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(remmert theory of complex function)

I am trying to solve this exercise, however it seems impossible because I don't know how to map a circle, and I will be very thankful if somebody points out to me:

Given a Circle $C$, is it possible to show that for $a_{1},b_{1}\in \mathbb{C}\backslash C $ there exists a Möbius transformation $$M(z)= \frac{az+b}{cz+d}$$

which fulfills: $$M(C)=C ; M(a_{1})=b_{1}$$

So we can write conditions: $M(a)=b \Rightarrow aa_{1}+b= b(ca_{1}+d)$

How does one map a circle? I thought that a circle in $\mathbb{C}$ needs 3 points to be uniquely determined , so we can put: $z_{1},z_{2},z_{3} \in C$

and the condition is the same as for the a to b mapping: $$az_{k}+b=z_{k}(cz_{k}+d), k=1,2,3 \in \mathbb{N}$$

It seems that there isn't much more one can do with this approach, so I think it is not the right one. What is the right approach?

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That last equation you've got - where does that come from? –  anon Feb 13 '12 at 11:16
It should write $az_{k}+b=z_{k}(cz_{k}+d)$ instead of $az_{k}+b=b(cz_{k}+d)$ –  VVV Feb 13 '12 at 11:22
VVV: Just because $M$ maps the circle $C$ to itself doesn't mean that it coincides with the identity function on that circle. For example, if you take the unit circle and consider the transformation of rotating it by $\pi/2$ radians (given by $z\mapsto (e^{i\pi/2}z+0)/(0z+1)$), for no $z\in C$ will $M(z)=z$, but it still maps the unit circle to itself. –  anon Feb 13 '12 at 11:24
@anon I don't understand, with only rotation but no lengthening or translation, isn't it impossible to map a point arbitrairy to a another ? –  VVV Feb 13 '12 at 13:17

2 Answers 2

up vote 1 down vote accepted

In your question you've taken the condition "$M(C)=C$ " and misunderstood it to mean that it leaves every single point on the circle unchanged, so that $M(z_{1,2,3})=z_{1,2,3}$ respectively. This interpretation is false. I provided a counterexample in my comment, but you misunderstood the point of the comment: the map $z\mapsto iz$ is a Möbius transformation that maps the unit circle to the unit circle (hence it satisfies the condition $M(C)=C$) however $M(z)=z$ is false for every single particular element $z$ on the unit circle. The point is that your interpretation of the condition does not make sense so you cannot build an approach off of it. I am not saying that pure rotations are the only transformations preserving circles; they are not.

The key to understanding this problem is geometry - more specifically, spherical geometry. We will first prove a particular case of the proposition, and then show this entails the full proposition.

Lemma. For any $u,v\in\mathbb{C}$, there is a Möbius transformation $\varphi$ such that $\varphi(u)=v$ and $\varphi(\mathbb{R})=\mathbb{R}$.

Visual Proof. The isometries of the real line are affine transformations $x\mapsto ax+b$, so we need to see that such a transformation is sufficient. Translations have an obvious interpretation of moving everything horizontally but not vertically, while real dilations will act on complex $z$ as dilations of $z$ on the ray extending from the origin to $z$. The line between the origin and $u$ will take on every possible $y$ (imaginary) value, so we can perform a real dilation on $u$ until it matches $v$ in height; on top of that we just need to translate the appropriate horizontal distance to get to $v$. With these two actions composed together we have our affine transformation, as desired.

Proposition. For any (generalized) circle $C$ and $u,v\in\mathbb{C}-C$, there is a Möbius transformation $\varphi$ such that $\varphi(u)=v$ and $\varphi(C)=C$.

Proof. Fix a Möbius transformation $\psi$ that maps $C\to\mathbb{R}$. By our lemma there is a transformation that preserves the real line but maps $\psi(u)$ to $\psi(v)$; call it $\rho$. Then our desired map is $\psi^{-1}\circ\rho\circ\psi $.

This write-up requires a bit of expansion in order to be as comprehensive as I'd like it to be, and I'll get to it next thing I do on MSE, but right now I have to go to sleep and catch some ZZZ's! (Also, this is basically of an elaboration on Blatter's answer and clarification of my comment.)

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Thank you, anon! –  VVV Feb 14 '12 at 10:22

A hint: You are only asked to show that such an $M$ exists. Therefore you may assume that $C$ is some special circle to begin with. When $C$ is the real axis it is easy to write down a suitable $M$.

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Thanks, Christian Blatter! –  VVV Feb 14 '12 at 10:22

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