TITLE: Why this determinant is conformally invariant?
QUESTION [6 upvotes]: While I was reading a paper about random analytic function I found a statement that I was not able to prove and after try brute force and search for some references I decided to ask for a help here. The statement is the following: Denote by $\mathbb{D}$ the unit dic on the complex plane, and consider the function $p:\mathbb{D}^n\to\mathbb{C}$ given by 
$$
p(z_1,\dots,z_n)\, =\, 4^{-n}\, \det\left(\frac{(1-|z_i|^2)(1-|z_j|^2)}{|1-z_i\overline{z}_j|^2}\right)_{i,j=1}^{n}\,
$$ 
then for any fixed $u\in\mathbb{D}$ the function $p$ is invariant for the Möbius transformation
$$
\phi(u,z)=\frac{u+z}{1+\overline{u}z},
$$
in the sense that
$$
p(z_1,\dots,z_n)\, = p(\phi(u,z_1),\ldots,\phi(u,z_n)).
$$
I appreciate any reference or hint to prove this fact.
Edition: The denominator was modified as point out for David and Anon.

REPLY [6 votes]: More pointedly, each entry in the matrix is a cross-ratio:
$$\frac{(1-|z|^2)(1-|w|^2)}{|1-z\bar{w}|^2}=\frac{(1-z\bar{z})(1-w\bar{w})}{(1-\bar{z}w)(1-z\bar{w})}=(z^{-1},w^{-1};\bar{z},\bar{w}).$$
This can be seen by dividing both numerator and denominator by $zw$ appropriately. Cross-ratios are invariant under Möbius transformations, in that sense that if $f$ is one, then
$$(a,b;c,d)=(f(a),f(b);f(c),f(d)).$$
EDIT: Ah, the last ingredient is provided by Theo Buehler. Since $\phi(u,z)^{-1}=\phi(\bar{u},z)$ and similarly we have $\overline{\phi(u,z)}=\phi(\bar{u},z)$, we can say that $\phi(u,\cdot)$ acting on both $z$ and $w$ is equivalent to the Möbius transformation $\phi(\bar{u},\cdot)$ acting on each individual component the $4$-tuple $(z^{-1},w^{-1},\bar{z},\bar{w})$, hence the cross-ratio is preserved under it.