Let \(f(z) = \sum_{n=-\infty}^\infty c_n z^n\) be analytic and one-to-one (univalent) in the open annulus \(r_0 < |z| < R_0\). For \(r_0 < r < R < R_0\), let \(D(r, R) = \{z \in \mathbb{C} \mid r < |z| < R\}\) be the concentric sub-annulus. Prove that the area \(S\) of the image domain \(f(D(r, R))\) is finite and given by the series formula: \[S = \pi \sum_{n=-\infty}^\infty n |c_n|^2 (R^{2n} - r^{2n}).\]
Solution
Since \(f\) is univalent and holomorphic on \(D(r,R)\), the real Jacobian is \[ J_f(z)=|f'(z)|^2, \] and change of variables gives \[ S=\operatorname{Area}(f(D(r,R))) =\iint_{D(r,R)}|f'(z)|^2\,dA(z). \]
The Laurent series may be differentiated termwise on compact subannuli: \[ f'(z)=\sum_{n\in\mathbb Z}n c_n z^{n-1}. \] For \(z=\rho e^{i\theta}\), Parseval’s identity gives \[ \int_0^{2\pi}|f'(\rho e^{i\theta})|^2\,d\theta =2\pi\sum_{n\in\mathbb Z}n^2|c_n|^2\rho^{2n-2}. \] Hence, by Tonelli’s theorem (all terms are nonnegative), \[ \begin{aligned} S &=2\pi\sum_{n\in\mathbb Z}n^2|c_n|^2 \int_r^R\rho^{2n-1}\,d\rho\\ &=\pi\sum_{n\in\mathbb Z}n|c_n|^2(R^{2n}-r^{2n}), \end{aligned} \] where the \(n=0\) term is \(0\).
The area is finite because \(\overline{D(r,R)}\) is a compact subset of the annulus of holomorphy, so \(f'\) is bounded there and \[ S\le \operatorname{Area}(D(r,R))\,\sup_{\overline{D(r,R)}}|f'|^2<\infty. \]