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Math Help - beginner cauchy integral equations

  1. #1
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    beginner cauchy integral equations

    all circles positively oriented. Use Cauchy's Integration Formulas to compute each of the following integrals. HINT: use partial fractions where appropriate.

    a) \int_{|z-2|=2} \frac{cosz}{(z-3)^2(z+1)}dz
    b) \int_{|z+2|=1} \frac{sinz}{z^2(z-1)^2}dz
    c) \int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz , t>0

    The Cauchy integral formula : \int_C \frac{f(z)dz}{z-z_o}=2\pi i f(z_o)

    NOW FOR c) in the denominator -1 is interior to C so i go as follows,

    \int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz=\int_{|z+2|=2} \frac{ze^{zt}}{(z-(-1))^{2+1}}dz so n=2 in the extension of Cauchy's integral formula given as \int_C \frac{f(z)dz}{(z-z_0)^{n+1}}=\frac{2\pi i}{n!}f^{(n)}(z_0)

    so i let f(z) = ze^{zt} then f'(z) = e^{zt}+zte^{zt} and f ''(z) = 2te^{zt}+zt^2e^{zt} = e^{zt}(2t+zt^2)
    hence f '' (-1) =  e^{-t}(2t-t^2) and then by the extended Cauchy integral formula i get
    \int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz = \frac{2\pi i}{2} e^{-t}(2t-t^2)= \pi i e^{-t}(2t-t^2) , is this correct ?

    OK, things i don't know :

    Now in a) neither 3 nor -1 are interior to C so neither of them are my Zo point. The same goes for b) in the denominator neither 0 nor 1 are interior to C, so where do i find Zo ? Now if the answer is to use partial fraction decomposition, my attempt :
    for a) \frac{cosz}{(z-3)^2(z+1)}= \frac{A}{z-3}+\frac{B}{(z-3)^2}+\frac{C}{z+1}
    cosz = A(z-3)(z+1)+B(z+1)+C(z-3)^2
    cosz=z^2(A+C)+z(-2A+B-6C)+(-3A+B+9C) and now what ? How do i equate anything??? I have the same confusion for b). I aprreciate any help.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: beginner cauchy integral equations

    Quote Originally Posted by punkstart View Post
    Now in a) neither 3 nor -1 are interior to C
    That is not true, 3 is (geometrically) interior to C and -1 is (geometrically and topollogically) exterior to C . So, choosing f(z)=\cos z/(z+1) (analytic in C and its geometrical interior) we have:

    \int_{|z-2|=2}\dfrac{\cos z}{(z-3)^2(z+1)}dz=\int_{|z-2|=2}\dfrac{f(z)}{(z-3)^2}dz=\dfrac{2\pi i}{1!}f'(3)=\ldots
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  3. #3
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    Re: beginner cauchy integral equations

    I'm such a fool!! I apologise. But for b) i do believe i have that same problem, finding Zo? And how do i get partial fractions ?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: beginner cauchy integral equations

    Quote Originally Posted by punkstart View Post
    I'm such a fool!! I apologise. But for b) i do believe i have that same problem, finding Zo? And how do i get partial fractions ?
    f(z)=\dfrac{\sin z}{z^2(z-1)^2} is analytic on C\equiv |z+2|=1 and in its geometrical interior. So, by Cauchy's Theorem \int_{C}f(z)=0 .
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  5. #5
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    Re: beginner cauchy integral equations

    I consider myself a fool TWICE ! Much appreciated.
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