# Image of shape under Möbius transformation

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 $$i$$.

• The generalized circles $C_1 = \{z: \operatorname {Re} z = 0\}$ and $C_2 = \{z: |z - i| = 1\}$ get mapped to straight lines through the origin because $f(0) = \infty$ and $f(2 i) = 0$. $f(C_1)$ is the real line because $f(i t)$ is real. $f(C_2)$ is orthogonal to $f(C_1)$. To choose the correct quadrant, take a sample point. Dec 7, 2019 at 1:54

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 $$0$$.
• I'm guessing that by 'vertical line' you mean the non-negative part of the line $i\mathbb{R}$. But, given our 3 points $\infty, i,$ and $0$ how do we know that this line ranges from $0$ to positive $\infty$ instead of $i$ to negative $\infty$? Dec 3, 2019 at 19:06
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 $$\mathrm{i}\infty$$.