Let
$$f(z) = \sum_{n=0}^\infty c_nz^n$$
be analytic in the disc $\mathbb{D} \ \{ z \in \mathbb{C} | |z| < 1 \}$. Assume $f$ maps $\mathbb{D}$ one-to-one onto a domain $G$ having area $A$. Prove
$$A = \pi \sum_{n=1}^\infty n |c_n|^2$$
Looking at the solution - I lack some fundamental understanding. Why is the Jacobian of the transformation $|f'(z)|^2$, giving us that $A = \int_{\mathbb{D}}|f'(z)|^2 dx dy$. Can someone help me see this by walking through the definition?
If we write $f(x+iy) = u(x,y)+iv(x,y) $ where $u,v$ are real, we can compute the Jacobian in the usual way as $$ \begin{vmatrix} u_x & u_y \\ v_x & v_y \end{vmatrix} = u_x v_y-u_yv_x. $$ The Cauchy–Riemann equations give that $u_x=v_y$ and $u_y=-v_x$, so this is equal to $u_x^2+u_y^2$.
On the other hand, the formula for coordinate transformations gives $\partial_z = \frac{1}{2}(\partial_x - i\partial_y)$, so $$ f'(x+iy) = \frac{1}{2}\left( u_x+v_y+i(u_y-v_x) \right) = u_x+iu_y, $$ by the Cauchy–Riemann equations again, and then $\lvert f'(x+iy) \rvert^2 = u_x^2+u_y^2 $. So the Jacobian is $\lvert f'(z) \rvert^2$ as required.
This is essentially equivalent to the beautiful answer by Chappers but avoids going to $(x,y)$ coordinates as much as possible.
The metric on the Argand plane is $$ds^2=dx^2+dy^2=(dx+idy)(dx-idy)=dz d\bar{z}\tag{1}$$ Let $w \to z=f(w)$, then holomorphicity of the map $f:\mathbb{C} \to \mathbb{C}$ implies existence of $f'(w)$ and in turn $dz=\frac{\partial z}{\partial w}dw=f'(w) dw$ and similarly for the antiholomorphic piece $d\bar{z}=\overline{f'(w)} d\bar{w}$.
So by $(1)$ $$ds^2=|f'(w)|^2 dw d \bar{w}$$
This proves that the Jacobian of the holomorphic maps are indeed $|f'(w)|^2$ implying that it is a conformal transformation.
