The image of a specific Mobius transformation

Question

Let $f:D\rightarrow \mathbb{C} :f(z)=\frac{z}{z-1}$ and $D= \{ z:|z|=1\}$ ,what is the image of $f$, $f(D)$?

Can one elaborate on some general methods of dealing with these kind of questions?

Answer 1

There are two things to note here:

1. A Mobius transformation maps a circle to either a line or a circle.
2. A Mobius transformation is uniquely determined by the images of three points.

Now, in your case, we have

• $f(1)=\infty$
• $f(-1)=1/2$
• $f(i) = (1-i)/2$

Now, since $f(1)=\infty$, the image must be a line, rather than a circle. By the other two computations, that line must go through $1/2$ and $(1-i)/2$. If you happen to be interested in the interior of the circle, then you simply check the image of some simple point in the interior. In this case, $f(0)=0$. Thus, the interior maps to the side of the line that contains the origin.