Map conformally $D(-2,2)\setminus \overline {D (-1,\frac12)}$ to the annulus $\{1 < |z| < 2\}$ This is coming from this question: 
https://math.stackexchange.com/questions/1332123/moving-around-a-circle-inside-a-different-circle-conformally
I will delete that question. This will be the same question. The reason is that it's not a hard question, but I can't get an answer, and that question has been edited and re-edited to the point where it's waaay too long. 
So the question:
Map conformally $D(-2,2)\setminus \overline {D (-1,\frac12)}$ to the annulus $\{1 < |z| < 2\}$ or explain to me why it can't be done. $D(a,b)$ is the disk with radius $b$ and center $a$, and $\overline D$ is its closure.
I made a picture in MS Paint! :)

The point is learning to move a circle within a different circle conformally, without messing up the outer one, and I'm looking for a methodology for this, not an ad-hoc solution to this particular problem.
What's a good strategy?
 A: It is almost always a good idea when dealing with disks or circles to move the centre to the origin, since that usually simplifies the notation. In this case, the centre of the larger disk. If, like I, you have difficulties to scale things you know from the unit disk to disks of other radii, then it is also a good idea to normalise the radius to $1$. So let's apply $f_1 \colon z \mapsto \frac{1}{2}(z+2)$ to the original configuration. That is a conformal map, and we now have to see whether we can conformally map
$$\Omega = D_1(0) \setminus \overline{D\bigl(\tfrac{1}{2}, \tfrac{1}{4}\bigr)}$$
to the annulus $\{ z : 1 < \lvert z\rvert < 2\}$.
To do that, or to see that it is impossible, it would be helpful to move the smaller disk so that its centre is the origin too. We can do that with an automorphism of the unit disk.
All automorphisms of the unit disk are Möbius transformations and hence map circles (straight lines are considered circles passing through $\infty$) to circles. As conformal maps, they preserve angles. So a diameter of $D\bigl(\frac{1}{2},\frac{1}{4}\bigr)$ will be mapped to a diameter of the image disk, or a circle intersecting the image circle at right angles. It will be a diameter of the image, if the diameter of the original passes through the hyperbolic centre of the small disk(1). We aim for that. By symmetry, the hyperbolic centre lies on the diameter of the unit disk passing through the Euclidean centre of the small disk. Here, that is the (intersection of the unit disk with the) real line.
The general form of automorphisms of the unit disk is
$$z \mapsto e^{i\varphi} \frac{z - w}{1 - \overline{w}z},$$
where $\varphi \in \mathbb{R}$ (we can normalise to $0 \leqslant \varphi < 2\pi$ if we wish) and $w \in D_1(0)$. If $\varphi$ is an integer multiple of $2\pi$, the automorphism maps the diameter of the unit disk passing through $w$ to itself. So we try to move the small disk so that its centre is $0$ with a map
$$T\colon z \mapsto \frac{z - r}{1 - rz}$$
where $r$ is real. Since $r$ is mapped to $0$, it must be the hyperbolic centre of the small disk, and hence $\frac{1}{4} < r < \frac{3}{4}$.
If the centre of the image is $0$, the two points where the diameter intersects the boundary circle of the small disk will be negatives of each other, so we look for $-T\bigl(\frac{1}{4}\bigr) = T\bigl(\frac{3}{4}\bigr)$:
\begin{align}
&& -T\bigl(\tfrac{1}{4}\bigr) &= T\bigl(\tfrac{3}{4}\bigr)\\
\iff && \frac{r-\frac{1}{4}}{1-\frac{r}{4}} &= \frac{\frac{3}{4}-r}{1-\frac{3r}{4}}\\
\iff && \frac{4r-1}{4-r} &= \frac{3 - 4r}{4 - 3r}\\
\iff && (4r-1)(4-3r) &= (3-4r)(4-r)\\
\iff && -12r^2 + 19r - 4 &= 4r^2 - 19r + 12\\
\iff && 16r^2 - 38r + 16 &= 0\\
\iff && (16r - 19)^2 &= 19^2 - 16^2\\
\iff && r &= \frac{19 - \sqrt{105}}{16}. \tag{$r < 1$}
\end{align}
So with that $r$, our automorphism maps the small disk to a disk with centre $0$ and radius
$$\frac{3-4r}{4-3r} = \frac{48 - 4(19-\sqrt{105})}{64 - 3(19-\sqrt{105})} = \frac{4\sqrt{105} - 28}{7 + 3\sqrt{105}}.$$
This radius is different from $\frac{1}{2}$ (it is irrational, but that's not the point here). That shows that it is impossible to map the domain $\Omega$ conformally to the annulus $\{ z : 1 < \lvert z\rvert < 2\}$, since the ratio of outer and inner radius is a conformal invariant of annuli:
Two annuli $\{ z : \rho_1 < \lvert z - a\rvert < \rho_2\}$ and $\{ z : r_1 < \lvert z-b\rvert < r_2\}$ are conformally equivalent if and only if $\dfrac{\rho_2}{\rho_1} = \dfrac{r_2}{r_1}$.
It is clear that two annuli with the same ratio between outer and inner radius are conformally equivalent - a conformal map can be written down immediately, a translation  (moving the centre of the first annulus to the centre of the second) followed by a dilation to scale the first annulus to the dimension of the second [or we can first scale, then translate]. It is not as easily seen that the equality of the ratios is necessary. If $f\colon A_1 \to A_2$ is a biholomorphic map between two annuli with centre $0$, by (repeated) reflection in the boundary circles, it extends to a biholomorphism of $\mathbb{C}\setminus \{0\}$ with itself. Hence it must have the form $z \mapsto \alpha\cdot z$ or $z \mapsto \dfrac{\alpha}{z}$ for some $\alpha \in \mathbb{C}\setminus \{0\}$. That such maps preserve the ratio of outer and inner radius of annuli is readily verified.
Corollary: Every biholomorphic map between annuli is a Möbius transformation.

(1) Euclidean circles and disks in the unit disk are also circles resp. disks with respect to the hyperbolic metric on the unit disk, but the hyperbolic and Euclidean centre coincide only when the centre is $0$.
