Expressing the area of the image of a holomorphic function by the coefficients of its expansion

I have the following problem.

Let $f:D\to \mathbb C$ be a holomorphic function, where $D=\{z:|z|\leq 1\}.$ Let $$f(z)=\sum_{n=0}^\infty c_nz^n.$$ Let $l_2(A)$ denote the Lebesgue measure of a set $A\subseteq \mathbb C$ and $G=f(D).$ Prove that $$l_2(G)=\pi\sum_{n=1}^\infty n|c_n|^2.$$

After a long struggle I managed to come up with the following formula $$l_2(G)=\iint_D |f\,'(z)|^2dxdy.$$ It looks like the right thing to use because $$f\,'(z)=\sum_{n=1}^\infty nc_nz^{n-1},$$ which is similar to what I have to prove. I understand that the $\pi$ will appear when I integrate something over the angle $\phi$ in polar coordinates. But I don't know how to find the square of the absolute value of the right-hand side to even start integrating...

EDIT: The function is supposed to be univalent (one-to-one). (I'm not perfectly sure I'm translating the term correctly. The (Polish) word in the statement of the problem was "jednolistna". I have not met it before.)

Solution 1:

Using your formula (which is correct for univalent $f$) and going to polar coordinates gives

$$ \begin{align*} \iint_D |f'(z)|^2 ~ dx\,dy &= \int_0^1 \int_0^{2\pi} r\left|f'(re^{i\theta})\right|^2 ~ d\theta \, dr \\ &= \int_0^1 \int_0^{2\pi} r \left(\sum_{n = 1}^\infty nc_nr^{n-1} e^{i(n-1)\theta}\right)\left(\sum_{n = 1}^\infty n\overline{c_n}r^{n-1} e^{-i(n-1)\theta}\right) ~ d\theta \, dr \\ &= \int_0^1\left( \sum_{n=1}^\infty 2\pi n^2|c_n|^2 r^{2n-1} \right)~dr \\ &= \sum_{n=1}^\infty \frac{2\pi n^2}{2n} |c_n|^2 \\ &= \pi \sum_{n=1}^\infty n|c_n|^2 \end{align*} $$

where I have made use of the orthogonality relation

$$\int_0^{2\pi} e^{ik\theta} ~d\theta = \begin{cases} 2\pi & k = 0\\ 0 & k\ne 0\end{cases}$$

on line 3.

Solution 2:

The formula you give is false.
For example if $f(z)=z^{17}$, then $f(D)=D$, which has area $\pi$, whereas the formula gives $17\pi$.
(Needless to say, $17$ denotes an arbitrary positive integer)

Edit
Since you have added the condition that $f$ is univalent, the formula becomes correct. Here is a sketch of proof.

The area $l_2(G)$ is given by $\int _G 1dudv=\int_DJac(u,v)dxdy$ if you write $f(z)=u(x,y)+iv(x,y)$.
We compute $Jac(u,v)=u_x^2+v_x^2=|f'(z)|^2$ and get $l_2(G)=\int_D |f'(z)|^2 dxdy$, just as you said.

You may now use polar coordinates $z=r \cdot exp(i\theta)$ and compute $l_2(G)=\int_{r=0} ^1\int_{\theta =0}^{2\pi}|f'(rcos \theta, rsin \theta)|^2 rdrd\theta $.
The integrand is $\Sigma _{n,m}nmc_n\bar c_mr^{n+m-1}exp(i(n-m)\theta)$.
You should only consider the terms with $n=m$ because the others will become zero in the $\theta$-integral.

The formula then drops out. (I'm confident you can fill in the technical details of this sketch: permuting integrals and infinite sums, etc.)

Solution 3:

$$|f\,'(z)|^2=\sum_{n=1}^\infty\sum_{m=1}^\infty nmc_n\overline{c_m}z^{n-1}\overline{z^{m-1}}$$

$$=\sum_{n=1}^\infty n^2|c_n|^2r^{2n}+2\sum_{n\ge m} nmc_nc_m r^{2m-1}z^{n-m}.$$

Now $\iint z^k dxdy=0$ when $k\ge1$ by symmetry (either $(-z)^k=-z^k$ or $(iz)^k=-z^k)$, so the latter term vanishes upon integration. Then we can change the integral to polar coordinates:

$$\int_0^{2\pi} \int_0^1 \sum_{n=1}^\infty n^2|c_n|^2r^{2(n-1)}\cdot rdrd\theta=\pi\sum_{n=1}^\infty n|c_n|^2. $$