Define Hyperbolic area of a subset $E$ of the unit disk $D$ to be $\displaystyle 4\int \int_E \frac{dx dy}{(1-|z|^2)^2}$. Show that the hyperbolic area is invariant under conformal self maps of $D$.
I am new to hyperbolic geometry. So I am confused with some concepts. This problem seems like a standard one but I couldn't find it in MSE. So if you find a duplicate please be kind enough not to downvote this and instead please just copy a link, then I will delete this problem.
I know that hyperbolic lengths are invariant under conformal self maps of $D$. So I tried a similar technique that has been used in the proof of that fact.
So if $w=f(z)$ then by Pick's lemma we have $\displaystyle \frac{|dw|}{1-|w|^2} =\frac{|dz|}{1-|z|^2}$.
By squaring both sides, $\displaystyle \frac{|dw|^2}{(1-|w|^2)^2} =\frac{|dz|^2}{(1-|z|^2)^2}$. Hence, $\displaystyle \frac{dw\overline{dw}}{(1-|w|^2)^2} =\frac{dz\overline{dz}}{(1-|z|^2)^2}$
After that I was stuck since I don't know how to transform into $dxdy$ form. Can somebody please help me?
Take a Möbius transformation $T(z)=(az+\overline c)/(\overline c z+a)$ with $|a|^2-|c|^2=1$ (all orientation-preserving isometries are of this form) and compute. Namely, if $T(z)=u+iv$, then the Jacobian is $$ \det\frac{\partial(u,v)}{\partial(x,y)}=\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}=\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial v}{\partial x}\right)^2=|T'(z)|^2=\frac{1}{|\overline cz+a|^4} $$ (using Cauchy-Riemann in the second identity). Other then this you need to show that $$ |T(z)|^2=\frac{(az+\overline c)(\overline a \overline z+c)}{(\overline c z+a)( c \overline z+\overline a)} $$ satisfies $$ \frac{1}{(1-|T(z)|^2)^2}=\frac{|\overline cz+a|^4}{(1-|z|^2)^2}. $$ The two powers cancel and give the desired result.