Question about a paragraph in the book complex analysis by Ahlfors.

Question

By $C_1$, we denote family of circles passing through $a,b$ and by $C_2$ we denote family of Appolonius circles with limit point $a,b$. In section $3.5$ entitled Families of circles, in one paragraph it is written that If a transformation $w=Tz$ carries points $a,b$ into $a',b'$, then it can be written in the form $$\frac{w-a}{w-b}=k\frac{z-a}{z-b}.$$ (To write it in this form i used cross ratios) Then clearly, this map takes the circles $C_1$ and $C_2$ into the family $C_1'$ and $C_2'$ with limit point $a',b'.$ In next paragraph it says that when $a$ and $b$ are fixed points of $T$, then the family $C_1$ and $C_2$ are mapped onto itself (this is also quite obvious). But then it says $c_1=c_1'$ for all $c_1\in{C_1}$ if $k>0$ (HOW?). And $c_2=c_2'$ for all $c_2\in{C_2}$ if $|k|=1$. For $C_2$ it is very much clear from the definition of $T$. Can anyone help me with the part how each circle is fixed when $k>0.$

Answer 1

A general description in terms of complex dynamical systems in the plane is easy to give. We are starting with the linear fractional case rather than the quadratic polynomial that gives rise to more complicated dynamics. Here is a general discussion of a hyperbolic transformation.

One first notices that fixed points a, and b are attractive or repulsive in pairs . Attractive if the absolute value of the derivative is less than 1 at a , repulsive if the absolute value of the derivative is greater than 1 at a. The value at the point b will be the reciprocal of the value at the point a. That will turn out to be k, or $ \frac{1}{k}$ .

The transformation $T$ given by $$\frac{w-a}{w-b}=k\frac{z-a}{z-b}.$$ will take any point $z_1$ in the plane and move it along the Steiner circle connecting a and b , hence $T(z_1)$ lies on the very same Steiner circle, I believe this was your question. In general the following surprising expression for forward iteration for positive n , motion toward the attractive fixed point holds. A similar expression under backward iteration also holds, in this case moving the iterates toward the repulsive fixed point.

$$\frac{w-a}{w-b}=k^n\frac{z-a}{z-b}.$$