How to find the map of an inversion I want to find the image of the unit disk $D:= \{z: |z|\lt1 \}$ under the Möbius transform.
$$f(z)=\dfrac{iz+3}{iz-1}=1+ \dfrac{4}{iz-1}$$
Now, $f$ can be decomposed into $f= f_5 \circ f_4 \circ f_3 \circ f_2 \circ f_1$, where
$$f_1(z)=iz, \ f_2(z)=z-1, \ f_3(z)=\frac{1}{z}, \ f_4(z)=4z, \ f_5(z)=z+1$$
I understand all of them except $f_3$. It says that the shifted disk is mapped onto the half space $\{ z: \mathfrak{R}(z) \lt \frac{1}{2} \}$ under the inversion. How do I find the map of the inversion in general?
Thank you for your help.

The image of $f_2(f_1(D))$ is the disk $U = \{ z: |z+1| < 1 \} $ with center $-1$ and radius $1$.
The boundary of this disk is a circle $C$ through $0$. Möbius transforms map circles to circles or (extended) lines.  Since $f_3(0) = \infty$,  this circle is mapped to a line $L = f_3(C)$.
$f_3$ preserves the 
(extended) real axis and conformal mappings preserve angles, therefore
$L$ intersects the real axis at a right angle: It is a line parallel
to the imaginary axis.
$C$ contains the point $-2$ and $f_3(-2) = -\frac 12$, therefore $L = \{ z: \mathfrak{R}(z) = -\frac{1}{2} \}$.
The disk $U$ is one of the connected components of the complement of $C$,
therefore $f_3(U)$ is one of the connected components of the 
complement of $f_3(C)$, i.e. either the halfplane left or right of $L$.
Since $f_3(-1) = -1$ , $f_3(U) = \{ z: \mathfrak{R}(z) < -\frac{1}{2} \}$  .
