Determine the image of the band 
Determine the image of the band $\Omega=\{(x,y)\in \mathbb{R}^2; x>0, 0<y<1\}$ under the transformation $$f(z)=\dfrac{i}{z},z\in\mathbb{C}\setminus\{0\}, z=x+iy$$

If $f(z)=w=\dfrac{i}{z}$, then $w^2=\dfrac{1}{z\overline{z}}=\dfrac{1}{\vert z\vert^2}\implies \vert z\vert^2=\dfrac{1}{w^2}$, therefore $\vert z\vert=\sqrt{x^2+y^2}=\dfrac{1}{w}$, I don't know is this correct, but how I can use the hypotesis $x>0, 0<y<1$?? Because, It makes no sense to rewrite $z=re^{i\theta}$ since I don't define a rectangle with this. regards!
 A: There are several possible approaches. Note that
$$f(x+iy) = \frac{i}{x+iy} = \frac{i(x-iy)}{x^2+y^2} = \frac{y+ix}{x^2+y^2}$$
In particular, if $x > 0$ and $y > 0$, then $f(x+iy)$ has positive real and imaginary parts, so the image is contained in the first quadrant. 
Next, let us see what happens to a horizontal line: $z = x+ic$. We have
$f(x+ic) = a+ib$, where $a=c/(x^2+c^2)$, $b=x/(x^2+c^2)$. You may already know that Möbius mappings, in particular your $f$, map the set of lines and circles to the set of lines and circles. If not, notice that
$$
(a-\frac1{2c})^2 + b^2 = \frac{1}{4c^2}
$$
so a horizontal line maps to circle with centre at $1/(2c)$ and radius $1/(2c)$. Do some tedious algebra to show this. (The quick way is to see that a horizontal line is mapped to a circle through $0$: the "point at $\infty$" goes to $0$. The circle also contains $f(ic) = \frac1c$ and $f(c+ic) = \frac{1}{2c}(1+i)$ from which we see that the center and radius is $1/(2c)$.
Varying $0 \le c < 1$, we see that the image contains the first quadrant except the half-disc: $(a-\frac12)^2+b^2 \le \frac14$. Graphically:

The blue curves are the semi-circles described above, the brown curves the image of vertical lines in the strip.
