Mapping a common part of the disks Map the common part of the disks $|z|<1$ and $|z-1|<1$ on the inside of the unit circle. Choose the mapping sot hat the two symmetries are preserved.
I don't really know how to approach this??
Any suggestions on how to start constructing such a linear transformation??
Thanks in advance!
 A: Hint:  What sort of mapping is allowed?  Later in the question you refer to a linear transformation, but the region you start with has two corners:  $\frac 12+\frac {\sqrt 3}2i$.  These will need to stay on the circumference, but the angle needs to be straightened out.  Where is the natural place to send them?  The other special points on the region to be mapped are $0$ and $1$ because they are on the other axis of symmetry.  Where should they go?  You still have continuum many points to go, but maybe you can deal with them wholesale now.
A: First of all, you should identify the points of intersection, which are: $e^{(-\frac {\pi}3)i}$ and $e^{(\frac {\pi}3)i}$. To simplify the answer, I'll call them $a$ and $b$ respectively.
So, we will start with the following mapping: $z_1 =\frac {z-a}{z-b}$. This will map our original region to a wedge from $2\frac {\pi}3$ to $4\frac {\pi}3$.
Our second map will rotate this new region, so that it will map it into a wedge from $0$ to $2\frac {\pi}3$. 
$z_2 = e^{(-\frac {2\pi}3)i} \times z_1$.
Our third map will expand this wedge so that it will be the upper half plane.
$z_3 = (z_2)^{3/2}$.
Now, we will map this last region into the unit circle. $w =\frac {z_3-i}{z_3+i}$.
So, our global map will be: $w = z_3(z_2(z_1(z)))$. Good luck!
