The title pretty much says it all. I am also looking for a concrete example if possible. I have looked at the proof, but I'm not exactly sure what it means because I am kind of confused on what the theorem actually says. Does it mean that if g(z) is a circle and f(z) is a Möbius transformation then f(g(z)) is a circle as well?
Complex Analysis – What Does It Mean That a Möbius Transformation Maps Circles and Lines to Circles and Lines?
analysiscomplex-analysistransformation
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In the sciences (as opposed to in mathematics) people are often a bit vague about exactly what assumptions they are making about how "well-behaved" things are. The reason for this is that ultimately these theories are made to be put to the test, so why bother worrying about exactly which properties you're assuming when what you care about is functions coming up in real life which are probably going to satisfy all of your assumptions.
This is particularly ubiquitous in physics where it is extremely common to make heuristic assumptions about well-behavior.
Even in mathematics we do this sometimes. When people say something is true for n sufficiently large, they often won't bother writing down exactly how large is sufficiently large as long as it's clear from context how to work it out. Similarly, in an economics paper you could read through the argument and figure out exactly what assumptions they need, but it makes it easier to read to just say "well-behaved."
Yes, the conjecture is true. A rational function which maps circles and lines to circles and lines is necessarily a Möbius transformation.
For the proof we can without loss of generality assume that $f(\infty) = \infty$ and that $f$ has a simple zero at $z=0$. (Otherwise consider $\tilde f = T \circ f \circ S$ with suitably chosen Möbius transformations $T$ and $S$.) Then every line through the origin is mapped to another line through the origin.
Choose two lines $l_1, l_2$ through the origin and let $L_1, L_2$ be the corresponding images. The angle between $L_1$ and $L_2$ is the same as the angle between $l_1$ and $l_2$ (here we use that $f$ has a simple zero, so that it preserves angles locally). Let us denote that angle between the lines with $\alpha$.
If we denote the reflection at the lines $l_1, l_2, L_1, L_2$ with $r_1, r_2, R_1, R_2$ respectively, then the Schwarz reflection principle tells us that $$ f \circ r_j = R_j \circ f \, , \, j=1, 2 $$ and therefore $$ f \circ r_1 \circ r_2 = R_1 \circ R_2 \circ f \, . $$ But the composition of reflections at intersecting lines is a rotation, and it follows that $$ f(e^{i2\alpha} z) = e^{i2\alpha} f(z) \, . $$ This holds for all $z$ and all real $\alpha$. Substituting this in the power series $f(z) = \sum_{n=1}^\infty a_n z^n$ at the origin gives $$ a_n (e^{i2n\alpha} - e^{i2\alpha}) = 0 $$ for all $n \ge 1$. If we choose $\alpha$ such that $e^{i2\alpha}$ is not a root of unity then $a_n = 0$ for $n \ge 2$ follows.
We have therefore shown that $f$ (after some composition with Möbius transformation) is a linear function, and that concludes the proof.
Best Answer
Consider the image of a line in $\mathbb{C}$ projected onto the Riemann sphere. As you follow the line in one direction out towards infinity, its projection onto the sphere tends to the north pole.
As you follow the line in the opposite direction, its projection also tends to the north pole!
If you take any line or circle in $\mathbb{C}$, and transform it using a Möbius transformation, then the resulting set will also be a line or a circle in $\mathbb{C}$.
All we really need to consider is the case $w = \frac{1}{z}$. Immediately, we should expect a problem at $z=0$.
Let $S$ be a circle of radius $r$ centered at $\alpha$. We know that $(z-\alpha)(\overline{z}-\overline{\alpha}) = r^2$ (this is just the equation for a circle), so $$ z\overline{z}-\alpha\overline{z}-\overline{\alpha}{z} = r^2-|\alpha|^2 \\ \frac{1}{w\overline{w}}-\frac{\alpha}{\overline{w}}-\frac{\overline{\alpha}}{w} = r^2-|\alpha|^2 $$
If there is no point in $S$ that goes through the origin, then the resulting image is going to just be a circle!
But if a point in $S$ does go through the origin, then $r = |\alpha|$, so $1-\alpha w - \overline{\alpha}\overline{w} = 0$.
This means that $\textrm{Re}(\alpha w) = \frac12$. Solving for this, we find that $\textrm{Re}(w)\textrm{Re}(\alpha)-\textrm{Im}(w)\textrm{Im}(\alpha) = \frac12$, which we should recognize as the parametric form of a line.
Therefore, $f(z) = \frac{1}{z}$ maps a circle to a circle, unless that circle goes through the origin, in which case it becomes a line.
Then, $F(z) = \frac{az+b}{cz+d}$ is just the composition of $\frac{1}{z}$ on the left and right by linear functions, which are just scalings and translations!