 Solution

For the Möbius transformation $z^{\prime }=(az+b)/(cz+d)$ with $ad-bc=1$,The power of $|c{z}_i+d|$ contributed by every pair containing $i$ iswhere momentum conservation and the mass-shell condition $p_i^2=4/{\alpha }^{\prime }$ were used. The Koba-Nielsen factor therefore contributes $|c{z}_i+d{|}^4$ at each insertion, exactly cancelling the transformed measure. Hence the remaining integral is $SL(2,\mathbb{C})$ invariant, and division by its volume removes the residual conformal-gauge redundancy.