how to show this is a conformal map

Quote:

Originally Posted by szpengchao

$f(z)=\frac{z}{(z-1)^{2}}$ maps disc |z|<1 to $C\ \{x+yi: x\leq -\frac{1}{4}, y=0 \}$

Let $f_1$ be the map $z\mapsto \tfrac{1+z}{1-z}$ .
Let $f_2$ be the map $z\mapsto z+1$ .
Let $f_3$ be the map $z\mapsto -\tfrac{1}{2}z$ .
Let $f_4$ be the map $z\mapsto z^2$ .
Let $f_5$ be the map $z\mapsto z+1$ .

Then, $z\mapsto \tfrac{z}{(z-1)^2}$ is $f_5\circ f_4\circ f_3 \circ f_2 \circ f_1$ .

1)The conformal map $f_1$ maps the disk $|z|<1$ to the half-plane $\Re (z) > 0$ .
2)The conformal map $f_2$ maps the half-plane $\Re(z) > 0$ to $\Re(z) > 1$ .
3)The conformal map $f_3$ maps the half-plane $\Re(z) > 1$ to $\Re(z) < - \tfrac{1}{2}$ .
4)The problem is that $f_4$ is not conformal on this half-plane. (Surprised)