In my current homework, there's the following task:
"The area outside of these two circular discs
$K_1=\{z{\in}\mathbb{C}:|z-\frac{5}{2}|\le\frac{3}{2}\}$, and
$K_2=\{z{\in}\mathbb{C}:|z+\frac{5}{2}|\le\frac{3}{2}\}$
is to be mapped onto the interior of a circular ring around zero. Find an appropriate transformation."
The only idea I have so far is to try to find a transformation such that both circles are mapped onto that mentioned circular ring (which is something that I don't really know how to do either), but I don't know if that even leads to the area outside of those two circles being inside that new ring... Either way, I'm thoroughly confused and really do not know how to do this. Any help would be appreciated!
Let $T$ be a Möbius transformation which maps $\Bbb C \setminus (K_1 \cup K_2)$ onto an annulus $r_1 < |w| < r_2$. Without loss of generality assume that for $j=1,2$ the boundary of $K_j$ is mapped onto the circle $C_j$ with radius $r_j$ and center zero.
The points $w_1 = 0$ and $w_2 = \infty$ are “conjugate” or “symmetric” with respect to both $C_1$ and $C_2$. It follows that their pre-images $$ z_1 = T^{-1}(0) \in K_1 \, , \, z_2 = T^{-1}(\infty) \in K_2 $$ are conjugate with respect to both $\partial K_1$ and $\partial K_2$. The general symmetry of the problem suggests to find a solution with $$ z_1 = -a \, , \, z_2 = a $$ for some real $a > 0$. $a$ and $-a$ are conjugate with respect to $\partial K_j$ if the product of their distances to the center of $\partial K_j$ is equal to the square of the radius of $\partial K_j$: $$ \left(\frac 52 -a \right)\left(\frac 52 +a \right) = \left(\frac 32 \right)^2 $$ which has the (positive) solution $a=2$.
Therefore $T(z) = \frac{z+2}{z-2}$ is a candidate, and I leave it to you to verify that this is actually a Möbius transformation with the desired properties.