Let me consider the map $$T:\Bbb{C}\setminus \{-i\}\rightarrow \Bbb{C}, ~z\mapsto \frac{z-i}{-iz+1}$$ I want to look at it as a map from the upper half plane to the unit disk, because there I know that it is an isometry if I endow the upper half plane with the Riemannian metrig $g_\Bbb{H}= \frac{1}{y^2} \left(\begin{matrix} 1 & 0\\ 0 & 1\end{matrix}\right)$ and where I endow the unit disk with $g_\Bbb{D}= \frac{1}{(1-x^2-y^2)^2}\left(\begin{matrix} 1 & 0\\ 0 & 1\end{matrix}\right)$. I want to find out where the coordinate system is sent around the origin. In other words, where horizontal and vertical coordinate lines are sent to.
What I know is that Fractional linear transformations are conformal, i.e. it respects angles. So I first considered the real line $\Bbb{R}$ then I remarked that $T(\Bbb{R})\ni -1,-i,1$ since fractional linear transformations maps generalised circles to generaliced circles we know that $T(\Bbb{R})$ is the half circle passing through $-1,-i, 1$. Now I thougth about taking the imaginary line passing, i.e. the line passing through $0$ orthogonal to $\Bbb{R}$. I remark that $T(I)\ni -i, 0,i$, but this would mean that $T(I)$ is the line passing through $-i,0,i$ which is a bit suspect for me since i don't think that a vertical line is mapped to a the same vertical line.
Can someone help me where my mistake is?
The imaginary line is the set $I=\{it \mid t \in \mathbb R$}. For $t \neq -1$, you have
$$T(it) = \frac{it - i}{t+1}= i\frac{t-1}{t+1}$$
... and I don't see how this can be equal to $i$. So you made an error in your computations.