Consider the standard mapping $w=\frac{1}{z}$. What is the image of the "half" plane above the line whose imaginary part is $c$, for the three cases of $c\gt 0 , c=0 , c\lt 0$?
For $c=0$ obviously the image and preimage are the same. For the other two I think I get some sort of sector of a circle with $radius=\frac{1}{c}$ I think, but I don't know how to calculate the rest rigorously. Any pointers would be appreciated.
$f(z)=\frac{1}{z}$ is a Mobius transformation, which maps 'circles' to 'circles' and connected regions to connected regions. For $c=0$ we can see that it maps $\mathbb{R}$ (which is a circle of radius $\infty$) onto itself. $f(i)=\frac{1}{i}=-i$, and so the image of the upper half plane is the lower half.
Consider now $c>0$, the trick is to pick three points on the line that you can easily see where they go to. $f$ does not map any part of the line to $\infty$ and so $\mathrm{Im}(z)=c$ will be mapped to a 'conventional' circle. This circle includes the points:
$$f(ci)=\frac{-i}{c}$$ $$f(\infty)=0$$
For the last piece of information, note that $\mathrm{Im}(z)=c$ and $\mathrm{Im}(z)=0$ meet parallel at $\infty$, to see this imagine the preimages of $\mathrm{Im}(z)=0$, $\mathrm{Im}(z)=c$ on the Riemann sphere and note that unless they meet parallel they will intersect in two places. Thus the image of $\mathrm{Im}(z)=c$ is a circle in the lower half plane passing through $0$ and $\frac{-i}{c}$ which is tangent to the real axis. To see this visualised, pay around with this applet. To see whether the region above the line is mapped inside or outside this circle you could pick a point in the upper half plane and see where it is mapped to.