I am trying to describe the image of $\{z: |z-i|<1,Re(z)<0\}$ under the Möbius transformation $f(z)=\frac{z-2i}{z}$. I would usually know how to describe this image by first considering the image of its boundary, and then a point on its interior. But I am having difficulty finding the image of the boundary.
I know that line segments map to line or circle segments, and $f(0)=-\infty, f(i)=-1,f(2i)=0$. So I think the line segment from $0$ to $2i$ maps to the line segment from $-\infty$ to $0$, or $\mathbb{R}_{\leq0}$.
But I can't figure out what the circular arc from $0$ to $2i$ maps to. My only guess is to consider some point on this arc, for example $-1+i$, which maps to $i$ under $f$. But I don't think this means the arc maps to the line segment from $-\infty$ (in the imaginary direction) to $i$.
Best Answer
Your method is mostly sound. The one improvement I can suggest is to think of $\infty$ as the limiting value as $|z|$ becomes large in any direction; that is, in the (extended) complex plane there is no difference between $\infty$ and $-\infty$. Therefore a "circle" through $\infty$ is just a line, but its direction is determined by the other two points, not by the "sign/phase of $\infty$".
Three points on the circle are $0$, $-1+i$, and $2i$, which are mapped respectively to $\infty$, $i$, and $0$. Therefore the image is the "circle" through those three points, which is the (vertical) line through $i$ and $0$.