Mapping the region $\Gamma_{z}$ using the conformal map $ \omega=\frac{-2z}{z^{2}+1}$ Suppose we have an analytic function $$ \omega=\frac{-2z}{z^{2}+1}$$
and the region $\Gamma_{z}$ given by $$\Gamma_{z}:=\left \{ z \in \mathbb{C}| \Im \left ( z \right )\geq 0 \wedge \left | z \right |\geq 1\right \}$$
As with many other problems of this type I tried to break it down into several conformal mapping compositions. For example, it would be very elegant to algebraically manipulate the expression for $\omega$ in such a way that it only has  $z^{2}$, and then first map $z \mapsto \omega_{1}=z^{2}$, and at last do the map $\omega_{1} \mapsto \omega$, where $\omega$ is a Möbius transform, ideally. However, I wasn't able to do so, and perhaps someone can suggest a different approach here.
 A: 
As with many other problems of this type I tried to break it down into several conformal mapping compositions. For example, it would be very elegant to algebraically manipulate the expression for $\omega$ in such a way that it only has  $z^{2}$, and then first map $z \mapsto \omega_{1}=z^{2}$, and at last do the map $\omega_{1} \mapsto \omega$, where $\omega$ is a Möbius transform, ideally.

That's the right idea, however the $z \mapsto z^2$ step can't be our first step.
The starting point is to recognise that we can write the map as a composition of $T \colon w \mapsto -\frac{1}{w}$ and $f \colon z \mapsto \frac{1}{2}\bigl(z + \frac{1}{z}\bigr)$.
$T$ is a Möbius transformation, and a particularly simple one.
You can either directly work with $f$, it's not a very complicated function, or write $f$ as the conjugation of $z \mapsto z^2$ with a suitable Möbius transformation. To find a suitable Möbius transformation, it is useful to look at some distinguished points of $f$. The most interesting points are fixed points and critical points. We see that $f$ has three fixed points, $1, -1, \infty$, and that $\pm 1$ are also critical points. The map $q \colon z \mapsto z^2$ has similar properties, it has three fixed points, $0,1,\infty$, and the two fixed points $0$ and $\infty$ are attained with multiplicity $2$, so are critical points. Thus to write $f$ in the desired form, we need a Möbius transformation $S$ that maps the two attractive fixed points of $q$ to the two attractive fixed points of $f$, and the third fixed point of $q$ to the third fixed point of $f$, then we will have $f = S\circ q \circ S^{-1}$. The requirement that $S(1) = \infty$ gives us the denominator of $S$, and then we need to choose which of $0$ and $\infty$ is mapped to $1$ and which to $-1$. We find that we can use $S(x) = \frac{1+x}{1-x}$ if we map $0 \mapsto 1$.
I'll leave it to you to assemble the pieces.
