Construct a Mobius map that maps the strip $\{z\in \mathbb C:0<\Im (z)<1\}$ onto the area between the circles $|z-1|=1$ and $|z-2|=2$. 
Construct a Mobius map that maps the strip $\{z\in \mathbb C:0<\Im
 (z)<1\}$ onto the area between the circles $|z-1|=1$ and $|z-2|=2$.

(The question earlier asked for a map from $|z-1|<1$ onto $|z|>2$ for which is got $g(z)=\frac 2{z-1}$ but I can't really see how to use this)
I can't get anywhere useful, I just have pages of algebra getting me nowhere.
Any help is appreciated, thank you
 A: For me, it's easier to move a pair of parallel lines to a desired destination than doing it for a pair of nested circles touching at one point. Therefore I suggest first constructing the map in the other direction, and then inverting the resulting Möbius transformation to obtain the map in the required direction.
One transforms a pair of nested circles touching at $p$ into a pair of parallel lines by applying a Möbius transformation that maps $p$ to $\infty$. Here, the two circles touch at $0$, and the simplest Möbius transformation mapping $0$ to $\infty$ is $\rho \colon z \mapsto 1/z$. $\rho$ maps $\mathbb{R}\cup \{\infty\}$ to itself, and the two circles intersect the real axis at right angles, so the image of each of the two circles under $\rho$ is a straight line intersecting $\mathbb{R}$ at a right angle, i.e. a line of the form $\{ w : \operatorname{Re} w = c\}$.
The smaller circle intersects $\mathbb{R}$ at $0$ and at $2$, and $\rho(2) = \frac{1}{2}$, so the image of $\{ z : \lvert z-1\rvert = 1\}$ under $\rho$ is $\bigl\{ w : \operatorname{re} w = \frac{1}{2}\bigr\}$. The larger circle intersects $\mathbb{R}$ at $0$ and at $4$, so the image of $\{ z : \lvert z-2\rvert = 2\}$ is $\bigl\{ w : \operatorname{Re} w = \frac{1}{4}\bigr\}$.
The point $3$ lies between the two circles ($\lvert 3-1\rvert > 1$ and $\lvert 3-2\rvert < 2$), and $\frac{1}{4} < \rho(3) = \frac{1}{3} < \frac{1}{2}$, so the region between the two circles is mapped to the strip $\bigl\{ w : \frac{1}{4} < \operatorname{Re} w < \frac{1}{2}\bigr\}$.
We now have a vertical strip of width $\frac{1}{4}$, and we want a horizontal strip of width $1$, thus we need a scaling by a factor of $4$, and a rotation by $\pm \frac{\pi}{2}$. Then we need a translation to move the strip to the required position. All that is easily achieved by
$$\sigma \colon w \mapsto i(4w-1),$$
and hence a Möbius transformation doing the ultimate goal is $(\sigma\circ \rho)^{-1}$. All other such transformations are obtained by composing it with an automorphism of the strip.
To find $(\sigma\circ \rho)^{-1}$, we isolate $z$ in
$$w = i\biggl(\frac{4}{z} - 1\biggr),$$
yielding
$$z = \frac{4}{1-iw} = \frac{4i}{w+i} =: g(w).$$
So you did get the (well, a, it's not unique) right map, and you just have miscalculated when trying to verify.
Let's make a couple of checks. The image of the line $\{w : \operatorname{Im} w = 0\}$ under $g$ is the circle through $g(\infty) = 0$, $g(0) = 4$, and $g(1) = \frac{4}{1-i} = \frac{4(1+i)}{2} = 2(1+i)$. It is easily seen that this is the circle $\{ z : \lvert z-2\rvert = 2\}$.
The image of the line $\{ w : \operatorname{Im} w = 1\}$ is the circle through $g(\infty) = 0$, $g(i) = \frac{4}{1-i^2} = 2$, and $g(2+i) = \frac{4}{1 - i(2+i)} = \frac{4}{2(1-i)} = 1+i$, which is $\{ z : \lvert z-1\rvert = 1\}$.
A point between the two lines is $\frac{i}{3}$, and $g(i/3) = \frac{4}{1-i^2/3} = \frac{4}{4/3} = 3$ lies between the two circles. Hence $g$ maps the strip $\{ w : 0 < \operatorname{Im} w < 1\}$ to the region between the two circles, as desired.
