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Math Help - Complex Analysis - Partial Fractions and Entire fucntion

  1. #1
    lpd
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    Complex Analysis - Partial Fractions and Entire fucntion

    Hi. I'm a bit stuck on this problem.

    a) Find the partial fraction decomposition of \frac{1}{(z-a)^2(z-b)}

    Let \frac{1}{(z-a)^2(z-b)} = \frac{A}{(z-a)} +\frac{B}{(z-a)^2}+\frac{C}{(z-b)}

    After some working and expanding and simplifying i get the following:
    A=-\frac{1}{(b-a)^2}=-\frac{1}{(a-b)^2}
    B=\frac{1}{a-b}
    C=\frac{1}{(b-a)^2}=\frac{1}{(a-b)^2}

    b) Hence show that, if f is an entire function and R> |a|,|b| then
    $ <br />
  \int_{|z|=R} \frac{f(z)}{(z-a)^2(z-b)}\,dz= 2\pi i(\frac{f(b)-f(a)}{(a-b)^2} +\frac{f'(a)}{a-b})

    I can show this. Not hard. Just applying Cauchy's Integral Forumlae.

    c) Calculate
    $ \int_{|z|=3} \frac{exp(sin\frac{\pi z}{2})}{(z-1)^2(z-2)}\,dz

    f(z)=exp(sin\frac{\pi z}{2})
    f(1)=exp(sin\frac{\pi}{2})=exp(1)
    f(2)=exp(sin\frac{2\pi}{2})=0
    f'(z)=\frac{1}{2}cos(\frac{\pi z}{2})exp(sin\frac{\pi z}{2})
    f'(1)=0

    So, $ \int_{|z|=3} \frac{exp(sin\frac{\pi z}{2})}{(z-1)^2(z-2)}\,dz=<br /> <br />
 2\pi i(\frac{0-exp(1)}{(exp(1)-0)^2} +\frac{0}{exp(1)-0})<br />
=-\frac{2\pi *i*exp(1)}{exp(2)}

    d) Calculate
    $ \int_{|z|=2} \frac{exp(sin\frac{\pi z}{2})}{(z-1)^2(z-3)}\,dz
    Does this just equal to zero because one of the roots is outside |z|=2? Did I put it in correct termiology?
    Last edited by lpd; September 14th 2010 at 01:11 AM.
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  2. #2
    lpd
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    edit: I solved it. don't worry. and d) doesn't equal to zero. thanks.
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