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.

• I didn't get your definition of $aera \ f(D)$ but what you want to obtain seems related to the Fourier series : $\int_0^{1} |f'(r e^{2i \pi t})|^2 dt = \sum_n |a_n|^2 n^2 r^{2n}$ and $\int_0^1 \frac1{r} \int_0^{1} |f'(r e^{2i \pi t})|^2 dt = \int_0^1 \sum_n |a_n|^2 n^2 r^{2n-1} dr = \frac{1}{2}\sum_n |a_n|^2 n$ Apr 30, 2016 at 1:32
• Do you want to say f is injective?
– zhw.
Apr 30, 2016 at 3:50
• Here are useful techniques. May 1, 2016 at 0:35

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 $$\text{area}\, f(D) = \int_{0}^{1} \int_{0}^{2 \pi} \lvert f'(re^{i\theta})\rvert^2 \, r d\theta dr \\ = 2 \pi \int_{0}^{1} \sum_{n=0}^\infty (n+1)^2 \lvert a_{n+1} \rvert^2 r^{2n+1} \, dr \\ = \pi \sum_{n=0}^\infty (n+1) \lvert a_{n+1} \rvert^2 = \pi \sum_{n=1}^\infty n \lvert a_{n} \rvert^2 \, .$$
(Note that the factor $\pi$ is missing in the formula in your question.)
• please explain why $$2 \pi \int_{0}^{1} \sum_{n=0}^\infty (n+1)^2 \lvert a_{n+1} \rvert^2 r^{2n+1} \, dr \\ = \pi \sum_{n=0}^\infty (n+1) \lvert a_{n+1} \rvert^2$$? Not really sure how you integrated that. Feb 12, 2021 at 22:03
• @jamesblack: Exchange the order of summation and integration, and then integrate $\int_0^1 r^{2n+1} dr = \frac{1}{2n+2}$. Feb 12, 2021 at 22:36
Alternatively, we can write the real and imaginary parts of $$f(z)$$ as $$u(r,\theta)$$ and $$v(r,\theta)$$, where $$z = r e^{i\theta}$$ and then compute the Jacobian determinant of $$F(r,\theta) = (u,v)$$. The Jacobian matrix has elements like $$\frac{\partial u}{\partial r} = \sum_{n \geq 1} n r_n r^{n-1} \cos(\varphi_n + n\theta),$$ where $$a_n = r_n e^{i\varphi_n}$$. Using the identity $$\cos(\varphi-\psi) = \cos\varphi \cos \psi + \sin\varphi \sin\psi$$ in the computation of the Jacobian matrix, and we get something like $$J_F(r,\theta) = \sum_{n \geq 1} n^2 |a_n|^2 r^{2n-1} + \sum_{n > m \geq 1} b_{m,n}(r) \cos(\varphi_n - \varphi_m + (n-m)\theta)$$ where $$b_{m,n}(r)$$ is a real function of $$r$$. Note that $$\cos(\varphi_n - \varphi_m + (n-m)\theta)$$ is a function of $$\theta$$ with $$2\pi$$ being a period. So, if we integrate the above sum in $$\theta$$ over $$[0,2\pi]$$, then the $$\cos$$'s vanish, and we get $$\int^{2\pi}_0 J_F(r,\theta) d\theta = \sum_{n \geq 1} 2\pi n^2 |a_n|^2 r^{2n-1}.$$ So $$\int^1_0 \int^{2\pi}_0 J_F(r,\theta) d\theta dr = \sum_{n \geq 1} \pi n |a_n|^2.$$