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Math Help - Area of a holomorphic function's range on the unit disk

  1. #1
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    Area of a holomorphic function's range on the unit disk

    Hello all.

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

    Write f(x,y)=u(x,y)+iv(x,y)

    The Jacobian of f is

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

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

    Using our change of variables

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

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

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

    =\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.


    =\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

    =\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 x=r\cos\theta,y=r\sin\theta.

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

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

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

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

    =\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 f.


    Any help/hint is greatly appreciated.
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  2. #2
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    Lemma:
    \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.
    Last edited by xxp9; March 13th 2011 at 09:57 PM.
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  3. #3
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    Lemma 2:
    if k \neq h
    \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= \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 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 2c_k \bar{c_h} z^{k-h}, thus u is harmonic and has the mean value property. Notice that u(0,0)=0
    = \int_0^1 r^{2h} r dr \int_0^{2\pi} u(r,\theta)d\theta
    = \int_0^1 r^{2h} r dr 0
    = 0
    Last edited by xxp9; March 13th 2011 at 09:53 PM.
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  4. #4
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    Back to your original question,
    \int_D |f'|^2 dxdy = \int_D f'\bar{f'} dxdy = 0.5 \int_D (f'\bar{f'}+\bar{f'}f') dxdy
     = 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

    For each term in the summeration, if k is not equal to h, the integrand holds the condition of lemma 2, so the integral of this term is 0.
    So the value of the summeratio equals to
     = \sum \int_D (k+1)^2 |c_{k+1}|^2 |z|^{2k} dxdy
    Just like to take the squre in term by term
    Last edited by xxp9; March 13th 2011 at 10:45 PM. Reason: some tex symbol
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  5. #5
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    Thanks

    Thank you very much. I think the idea I was missing was having the Mean Value Property in mind.
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