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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 $(0,\infty)$

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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.

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