Miyake proves this lemma (and subsequently uses it to show the area of a hyperbolic triangle is the angle deficit):
$$ (y^{-1}dz)\circ \alpha - y^{-1}dz = -2i d[log(j(\alpha,z)],$$
where $\alpha = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{R})$ acts on $H = \{z \in \mathbb{C}| Im(z)>0$ by fractional linear transformations ($\alpha z = \frac{az+b}{cz+d}$) and $j(\alpha,z)=cz+d$.
I can understand every step of the proof but the first, where he says
$$ (y^{-1}dz)\circ \alpha - y^{-1}dz = (\frac{|j(\alpha,z)|^2}{j(\alpha,z)}-1)y^{-1}dz$$
He references the identities
$$dz\circ \alpha = \frac{d(\alpha z)}{dz}dz = det(\alpha)j(\alpha,z)^{-2}dz$$
and
$$Im(\alpha z) = \frac{det(\alpha)Im(z)}{|j(\alpha,z)|^2}$$.
It seems to me like
$$(y^{-1}dz)\circ \alpha - y^{-1}dz = y^{-1}(dz\circ \alpha) - y^{-1}dz \\ = (\frac{det(\alpha)}{j(\alpha,z)^2} - 1)y^{-1}dz \\ = (\frac{|j(\alpha,z)|^2}{j(\alpha,z)}\frac{Im(\alpha z)}{Im(z)}-1)y^{-1}dz$$
I don't see any reason why $\frac{Im(\alpha z)}{Im(z)}$ should be 1, so what am I doing wrong? I suspect it's something about $(y^{-1} dz)\circ \alpha$, since I'm not sure why the parentheses are there, but presumably there's a reason.