Find this Mobius Transform which fixes the origin and takes inside the circle Find a Mobius Transform which fixes the origin and takes the inside of the circle C1 = ${z : |z-1| = 1}$ to the outside of the circle C2 = ${z: |z+1| = 1}$ in such a way as the points where the lines L+- = ${z :Im(z) = +-1}$ intersects C1 are mapped to the points where L+- intersect C2. What is the image of the line ${z : Re(z) = 1/2}$ under this transformation?
I have no idea how to start this question, I'm very familiar with simpler Mobius Transform questions not involving circles. Could someone please point me in the right direction or help solve it? Revising for finals that are in January.
 A: Any Mobius transformation decomposes into a composition of translations, spiral transformations and inversion, where it is enough to use inversion in the unit circle given by$z \mapsto \frac{1}{z}$. It is trivial to see that this basic inversion maps the unit circle to itself and swaps its inside and outside.
Now we can easily find the desired transformation as a composition of steps. Firstly translate C1 to the origin and then apply the basic inversion and then translate the result to C2. Note that the two points on C1 map to the desired place on C2 already. (And I don't know why your question is so complicated since it's just requiring $\{1+i,1-i\} \mapsto \{-1+i,-1-i\}$.) But the origin is not preserved. Note that we can easily fix that by inserting a rotation on the intermediate unit circle $180^\circ$, since it simply swaps the two points we are interested in and makes the origin go where it is wanted.
Then we can easily find out where the line $\{ z : Re(z) = \frac{1}{2} \}$ goes.
A natural question is, how do we know that this is the only way? This is by the theorem that given any $a,b,c,p,q,r$ in the extended complex plane, there is a unique Mobius transformation mapping $(a,b,c) \mapsto (p,q,r)$ in that order. This theorem is not that hard to prove, and is very useful. Here for example we want $0 \mapsto 0$ and some pair of points to some pair, so we know that there are exactly $2$ transformations, both of which we have already found.
