Image of shape under Möbius transformation

Question

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 $i$.

Answer 1

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 $0$.

Answer 2

One way forward is to define a path on your circular arc. $$ p(\theta) = \mathrm{i} + \mathrm{e}^{\mathrm{i}\theta}, \pi/2 \leq \theta \leq 3\pi/2 \text{.} $$ Then look at $f(p(\theta))$, obtaining the nonnegative half of the imaginary axis.

Another way, which is a small extension of your method is to study how $f$ sends the near-$0$ end of the semicircle to complex infinity, i.e., from what direction $f(-\varepsilon)$ approaches complex infinity as (real) $\varepsilon \rightarrow 0$. We see $$ f(-\varepsilon) = \frac{-\varepsilon - 2\mathrm{i}}{-\varepsilon} = 1 + \frac{2}{\varepsilon}\mathrm{i} \text{.} $$ As $\varepsilon \rightarrow 0$, $f(-\varepsilon)$ goes to $1 + \mathrm{i}\infty$. We should perhaps be unsurprised then, that the image of the near-$0$ end of the semicircle descends from $\mathrm{i}\infty$.