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Math Help - [SOLVED] Application of Cauchy's Integral Formula

  1. #1
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    [SOLVED] Application of Cauchy's Integral Formula



    Here's what I got (since \frac{1}{2 - z} = \frac{-1}{z-2}).

    The book says the answer is 2\pi i, not -2\pi i. Who is right?
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  2. #2
    Member mabruka's Avatar
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    The book is right. Check you have got the right singularity.



    With "right" i mean the one who is inside your curve and apply the theorem as you just did.
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  3. #3
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    Quote Originally Posted by mabruka View Post
    The book is right. Check you have got the right singularity.



    With "right" i mean the one who is inside your curve and apply the theorem as you just did.
    Oh, because the curve is |z+1|= 2, a = -2 is inside the curve, and a = 2 is not. Therefore, I get \int \! \frac{z^2 (\frac{1}{2 - z}) \, dz}{z - (-2)}

    = f(-2) = 2\pi i(-2)^2(\frac{1}{2 - (-2)}) = 2\pi i

    Cool, thanks.
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    Member mabruka's Avatar
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