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Math Help - One More Complex Analysis Problem

  1. #1
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    One More Complex Analysis Problem

    Here is another Complex problem I can't figure out

    If f is analytic for │z│≤1 show that

    ∫D f(x+iy) dx dy = f(0)pi

    D = {(x,y) : x² + y² ≤ 1}

    Thanks for the help.

    Taypez
    Last edited by taypez; December 18th 2006 at 01:08 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by taypez View Post
    Here is another Complex problem I can't figure out

    If f is analytic for │z│≤1 show that

    ∫D f(x+iy) dx dy = f(0)pi

    D = {(x,y) : x² + y² ≤ 1}

    Thanks for the help.

    Taypez
    Let:

    f(z)=\sum_{n=0}^{\infty}a_n\,z^n,\ \ z \in D

    then:

    \int_D f(x+iy) dxdy=\sum_{n=0}^{\infty}a_n\,\int_D (x+iy)^n dxdy

    (the change in order of integration is permitted here).

    So we now need only consider the following integrals:

    a_n \,\int_D (x+iy)^n\, dxdy.

    Now change to polars:

    a_n \,\int_D (x+iy)^n\, dxdy=a_n \,\int_D r^n e^{i\,n \theta}\, r\,d\theta dr=a_n \,\int_0^{1}r^{n+1}\int_0^{2 \pi} e^{i\,n \theta}\, d\theta dr.

    This last term is a_0\, \pi when n=0 and 0 otherwise, which is sufficient to prove the required result:

    \int_D f(x+iy) dxdy=\pi\,a_0=\pi \, f(0).

    RonL
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