Let \(f(z) = \sum_{n=0}^\infty c_n z^n\) be analytic and one-to-one in \(|z| < 1\). For \(0<r<1\), let \(D_r\) be the disk \(|z|<r\). Show that the area of \(f(D_r)\) is finite and is given by \[ S = \pi \sum_{n=1}^\infty n |c_n|^2 r^{2n} .\]
Note that in general the area of \(f(D_1)\) is infinite.
Solution
Write \(f=u+iv\). By the Cauchy–Riemann equations the Jacobian determinant of \(f\) is \(u_xv_y-u_yv_x=u_x^2+v_x^2=\abs{f'}^2\). Since \(f\) is injective on \(\DD_r\), the change of variables formula gives, in polar coordinates \(z=\rho e^{it}\) on \(\DD_r\), \[ \mu(f(\DD_r)) &= \int_{\DD_r} \abs{f'(z)}^2\,dA \\ &= \int_{\DD_r} \sum_{j, k\geq 1} jk c_j \bar{c_k} z^{j-1}\bar{z}^{k-1}\,dA \\ &= \int_0^r\int_0^{2\pi} \sum_{j, k\geq 1} jk c_j \bar{c_k} \rho^{j+k-1} e^{i(j-k)t} \,dt\,d\rho \\ &= \int_0^r 2\pi\sum_{k\geq 1} k^2 \abs{c_k}^2 \rho^{2k-1} \,d\rho \\ &= \pi \sum_{k\geq 1} k \abs{c_k}^2 r^{2k} .\] The double series converges uniformly on \(\overline{\DD_r}\), which justifies integrating term by term, and only the terms \(j=k\) survive the \(t\)-integration. The area is finite because \(\abs{f'}^2\) is bounded on the compact set \(\overline{\DD_r}\).