Image of koebe map

Question

Let $f(z)=\frac {z}{(1-z)^2}$ how can I find the image of unit disk under this transformation? Also is it possible to find an explicit expression for the inverse of this transformation?

Answer 1

Note that $$\frac{z}{(1-z)^2} = \frac{1}{4} \left( \frac{1+z}{1-z} \right)^2 - \frac{1}{4}$$ and that $$z \mapsto \frac{1+z}{1-z}$$ maps the unit disc to the right half plane. The image of the Koebe map is therefore $\mathbb{C} \setminus (-\infty, -\frac{1}{4}]$.