Suppose that f is a one-to-one analytic function mapping the disc $|z|<1$ onto a bounded domain D. Show that the area of D is given by $$A(D)=\int \int_{|z|<1} {|f'(z)|}^2 dxdy$$
This is a practice problem from Fisher's complex variables chapter 3.5, which is on Riemann's mapping thm and Shwarz tranformation.
I don't see how i can apply either concept to this problem however–the double integral looks more like Green's thm to calculate domain areas but as a complex variable z. I am stuck on deriving the actual result (which seems somewhat intuitive); any help or guidance is appreciated.
Best Answer
If we identify $f=u+iv$ with the corresponding mapping $(x, y) \mapsto (u(x, y), v(x,y))$ then $$ A(D) = \int_{x^2+y^2<1} | \det J_f(x, y)| \, dx dy $$ where $J_f$ is the Jacobian matrix of $f$. Now $$ \det J_f(x, y) = \begin{vmatrix} u_x & u_y \\ v_x & v_y \end{vmatrix} = \begin{vmatrix} u_x & -v_x\\ v_x & u_x \end{vmatrix} = u_x^2 + v_x^2 $$ using the Cauchy-Riemann equations. On the other hand, $$ |f'(z)|^2 = |u_x + iv_x|^2 = u_x^2 + v_x^2 \, . $$