Let $f$ be an analytic isomorphism on the unit disc $D$, find the area of $f(D)$

Let $f$ have power series $f(z) = \sum_{n=1}^\infty a_n z^n$ in $D$, then prove that $\mathrm{area}\, f(D) = \sum_{n=1}^\infty n \,|a_n|^2$.

Note: We define $\mathrm{area}\, S = \iint_S \mathrm{d}x\,\mathrm{d}y$.

I presume the way to do this is to take the integral $$\iint_{f(D)} \mathrm{d}x\,\mathrm{d}y = \iint_D \mathbf{J}_f (x+iy) \,\mathrm{d}x\,\mathrm{d}y = \int_0^1 \int_0^{2\pi} r \mathbf{J}_f (r e^{i \theta}) \mathrm{d}\theta\,\mathrm{d}r,$$ and letting $\gamma_r : [0,2\pi]\to\Bbb{C}, \,\gamma_r (\theta) = r e^{i\theta},\, \gamma_r '(\theta) = i\gamma_r (\theta),$ then we get $$\mathrm{area}\,f(D) = \int_0^1 \int_{\gamma_r}\frac{\lvert f'(z)\rvert^2 \bar{z}}{ir}\mathrm{d}z\,\mathrm{d}r.$$ Unfortunately, this approach seems to be a dead end. I think I'm meant to use Cauchy's Formula somewhere, but I can't see where that might be useful in this kind of question.

Solution 1:

Applying Parseval's identity (see e.g. Proof of Parseval's identity) to $$ f'(z) = \sum_{n=0}^\infty (n+1)a_{n+1} z^n $$ gives $$ \int_{0}^{2 \pi} \lvert f'(re^{i\theta})\rvert^2 \, d\theta = 2 \pi \sum_{n=0}^\infty (n+1)^2 \lvert a_{n+1} \rvert^2 r^{2n} $$ for $0 \le r < 1$. If $f$ is injective on the unit disk $D$ then $$ \text{area}\, f(D) = \int_{0}^{1} \int_{0}^{2 \pi} \lvert f'(re^{i\theta})\rvert^2 \, r d\theta dr \\ = 2 \pi \int_{0}^{1} \sum_{n=0}^\infty (n+1)^2 \lvert a_{n+1} \rvert^2 r^{2n+1} \, dr \\ = \pi \sum_{n=0}^\infty (n+1) \lvert a_{n+1} \rvert^2 = \pi \sum_{n=1}^\infty n \lvert a_{n} \rvert^2 \, . $$

Exchanging the order of summation and integration can be justified for example by the monotone convergence theorem.

(Note that the factor $\pi$ is missing in the formula in your question.)