I'm trying to show $-\arg{j_{AB}(z)}+\arg{j_{A}(Bz)}+\arg{j_{B}(z)}$ does not depend on $z\in\mathbb{H}$ where $A,B\in SL_2{(\mathbb{R})}$ and $j_A(z)=cz+d$, $A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and the branch cut is made at $\pi$.
The book says it follows from the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ but I don't see why the statement above follows from the identity. I can see that the identity implies
$$-\arg{j_{AB}(z)}+\arg{j_{A}(Bz)}+\arg{j_{B}(z)}\equiv0\mod{(-\pi,\pi]}.$$
And I can also see that $-\arg{j_{AB}(z)}+\arg{j_{A}(Bz)}+\arg{j_{B}(z)}$ takes only 3 possible values, $-2\pi, 0, 2\pi$. However, I really don't see why this expression always takes the same value no matter what $z$ I choose. Can anybody give me some insights?