Möbius transformation $0

Question

I'm struggling a bit with the Möbius transformation below.

Describe the image of the region under the transformation

b) the strip $0<x<1$ under $w=\frac{z}{z-1}$

My solution is so far:

1. Check that it is in fact a valid M.transformation with $ad-bc \neq 0$.
2. Calculate transformation of 3 points on the strip.

$p_1=(0)$ $\Rightarrow w(p_1)=0$

$p_2=(\frac{1}{2})$ $\Rightarrow w(p_2)=-1$

$p_3=(1)$ $\Rightarrow w(p_3)=\infty$

My conclusion from this is that its a line since it contains $\infty$ and since we got no imaginary part i would like to answer that the image is $\{w:Re(w)<0, Im(w)=0\}$ which is not the correct answer.

The correct answer is $\{w:Re(w)<1, |w-\frac{1}{2}|>\frac{1}{2}\}$, how come?

Would anyone like to give me hint on how to proceed?

Answer 1

Write $z=\frac{w}{w-1}=\frac{u+iv}{(u-1)+iv}=\frac{u(u-1)+v^2-iv}{(u-1)^2+v^2}$
$\implies Re(z)=\frac{u^2+v^2-u}{(u-1)^2+v^2}$
Given $0<Re(z)<1$
$\implies 0<\frac{u^2+v^2-u}{(u-1)^2+v^2}<1$
$\implies 0<u^2+v^2-u<(u-1)^2+v^2$

From the first half inequality;

$0<u^2+v^2-u$
$\implies (1/2)^2<(u-1/2)^2+v^2$
which is the exterior of circle in $w-$plane centered at $(1/2,0)$ with radius $1/2$ i.e. $|w-1/2|>1/2$.
From the rest half inequality;

$u^2+v^2-u<(u-1)^2+v^2$
$\implies u<1$
$\implies Re(w)<1$
Thus, the required set is $\{w: Re(w)<1, |w-1/2|>1/2\}$