# Area of a holomorphic function's range on the unit disk

• Mar 12th 2011, 04:45 PM
user1
Area of a holomorphic function's range on the unit disk
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 $\displaystyle \pi\sum_{n=1}^{\infty}n|c_{n}|^{2}$.
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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.
• Mar 13th 2011, 08:42 PM
xxp9
Lemma:
$\displaystyle \int_0^{2\pi} u(r,\theta) d\theta = 0$ for any harmonic function u with u(0,0)=0, given a fixed r.

This is a direct application of mean value property of harmonic functions.
• Mar 13th 2011, 09:01 PM
xxp9
Lemma 2:
if $\displaystyle k \neq h$
$\displaystyle \int_D (c_k \bar{c_h} z^k {\bar{z}}^h+\bar{c_k} {c_h} {\bar{z}}^k {{z}}^h) dxdy = 0$

Proof of the lemma:
Suppose k>h, LHS=$\displaystyle \int_D z^h {\bar{z}}^h( c_k \bar{c_h} z^{k-h} + \bar{c_k} {c_h} {\bar{z}}^{k-h}) dxdy$
Let $\displaystyle u=c_k \bar{c_h} z^{k-h} + \bar{c_k} {c_h} {\bar{z}}^{k-h}$, u is the real part of an analytic function $\displaystyle 2c_k \bar{c_h} z^{k-h}$, thus u is harmonic and has the mean value property. Notice that u(0,0)=0
=$\displaystyle \int_0^1 r^{2h} r dr \int_0^{2\pi} u(r,\theta)d\theta$
= $\displaystyle \int_0^1 r^{2h} r dr 0$
= 0
• Mar 13th 2011, 09:58 PM
xxp9
$\displaystyle \int_D |f'|^2 dxdy = \int_D f'\bar{f'} dxdy = 0.5 \int_D (f'\bar{f'}+\bar{f'}f') dxdy$
$\displaystyle = 0.5 \sum \sum \int_D (k+1)(h+1) [c_{k+1} \overline{c_{h+1}} z^k {\overline{z}}^h + \overline{c_{k+1}} c_{h+1} {\overline{z}}^k z^h ]dxdy$
$\displaystyle = \sum \int_D (k+1)^2 |c_{k+1}|^2 |z|^{2k} dxdy$