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Cauchy Integral Extension Complex Integrals

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Allow D to be the circle lz+1l=1, counterclockwise. For all positive n, compute the contour integral.


    2. Relevant equations

    int (z-1/z+1)^n dz


    3. The attempt at a solution

    I know to use the extension of the CIF.

    Where int f(z)/(z-zo)^n+1 dz = 2(pi)i* (f^(n)(zo)/n!) ....

    However, I'm unsure how to execute the integral for my answer to depend on n.

    I made f(z)=(z-1)^n

    Then,

    Int ( f(z)/(z+1)^n) = 2(pi)i*(f^n(zo)/n!)

    Evaluating

    f(zo) at zo=-1

    = (-2)^n

    So..

    2(pi)i*((-2)^n/n!)

    I know I'm missing components of the derivative operator... but I'm not sure how to go about completing this.

    I appreciate any help.
     
  2. jcsd
  3. Oct 15, 2009 #2
    [tex]f^{(n)}(z_0)=\left.\frac{{\rm d}^n f}{{\rm d} z^n}\right|_{z=z_0}\neq \left(f(z_0)\right)^n[/tex]
     
  4. Oct 15, 2009 #3
    I'm trying to go back...

    Int ((z-1)/(z+1))^n dz

    If f(z) = (z-1)^(n)

    Then,
    Int (f(z)/(z+1)^n)

    Where zo=-1

    So,
    2(pi)i*f^(n)(-1)/n!

    For any n>0

    Is this sufficient to assume? Then for whichever n is used, f(z) can be differentiated the amount of times and evaluated at (-1) as needed.
     
  5. Oct 15, 2009 #4
    Good question!

    Your problem seems to be in this part

    [tex] \int = 2\pi i Res ( \frac {d^{n-1} } {dz^{n-1}} (z - 1)/n! ) [/tex]
    // I am not completely sure about how to write the Residue part

    So what is the derivate of (z-1)?
    The second derivate?
    ...
    The n'th derivate?
     
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