# Find a Möbius transformation such that a semi-circle in the upper half plane is mapped to $i\mathbb{R}_{>0}$

I am giving the semi-circle $$\ell = \{z\in\mathbb{H}:|z|^2=4\}$$ and I want to create a Möbius transformation $$M:\mathbb{H} \to \mathbb{H}$$ via $$M(z) = \frac{az+b}{cz+d}$$, where $$ac-bd > 0$$ such that the semi-circle gets mapped to the vertical line $$i\mathbb{R}_{>0}$$.

I know that any Möbius transformation is defined by how it maps any three points, after seeing a similar question, I thought about mapping the following equations: $$M(-2) = 0, M(2i) = i, M(2) = \infty$$ This will naturally give you a system of equations. However, my major concern is that the two points $$z=-2$$ and $$z=2$$ are not elements of $$\mathbb{H}$$. Intuitively I think it is okay as when I get infinitesimally close to $$z=-2$$, $$M(z)$$ gets infinitesimally close to $$0$$ and likewise, as I get infinitesimally close to $$z=2$$, $$M(z)$$ gets infinitesimally close infinity (aka just infinity).

Is this the correct way of thinking about Möbius transformation? Or would it be better to map three points that we know are a part of $$\ell$$. So for example: $$M(-2/\sqrt2)+ i2/\sqrt2) = 0, M(2i) = i, M(2/\sqrt2)+ i2/\sqrt2)=\infty$$ Im not 100% convinced because something like that would only map the selements between $$z=-2/\sqrt2+ i2/\sqrt2$$ and $$z=2/\sqrt2+ i2/\sqrt2$$ to $$(0,\infty)$$

Your approach is fine. A Möbius transformation with $$M(-2) = 0, M(2i) = i, M(2) = \infty$$ maps the full circle $$\{ |z|=2 \}$$ to the extended line $$i \Bbb R \cup \{ \infty \}$$.
The semi-circle in the upper half-plane is then mapped to an open connected subset of that extended line having $$0$$ and $$\infty$$ as boundary points and containing the point $$i$$, and that is the “upper” part of the imaginary axis.