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
Goal: For \(f(z) = \sum_{n=0}^\infty c_n z^n\) analytic and one-to-one in \(\abs{z} < 1\), show the area of \(f(D_r)\), \(0 < r < 1\), is finite and equals \(S = \pi \sum_{n=1}^\infty n\abs{c_n}^2 r^{2n}\).
1. The area of \(f(D_r)\) equals \(\iint_{D_r} \abs{f'(z)}^2\, dA(z)\).
\(f\) is injective and holomorphic on \(D_r\), so it is a \(C^1\) diffeomorphism onto its image with Jacobian \(\abs{\det J_f} = \abs{f'}^2\) (the Cauchy–Riemann equations make the Jacobian \(\abs{f'}^2\)); the change-of-variables formula gives \(\mathrm{Area}(f(D_r)) = \int_{f(D_r)} 1\, dA = \int_{D_r} \abs{f'}^2\, dA\).
2. \(f'(z) = \sum_{n=1}^\infty n c_n z^{n-1}\), converging uniformly on compact subsets of \(D_1\).
Term-by-term differentiation of the power series is valid inside the radius of convergence, which is at least \(1\) since \(f\) is analytic in \(\DD\).
3. \(\iint_{D_r} \bar{z}^{m} z^{n}\, dA(z) = 0\) for \(m \neq n\), and \(\iint_{D_r} \abs{z}^{2k}\, dA(z) = \frac{2\pi r^{2k+2}}{2k+2}\).
In polar coordinates \(z = \rho e^{i\theta}\), \(\iint_{D_r} z^n \bar{z}^m\, dA = \int_0^r \rho^{n+m+1} d\rho \int_0^{2\pi} e^{i(n-m)\theta} d\theta\), and \(\int_0^{2\pi} e^{ik\theta} d\theta = 0\) for \(k \neq 0\), \(2\pi\) for \(k = 0\).
4. \(\iint_{D_r} \abs{f'(z)}^2\, dA = \sum_{n=1}^\infty n^2\abs{c_n}^2 \cdot \frac{2\pi r^{2n}}{2n}\).
Expand \(\abs{f'}^2 = \qty(\sum n c_n z^{n-1})\qty(\sum m \bar{c_m} \bar{z}^{m-1})\); by 3 the cross terms with \(n \neq m\) integrate to zero, and \(\iint_{D_r} \abs{z}^{2n-2}\, dA = \frac{2\pi r^{2n}}{2n}\) by 3 with \(k = n-1\).
Uniform convergence on \(\overline{D_r}\) justifies integrating the (absolutely convergent) product term by term; the series is finite by 5.
5. The series \(\sum_{n=1}^\infty n\abs{c_n}^2 r^{2n}\) converges.
Since \(f\) is analytic in \(\DD\), \(\abs{c_n} s^n \to 0\) for every \(s < 1\); fixing \(s\) with \(r < s < 1\), \(\abs{c_n} \leq M s^{-n}\), so \(n\abs{c_n}^2 r^{2n} \leq M^2 n (r/s)^{2n}\), a convergent geometric-type series.
6. Q.E.D.
1, 4 and 5 give \(\mathrm{Area}(f(D_r)) = \sum_{n\geq1} n^2\abs{c_n}^2 \frac{2\pi r^{2n}}{2n} = \pi\sum_{n\geq1} n\abs{c_n}^2 r^{2n}\), finite as claimed.