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Math Help - complex analysis questionnn

  1. #1
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    complex analysis questionnn

    i have put up the question as a jpeg attachment since i dont know how to type all the math symbols.
    i would appreciate any help i am quite stuck it is one of the review questions for my monday exam.
    Attached Thumbnails Attached Thumbnails complex analysis questionnn-capture.jpg  
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    \displaystyle \int_{\gamma} f(z)\ dz = 2 \pi i \sum_{k} r_{k}

    ... where the r_{k} are the residues of the poles inside \gamma. In this case the poles are z=1 and z=2, both inside \gamma...

    Kind regards

    \chi \sigma
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  3. #3
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    Yes.
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  4. #4
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    what i was wondering was whether I could simply expand z = x+iy throughout the integral and then solve it that way by splitting the real and imaginary parts or whether there was a shortcut/better way to go about doing it?

    Thanks a lot!
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by shaheen7 View Post
    what i was wondering was whether I could simply expand z = x+iy throughout the integral and then solve it that way by splitting the real and imaginary parts or whether there was a shortcut/better way to go about doing it?

    Thanks a lot!
    As Chisigma mentioned in post number 2 Cauchy Residue theorem is the "shortcut way" of doing this problem. Have you learned this theorem?
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  6. #6
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    yes i have studied the theorem i am just a bit confused about how to go about getting the residue for this particular problem.
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by shaheen7 View Post
    yes i have studied the theorem i am just a bit confused about how to go about getting the residue for this particular problem.
    Since both of the poles are simple the resides can be calculated as follows

    \displaystyle f(z)=\frac{e^z}{(z-1)(z-2)}

    The residue at z=1

    \displaystyle \text{Res}(f,z=1)=\lim_{z\to1}(z-1)\frac{e^z}{(z-1)(z-2)}=\lim_{z \to 1}\frac{e^z}{(z-2)}=-e

    In general an nth order pole's residue can be calculated using the formula from post #4 in this link

    http://www.mathhelpforum.com/math-he...tml#post544571
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  8. #8
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    so the answer would simply be 2(pi)i multiplied by (-e)(e2) which are the residues? is that it?
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  9. #9
    Behold, the power of SARDINES!
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    Quote Originally Posted by shaheen7 View Post
    so the answer would simply be 2(pi)i multiplied by (-e)(e2) which are the residues? is that it?
    it would be the sum of the residues

    2\pi i ( -e+e^2)
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