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Laurent series

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    describe the laurent series for the function

    [tex] f(z) = z^3 cos(\frac {1}{z^2}) [/tex]

    b) use your answer to part a to compute the contour integral

    [tex] \int z^3 cos(\frac {1}{z^2}) dz [/tex]

    where C is the unit counter-clockwise circle around the origin.


    2. Relevant equations



    3. The attempt at a solution

    a)

    [tex] f(z) = z^3 * \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!} * ( \frac {1}{z^2} )^2^n [/tex]

    [tex] f(z) = \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!} * \frac {1}{z^n} [/tex]

    b) so would I just evaluate

    [tex] \sum_{n=0}^1 \frac {(-1)^n}{(2n)!} * \frac {1}{z^n} [/tex]
     
    Last edited: Sep 8, 2009
  2. jcsd
  3. Sep 8, 2009 #2

    gabbagabbahey

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    Ermmm...

    [tex] z^3\left( \frac {1}{z^2} \right)^{2n}=\frac{1}{z^{4n-3}}\neq\frac{1}{z^n}[/tex]

    Not quite, the Residue at [itex]z=0[/itex] will be given by the coefficient of the [itex]\frac{1}{z}[/itex] term in the Laurent series...what is that coefficient?...What does that make the integral?
     
  4. Sep 8, 2009 #3
    the coefficient is

    [tex] \frac {(-1)^n}{(2n)!} [/tex]

    so it would be 1?

    [tex] 1 + \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!} * \frac {1}{z^4^n^-^3} [/tex]

    ?

    sorry, a bit confused about te residue part.
     
    Last edited: Sep 8, 2009
  5. Sep 8, 2009 #4

    gabbagabbahey

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    [itex] \frac {(-1)^n}{(2n)!} [/itex] is the coefficient of the [itex]\frac{1}{z^{4n-3}}[/itex] term....you want to find the coefficient of the [itex]\frac{1}{z^{1}}[/itex] term....so, for what value of [itex]n[/itex] does

    [tex]\frac{1}{z^{4n-3}}=\frac{1}{z^{1}}[/tex] ?

    Plug that value of [itex]n[/itex] into [itex] \frac {(-1)^n}{(2n)!} [/itex] to get the coefficient of that term. That coefficient will be equal to the residue of [itex]f(z)[/itex] at [itex]z=0[/itex].
     
  6. Sep 8, 2009 #5
    hmm

    so for n = 1 [tex] \frac {1}{z^4^n^-^1} = \frac {1}{z^1} [/tex]

    then

    [tex] \frac {(-1)^1}{(2(1))!} = \frac {-1}{2} [/tex]

    then by the residue thrm.

    [tex] \int z^3 * cos(\frac {1}{z^2}) = 2*\pi*i * \frac {-1}{2} [/tex]

    [tex] = -\pi*i [/tex]

    is this the right answer?

    thanks
     
  7. Sep 8, 2009 #6

    gabbagabbahey

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    Looks good to me.:approve:
     
  8. Sep 8, 2009 #7
    thanks! that was very helpful.
     
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