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

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

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

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

    so i let f(z) = $\displaystyle ze^{zt}$ then f'(z) = $\displaystyle e^{zt}+zte^{zt}$ and f ''(z) = $\displaystyle 2te^{zt}+zt^2e^{zt} = e^{zt}(2t+zt^2)$
    hence f '' (-1) =$\displaystyle e^{-t}(2t-t^2) $and then by the extended Cauchy integral formula i get
    $\displaystyle \int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz = \frac{2\pi i}{2} e^{-t}(2t-t^2)= \pi i e^{-t}(2t-t^2)$ , 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) $\displaystyle \frac{cosz}{(z-3)^2(z+1)}= \frac{A}{z-3}+\frac{B}{(z-3)^2}+\frac{C}{z+1}$
    $\displaystyle cosz = A(z-3)(z+1)+B(z+1)+C(z-3)^2$
    $\displaystyle cosz=z^2(A+C)+z(-2A+B-6C)+(-3A+B+9C)$ 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, $\displaystyle 3$ is (geometrically) interior to $\displaystyle C$ and $\displaystyle -1$ is (geometrically and topollogically) exterior to $\displaystyle C$ . So, choosing $\displaystyle f(z)=\cos z/(z+1)$ (analytic in $\displaystyle C$ and its geometrical interior) we have:

    $\displaystyle \int_{|z-2|=2}\dfrac{\cos z}{(z-3)^2(z+1)}dz=\int_{|z-2|=2}\dfrac{f(z)}{(z-3)^2}dz=\dfrac{2\pi i}{1!}f'(3)=\ldots$
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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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  4. #4
    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 ?
    $\displaystyle f(z)=\dfrac{\sin z}{z^2(z-1)^2}$ is analytic on $\displaystyle C\equiv |z+2|=1$ and in its geometrical interior. So, by Cauchy's Theorem $\displaystyle \int_{C}f(z)=0$ .
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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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