Logic behind Möbius transformation So I am attempting to get a grasp on Möbius transformations. Specifically, I'm trying to understand the logic and reasoning behind using them to map certain subsets of the complex plane into others. The following problem is from Lars Ahlfors, "Complex analysis", 1966.(p.83).
Find the Mobius transformation which carries the circle $|z|=2$ into $|z+1|=1$, the point $-2$ into the origin, and the origin into $i$.
Graphically: 
Since i know where the two specific points $z_1=-2$ and $z_2=0$ are being mapped, I can set up the equation: 
$$ \frac{f(z)-f(z_1)}{f(z)-f(z_2)}=K\frac{z-z_1}{z-z_2} $$
Which yields: $\frac{f(z)}{f(z)-i}=K\frac{z+2}{z}$, or equivalently: 
$$[\ast] \quad \quad f(z)=\frac{(-iK)z+(-2iK)}{(1-K)z+(-2K)}$$
Now, If I knew where one more point was mapped, I would be able to determine the coefficient $K$, and thus $f$. This is the point where I do not know how to reason to get past. My thought was "try $f(2)=-2$", since this would preserve the diametrically opposite points. The resulting transformation does however not map the circle to the circle. 
Therefore, to figure out the answer, I put a complex number $K$ in geogebra, defined the function $f$ as in $[\ast]$, put a bunch of points $A,B,C,D,E$ evenly on the circle $|z|=2$, and varied the parameter $K$ until $f(A),f(B),f(C),f(D),f(E)$ all ligned up on the circle $|z+1|=1$. This gave the result $K=-i$, which means that the answer to the problem is $$f(z)=\frac{iz+2i}{2z+2}$$
I do not understand how I would have figured this out by calculation, since I seem to miss some crucial information... I did however figure out that if i choose $f(-2i)=-1+i$ for my third point, I get the correct transformation, but I do not know why as I found this by trial and error....
Thanx, Robin 
 A: You almost have the right idea with your "diametrically opposite" approach, but not quite. Instead, we want to use the idea of symmetry with respect to a circle--see Definition 13 on page 81--to give us a third point.
As mentioned after the development of $(11)$ on page 81, the symmetric point to the center of a circle is $\infty.$ Thus, by the Symmetry Principle (Theorem 15 on page 82), we need our transformation to send $\infty$ to the point symmetric to $i$ in the circle $|z+1|=1.$ Using $(11)$ with $a=-1,$ $z=i,$ and $R=1,$ we find that $$\bigl(i^*-(-1)\bigr)\left(\overline{i}-\overline{-1}\right)=1^2\\(i^*+1)(-i+1)=1\\(-i+1)i^*+1-i=1\\(-i+1)i^*=i\\i^*=\frac{i}{1-i}.$$
Since $$\frac{i}{1-i}=i^*=f(\infty)=\frac{-iK}{1-K}$$ from your previous work, whence $$i(1-K)=-iK(1-i)\\i-iK=-K-iK\\K=-i,$$ we obtain $$f(z)=\frac{-z-2}{(1+i)z+2i}$$ by substitution of $K=-i$ into your $[*]$ formula.

Added: As an alternative to the formulaic approach to finding $i^*,$ we can simply use the graphical approach from Fig. 3-2 on page 82. The tangent lines to $|z+1|=1$ through $i$ are the lines $\Re(z)=0$ and $\Im(z)=1,$ which intersect $|z+1|=1$ at $0$ and $-1+i,$ respectively. The segment from $i$ to $-1$ clearly intersects the segment from $0$ to $-1+i$ at $\frac{-1+i}2,$ and since $2=(1+i)(i-1),$ then $$\frac{-1+i}{2}=\frac{-1+i}{(1+i)(1-i)}=\frac{i(1+i)}{(1+i)(1-i)}=\frac{i}{1-i}=i^*,$$ as desired.
