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Math Help - The complex gamma function

  1. #1
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    The complex gamma function

    Hi Havent even got a clue where to start with this one:

    The complex gamma function-q3d.jpg

    If you can answer my previous 3 questions posted plus this one, i will be eternally grateful and quite possible stand a chance at passing my maths degree.

    Thanks again

    the moolmiester
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  2. #2
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    Here is the second one. We need to use the residue theorem here. Look at picture below. The red dots are the poles.

    Theorem: Let n be a non-negative integer then \mbox{res}(\Gamma(z),-n) = \frac{(-1)^n}{n!}.

    Thus, by the residue theorem we have,
    2\pi i \left( \frac{1}{2!} - \frac{1}{1!}+\frac{1}{0!} \right) = \pi i
    Attached Thumbnails Attached Thumbnails The complex gamma function-picture28.gif  
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by moolimanj View Post
    Hi Havent even got a clue where to start with this one:

    Click image for larger version. 

Name:	Q3d.JPG 
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ID:	3766

    If you can answer my previous 3 questions posted plus this one, i will be eternally grateful and quite possible stand a chance at passing my maths degree.

    Thanks again

    the moolmiester
    The first is:

    \frac{\Gamma(2+2i)}{\Gamma(-1+2i)}

    Now to do this you need a basic recurrence for \Gamma(z), this is:

    \Gamma(1+z)=z\Gamma(z).

    Using this we have:

    \Gamma(2+2i)= (1+2i)(2i)(-1+2i) \Gamma(-1+2i),

    so:

    <br />
\frac{\Gamma(2+2i)}{\Gamma(-1+2i)}= (1+2i)(2i)(-1+2i)<br />

    RonL
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