prove hyperbolic metric is independent of conformal map A conformal map $g(z)$ of a domain $D$ onto the open unit disk $\mathbb{D}$ induces the metric $\rho_D$ on $D$ defined by 
$d \rho_D (z) =\frac {2\vert g'(z) \vert}{1- \vert g(z)\vert ^2}$$ \vert dz\vert $ for $z\in D$. 
Show that $\rho_D$ is independent of the conformal map $g(z)$ of $D$ onto $\mathbb{D}$. 
I actually don`t understand the strategy to do that. What does it mean here to be independent of the conformal map? 

Let $f(z)$ be another conformal map of $D$ onto the open unit disk $\mathbb{D}$. Then $(g\circ f^{-1})(z)$ is a bijective mapping of $\mathbb{D}$ to $\mathbb{D}$, so it can be expressed as $$
(g\circ f^{-1})(z)=e^{i\theta }\frac{z-a}{1-\bar{a}z},$$
where $a(|a|<1)$ is some point in $\mathbb{D}$ and $\theta \in\mathbb{R}$.
Therefore $g$ is expressed as $$
g(z)=e^{i\theta }\frac{f(z)-a}{1-\bar{a}f(z)}.$$
It is easy to check that \begin{align}
&g^\prime(z)=e^{i\theta }\frac{(1-|a|^2)f^\prime(z)}{(1-\bar{a}f(z))^2},\quad|g^\prime(z)|=\frac{(1-|a|^2)|f^\prime(z)|}{|1-\bar{a}f(z)|^2}\tag{1},\\
&1-|g(z)|^2=1-\left| e^{i\theta }\frac{f(z)-a}{1-\bar{a}f(z)}\right|^2=\frac{(1-|a|^2)(1-|f(z)|^2)}{|1-\bar{a}f(z)|^2}.\tag{2}
\end{align}
$(1)$ and $(2)$ yields $$
\frac{|g^\prime(z)|}{1-|g(z)|^2}=\frac{|f^\prime(z)|}{1-|f(z)|^2}.
$$
