We learned about conformal mappings and various properties of locally one-to-one, analytic mappings of the unit disk. I am having trouble with the following problem, can anyone help?
Let $f(z) =\sum_{n=0}^\infty a_n z^n$ be analytic and locally one-to-one in the unit disk, $|z| \leq 1$, and suppose $f$ maps the unit disk onto a domain $D$ whose area is $A$. Show that $A = \pi \sum^{\infty}_{n=1} n |a_{n}|^2.$
Also, if it is known that f'(0)=1, can we determine whether $A \leq \pi$ or $A \geq \pi$?
EDIT: I have attempted to show this using Green's Theorem since we talked about this theorem in class and proved several important results related to area. Here is what I was able to come up with. Are there any mistakes?
Green's Theorem can be used to show that if $C$ is a simple closed contour and $S$ is a region bounded by $C$, then $ \mbox{Area of }S = \frac{1}{2i}\int_C \overline{z} dz.$
Since $f$ is locally one-to-one, $D$ is a simple closed curve. Therefore, $ A = \frac{1}{2i}\int_{\partial D} \overline{w} dw.$
By changing variables, we obtain
A = \frac{1}{2i}\int_{\partial D} \overline{f(z)}f'(z) dz.
Changing to polar coordinates and using the fact that $\partial D = \{e^{it} : 0 \leq t \leq 2\pi \}$ implies:
A = \frac{1}{2i}\int^{2\pi}_{0} \overline{ f(e^{it})} f'(e^{it}) dt = \frac{1}{2i}\int^{2\pi}_{0} \left(\sum^{\infty}_{n=0} \overline{a}_n e^{-int}\right)\left(\sum^{\infty}_{m=1} m{a_m} e^{-i(m-1)t}\right)ie^{it}dt = \frac{1}{2} \int^{2\pi}_{0} \left(\sum^{\infty}_{n=0} \overline{a}_n e^{-int}\right)\left(\sum^{\infty}_{m=1} m{a_m} e^{-imt}\right)dt. By expanding the product of sums and using the fact that $e^{ikt}$ integrates to 0 as $0 \leq k \leq 2\pi$, we have:
$A = \frac{1}{2}\int^{2\pi}_{0} \sum^{\infty}_{k=1}k\overline{a}_k a_k dt = \pi \sum^{\infty}_{k=1} k |a_k|^2.$
Also, if f'(0)=1, then $a_1 = 1$. Thus, $A = \pi + \pi \sum^{\infty}_{n = 2} n|a_n|^2 > \pi$. Equality holds when $a_k = 0$ for $k > 1$, that is, when $f(z) = a_0 + z$ for some constant $a_0$.