Question : Find the image of the strip $0<\operatorname{Im} z<1$ under the mapping $w=(z-i) / z$ ?
My Approach : Writing $z=x+iy$, the given mapping is $$w = u + i v = \frac{x^2+y(y-1)- i x}{x^2 + y^2}$$
When $y=0$, $w=1-i/x$ and when $y=1$, $w=x^2-ix/x^2+1$. Writing $w=u+iv$ and trying to get an expression in terms of $u$ and $v$, I wasn't successful. Any help in solving this question is much appreciated.
A linear fractional transformation (a.k.a., Möbius transformation) maps straight lines in the complex plane to lines or circles and maps the half-planes separated by a straight line to half-planes or to the interior or exterior of a circle as appropriate. You only need to look at three distinct values of the transformation on three points on a line to tell whether or not you have a line or a circle and then looking at the value on one point off the line will tell where the half-planes separated by the line are mapped. To understand this geometrically, it is very helpful to think of the Riemann sphere, the result of adding a point at infinity to the complex plane, which lets us think of straight lines in the planes as circles that pass through $\infty$ and lets us write $u/0 =\infty$, provided $u \neq 0$..
So let's look at the edges $\operatorname{Im}z = 0$ and $\operatorname{Im}z = 1$ of the strip.
$\mathbf{\operatorname{Im}z = 0{:}}$
We have $w(0) = \infty$, so the image of this edge of the strip is a straight line. We have $w(1) = 1 - i$ and $w(2) = 1 - i/2$. So the image of this edge is the vertical line $\operatorname{Re}z = 1$. Also $w(i) = 0$, so $w$ maps the half-plane above this edge to the half-plane to the left of this vertical line.
$\mathbf{\operatorname{Im}z = 1{:}}$
We have $w(i) = 0$, $w(1 + i) = 1/2 - i/2$ and $w(-1 + i) = 1/2 + i/2$. These three values are not collinear, so the image of the line is a circle, which we can see must be the circle of radius $1/2$ centred at $1/2$. Also $w(2i) = 1/2$ so $w$ maps the half-plane above this edge to the interior of the circle, and hence maps the half-plane below this edge to the exterior of the circle.
Conclusion:
The image of the strip is the intersection of the open half-plane to the left of the line $\operatorname{Re}z = 1$ with the exterior of the circle $|z - 1/2| = 1/2$.