As you see in the linked post, the conformal factor is $\frac{2}{1-\left\|x \right\|^2}$.
But I'm more interested in how the outcome turns out. According to what I've heard, this may be demonstrated by comparing the lengths of tangent vectors.
Length of a tangent vector:
Let $D$ denote the Poincare Unit Disk.
Given a smooth map $\zeta :[-\epsilon,\epsilon]\to D$ with $\zeta (0)=\zeta _0$ and $\frac{\mathrm{d} }{\mathrm{d} t}|_{t=0}\zeta (t)=\eta$, the length of the tangent vector $\eta$ in a metric $d$ is given by $\frac{\mathrm{d} }{\mathrm{d} t}|_{t=0}d(\zeta_0,\zeta(t))$.
I'm not sure how to set up the equation that compares the lengths of the tangent vectors to calculate the conformal factor. Can I get some help?
Lecture Note Definition of $D$:
Let $\Pi : H \to D = \{\zeta \in \mathbb{C}| |\zeta|<1\}$ s.t $\Pi (x,y,z) \mapsto \frac{x+iy}{1+z}$.
$\Pi^{-1}:\mathbb{C} \to H$ is defined by $\zeta \mapsto \left ( \frac{2\zeta}{1-|\zeta|^2},\frac{1+|\zeta|^2}{1-|\zeta|^2} \right )$.
The distance function is $d_D(\zeta_1,\zeta_2)=\cosh^{-1}\left ( 1+\frac{2|\zeta_1-\zeta_2|^2}{(1-|\zeta_1|^2)(1-|\zeta_2|^2)} \right )$
Equation of geodesic: $\frac{2\zeta}{1-|\zeta|^2}\eta + \frac{1+|\zeta|^2}{1-|\zeta|^2}z_0=0$.