched and since the latter becomes the former with $z \to \bar z$ which preserves all the other properties of $f$, we can assume they are fixed, hence the points $\frac{1}{2} \pm \frac{1}{2}i$ are fixed alongside $\frac{1}{2}$. But the same argument as above by halving radiuses and taking circles with horizontal tangents at fixed points, shows that that vertical segment $[\frac{1}{2}, \frac{1}{2}+i\frac{1}{2}]$ is pointwise fixed (induction and limiting argument), hence all the horizontal lines passing through them are fixed, hence the full square $[0, \frac{1}{2}]\times [0, \frac{1}{2}]$ is pointwise fixed for example. Since any point on the real axis is an intersection of the real axis with a line passing through two points in the above square, the real axis is pointwise fixed and similarly the imaginary axis is too, hence $f(z)=z$ since then all vertical and horizontal lines are fixed, hence their intersctions are so. Done!