I need to find the image of the strip $0<x<1$ if $w = \frac{z-1}{z-2}$.
To approach this problem, I followed the advice of David Quinn on this post, and solved $w = \frac{z-1}{z-2}$ for $z$. So, $z=\frac{1-2w}{1-w}$.
Then, since $w = u+iv$, I substituted $u+iv$ in for $w$ above, and multiplied top and bottom by the complex conjugate of the bottom to obtain $\frac{1-3u+2u^{2}+2v^{2}}{(1-u)^{2}+v^{2}} - i \frac{v}{(1-u)^{2}+v^{2}}$.
Now, $x = Re\,z =\frac{1-3u+2u^{2}+2v^{2}}{(1-u)^{2}+v^{2}} $, so what we need is $ 0 < \frac{1-3u+2u^{2}+2v^{2}}{(1-u)^{2}+v^{2}} < 1$.
For $0 < \frac{1-3u+2u^{2}+2v^{2}}{(1-u)^{2}+v^{2}}$, we really only need $0 < 1-3u+2u^{2}+2v^{2}$. So, by completing the square, I get that the area that should make that quantity $> 0$ is the exterior of a circle centered at $(3/4,0)$ with radius $1/4$, or $1 < \displaystyle \frac{(u-3/4)^{2}}{(1/4)^{2}} + \frac{v^{2}}{(1/4)^{2}}$.
For $1 > \displaystyle \frac{1-3u+2u^{2}+2v^{2}}{(1-u)^{2}+v^{2}}$, by completing the square again, I get that the area that should make that quantity $< 1$ is the interior of a circle centered at $\left(\frac{1}{2},0 \right)$ with radius $\frac{1}{2}$.
And also, we need to be sure to omit the point $(1,0)$ because that makes the denominator in $\displaystyle \frac{1-3u+2u^{2}+2v^{2}}{(1-u)^{2}+v^{2}}$ zero.
I know I went wrong somewhere, because these two areas together don't make any sense. It's almost like the $0>$ part was redundant. So, I'm assuming I made a mistake.
Can someone please help me fix it?
Sorry to say that you complicate too much.
The simplest approach is to note that Möbius transformations with matrix with real entries having determinant $\pm1$ (added after benji's comment) take vertical lines and circles centered at the real line to vertical lines or circles centered at the real line. In this case, for $x=0$ it gives the circle intersecting the real line at $1/2$ and at $$\lim_{y\to+\infty}\frac{iy-1}{iy-2}=1.$$ Similarly, for $x=1$ it gives the circle intersecting the real line at $0$ and at$$\lim_{y\to+\infty}\frac{1+iy-1}{1+iy-2}=\lim_{y\to+\infty}\frac{iy}{iy-1}=1.$$