Hello all.

Suppose $\displaystyle f\in H(D(0,1))$, where D(0,1) is the open unit disc, f is one-to-one in D(0,1), $\displaystyle \Omega=f(D(0,1))$, and $\displaystyle f(z)=\sum c_{k}z^{k}$. Prove that the area of $\displaystyle \Omega$ is $\displaystyle \pi\sum_{n=1}^{\infty}n|c_{n}|^{2}$.

...........................................

Here are my partial solutions:

Write $\displaystyle f(x,y)=u(x,y)+iv(x,y)$

The Jacobian of f is

$\displaystyle J(x_{0},y_{0})=|f'(x_{0},y_{0})|^{2}$

(I worked this out, but I'm not showing it here because I'm fairly confident it is correct)

Further $\displaystyle f'(z)=\sum_{k=1}^{\infty}kc_{k}z^{k-1}$

The area of $\displaystyle \Omega$ is $\displaystyle \intop\intop_{\Omega}dudv$

Using our change of variables

$\displaystyle =\intop\intop_{\Omega}|J(x,y)|dxdy$

$\displaystyle =\intop_{-1}^{1}\intop_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}|J(x,y)|dxdy$

$\displaystyle =\intop_{-1}^{1}\intop_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}|f'(x,y)|^{2}dxdy$

$\displaystyle =\intop_{-1}^{1}\intop_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}|\sum_{k=1}^{\infty}kc_{k}(x+iy)^{k-1}|^{2}dxdy$

We use the uniform convergence of the power series to swap integration and summation...

Here I made a mistake (we can't just take the square in term by term), but it seems to have led to the right answer.

$\displaystyle =\sum_{k=1}^{\infty}k^{2}|c_{k}|^{2}\intop_{-1}^{1}\intop_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}|(x+iy)^{k-1}|^{2}dxdy$

$\displaystyle =\sum_{k=1}^{\infty}k^{2}|c_{k}|^{2}\intop_{-1}^{1}\intop_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}|(x+iy)^{k-1}|^{2}dxdy$

Now we switch to polar coordinates via $\displaystyle x=r\cos\theta,y=r\sin\theta$.

This gives $\displaystyle J(r,\theta)=r$.

$\displaystyle =\sum_{k=1}^{\infty}k^{2}|c_{k}|^{2}\intop_{0}^{2\ pi}\intop_{0}^{1}|(re^{i\theta})^{k-1}|^{2}rdrd\theta$

$\displaystyle =\sum_{k=1}^{\infty}k^{2}|c_{k}|^{2}\intop_{0}^{2\ pi}\intop_{0}^{1}r{}^{2k-1}drd\theta$

$\displaystyle =\sum_{k=1}^{\infty}k^{2}|c_{k}|^{2}\intop_{0}^{2\ pi}\left(\frac{r^{2k}}{2k}\right)|_{0}^{1}d\theta$

$\displaystyle =\sum_{k=1}^{\infty}k^{2}|c_{k}|^{2}\left(\frac{2\ pi}{2k}\right)=\pi\sum_{k=1}^{\infty}k|c_{k}|^{2}$

So:

I don't understand how my incorrect simplification of the square of the sum led to this correct result.

When I have tried to proceed through properly applying the square, I don't know how to simplify to a result I can evaluate.

I'm not sure where I am/should be using the one-to-oneness of $\displaystyle f$.

Any help/hint is greatly appreciated.