What is the image of the right-half plane $\mathfrak H$ under the map $f$? Let $\mathbb{D} \subset \mathbb{C}$ be the unit disk. Let $\mathfrak{H}$ denote the right half-plane and let $\displaystyle f(z) = \frac{2z^2}{(1+z)^{2}}$. What is the image of $\mathfrak{H}$ under the map $f$?
A brief solution would help.
 A: You can simplify it, using $\displaystyle\frac z{1+z} = 1-\frac1{1+z}$. If we introduce $w:=1+z$ (this $w$ hence can be any point from the shifted version of $\mathfrak H$, by one to the right), this is $1-\displaystyle\frac1w$, this has to be squared and duplicated:
$$f(z)=2\left(1-\frac1w\right)^2$$
So, $w$ is from $\mathfrak H+1 =\{w\mid \mathfrak{Re}(w)\ge 1\}$. Where will $w\mapsto \displaystyle\frac1w$ take this?
This is going to be the image of that half plane under the inversion through the unit circle (the inversion in real coordinates is given by $(x,y)\mapsto\displaystyle\frac1{x^2+y^2}(x,y)$, that is, for complex numbers $z\mapsto\displaystyle\frac z{z\bar z}$, so one also needs the conjugation: reflection to the $x$-axis).
We will get that $w\mapsto \displaystyle\frac1w$ maps $\mathfrak H+1$ to the disc $B_{1/2}(\frac12)$, almost containing $0=\frac1\infty$ in its edge and tangent to the unit circle.
(This can also be verified directly.)
Then $z\mapsto 1-z$ is applied, it is the reflection on the point $\frac12$, so $B_{1/2}(\frac12)$ is invariant.
Then $z\mapsto z^2$ is applied...
