How to prove $ \bigl\lvert \frac{az+b}{cz+d}\bigr\rvert = \frac{\operatorname{Im}(z)}{\vert cz+d \vert^2}$? $z\in \mathbb{H}$

Question

$\Gamma(z, K):=\{g\in\Gamma \ \vert \ g(z) \in K \}$

How to show $\biggl\lvert \dfrac{az+b}{cz+d}\biggr\rvert = \dfrac{\operatorname{Im}(z)}{\bigl(\lvert cz+d \rvert\bigr)^2}$ for $z\in \mathbb{H}$ and
$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \operatorname{SL}(2, \mathbb{R})$?