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Cauchy's integral formula

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

    Use the Cauchy integral formula to compute the formula:

    [tex]\int_\gamma\frac{sinh(z)}{z^3}[/tex] for [tex]\gamma[/tex] any circle of radius >0 around the origin

    2. Relevant equations

    cauchy's integral formula:

    [tex]f(z_0)=\frac{1}{2i\pi}\int\frac{f(z)}{z-z_0}dz[/tex]

    3. The attempt at a solution

    I don't know how to start
    ?
     
  2. jcsd
  3. Nov 1, 2009 #2
    What you need here is the derivative of the Cauchy's integral formula w.r.t. z_0.

    If you differentiate the right hand side, you easily find that you need to differentiate the integrand w.r.t. z_0, i.e. you can bring the derivative inside the integral. (and if )

    Now, what you then need to do is recognize that the integral of
    sinh(z)/z^3 can be written in the form of that derivative of Cauchy's formula:

    sinh(z)/z * 1/(z-z_0)^2

    with z_0 = 0

    sinh(z)/z only has a removable singularity at z = 0, you can define this function to be 1 at z = 0 to make it a complex differentiable function everywhere.
     
  4. Nov 1, 2009 #3
    I didnt understand what you said about the derivative...why are we differentiating?

    what do you mean 'removable singularity'?

    thank you
     
  5. Nov 1, 2009 #4

    gabbagabbahey

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    Gold Member

    Just use the Laurent series for [tex]\sinh(z)[/itex], and the Cauchy integral formula for simple poles of order>1...
     
  6. Nov 1, 2009 #5
    There should be the general form of Cauchy's integral formula located in your text. It is
    [tex]f^{(n)}(w) = \frac{n!}{2\pi i} \int_\gamma \frac{f(z)}{(z-w)^{n+1}} \,dz [/tex]

    In your case, f(z)=sinh(z), w=0, n=2. Note that f is then entire, i.e. analytic everywhere. Then
    [tex]\int_\gamma \frac{\sinh z}{(z-0)^3} \,dz = \frac{2\pi i}{2!} f^{(2)}(0) [/tex]
     
  7. Nov 1, 2009 #6
    f(0)=sihn(0)=0
    ?

    why is it entire?
     
  8. Nov 1, 2009 #7
    Because
    [tex]\sinh z = \frac{e^z - e^{-z}}{2},[/tex]
    so it is the sum of two entire functions.

    And technically you should be evaluating the second derivative at z=0, not just f, but in this case, the second derivative is just f(2)(z)=f(z)=sinh(z). If you don't know what singularities or Laurent series are, then don't worry about them. It seems that if you are just now doing Cauchy's integral formula, then your course hasn't arrived at those other topics yet. Does your text have the general formula I posted?
     
  9. Nov 1, 2009 #8
    yes, i also should have put the formula you posted in my post.

    thanks to all
    :)
     
  10. Nov 1, 2009 #9
    Okay, that's good and no problem for not posting it. What Count Iblis was talking about when they were referring to taking the derivative, they were basically deriving the formula I posted.
     
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