General form of a Möbius transformation sending two points to two points and a circle to another.

Question

Suppose I am given a circle $C$ in $\Bbb C^*$ and two points $w_1,w_2$. Given another circle $C'$ and points $z_1,z_2$, what is the procedure to find a Möbius transformation that sends $C\to C'$, $w_i\to z_i,i=1,2$? Here $z_1\in C\not\ni z_2$; $w_1\in C'\not\ni w_2$. For example, take $|z|=2$, $w_1=-2,w_2=0$. Then, the transformation $T(z)=-\dfrac{z+2}{2}$ sends $|z|=2$ to $|z+1|=1$, $-2$ to $0$, and $0$ to $-1$. Hence, I need to find a transformation that fixes $|z+1|=1$ and $0$, and sends $-1$ to $i$. I know that if $|\alpha|\neq 1$, the transformation $$T(z)=\frac{z-\alpha}{1-z\bar \alpha}$$ fixes $|z|=1$, sends $\alpha$ to $0$ and has fixed points $\sqrt{\dfrac{\alpha}{\bar\alpha}}$. I obtained $T$ using $4$ successive transformations $T_1=-(z+2)$, $T_2=\dfrac{z}{z+2}$, $T_3=-z$ and $T_4=-\dfrac{z}{z+1}$, which seems a bit ineffective. How can I generally find $T$ given $(C,C',(z_1,z_2),(w_1,w_2))$?

Answer 1

First, take a Möbius transform $\phi$ that takes $C$ to $\mathbb R$ and $z_1$ to $i$. This $\phi$ always exists uniquely, provided $z_1\notin C$; it's not too hard to map a circle to the real axis, and the transforms fixing the real axis are easy to classify. Let $\psi$ be the corresponding map for $C'$ and $w_1$.

Now $T=\psi^{-1}\circ\phi$ maps $C$ to $C'$ and $z_1$ to $w_1$. If there was another Möbius transform $g$ with this property, then $\psi\circ g\circ\phi^{-1}$ is a Möbius transform fixing the real axis and $i$, so $\psi\circ g\circ\phi^{-1}$ is the identity and thus $g=T$.

That is, the data $(C,C',z_1,w_1)$ uniquely determines the mapping $T$ (assuming $z_1\notin C$ and $w_1\notin C'$). Therefore your problem is overdetermined, and it has a solution if and only if $\phi(z_2)=\psi(w_2)$. An easy way to see that some conditions are needed is to see that if $z_1$ is the reflection of $z_2$ across $C$, the same has to hold for $w_1$, $w_2$ and $C'$, or that if the points $z_i$ are on the same side of $C$, the points $w_i$ can't be on different sides of $C'$.

The above method gives a way to construct $T$ in terms of two Möbius transforms. Is this method sufficient for you?

Answer 2

Each instance of your problem involves a lot of data, whereby you haven't even indicated in which format the two circles are given. It follows that there is no simple "one formula suits it all" solution. Given an instance of the problem one can reduce it to the special case $z_1=w_1=0$, $z_2=w_2=\infty$. The Moebius transform you are asking for is then a euclidean similarity $$z\mapsto w=a z,\qquad a\in{\mathbb C}^*\ .$$ It exists only if the two circles are in "similar positions" with respect to the origin.