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Thread: Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

  1. #1
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    Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

    a) Use Cauchy's integral theorem for derivatives to evaluate the contour integral

    I = e^z/((z-2)^3) dz around the contour |z|=3.

    b) Verify that your answer is correct by finding a suitable Laurent series expansion of e^z/((z-2)^3) and integrating this series around the contour |z|=3.

    [Hint: e^z= e^2.e^z-2]

    a) I've got k=2, f(z)=e^z and z_0=2.

    f'(z) = e^z so f'(z_0)= e^2. Equally f''(z_0) = e^2. (Right?)

    Then I = (2*pi*i/2!)*f''(z_0) = (pi*i)*e^2
    = pi*i*e^2 as my final answer.

    For part b) I've wrote out the expansion of e_z first. Then divided through by ((z-2)^3). But I'm struggling on where to go next.

    The reason I'm so rusty is this paper is a rework and so its been a few months since I've done any calculus work. If anyone could give me a hand, I'd appreciate it.
    Last edited by mr fantastic; Aug 31st 2010 at 03:28 AM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by dude113 View Post
    a) Use Cauchy's integral theorem for derivatives to evaluate the contour integral

    I = e^z/((z-2)^3) dz around the contour |z|=3.

    b) Verify that your answer is correct by finding a suitable Laurent series expansion of e^z/((z-2)^3) and integrating this series around the contour |z|=3.

    [Hint: e^z= e^2.e^z-2]

    a) I've got k=2, f(z)=e^z and z_0=2.

    f'(z) = e^z so f'(z_0)= e^2. Equally f''(z_0) = e^2. (Right?)

    Then I = (2*pi*i/2!)*f''(z_0) = (pi*i)*e^2
    = pi*i*e^2 as my final answer.

    For part b) I've wrote out the expansion of e_z first. Then divided through by ((z-2)^3). But I'm struggling on where to go next.

    The reason I'm so rusty is this paper is a rework and so its been a few months since I've done any calculus work. If anyone could give me a hand, I'd appreciate it.
    The answer for (a) is correct. For (b), use the hint. You know that e^z = 1+z+\frac{z^2}{2!} + \frac{z^3}{3!} + \ldots, and so \displaystyle\frac{e^z}{z^3} = \frac1{z^3} + \frac1{z^2} + \frac1{2!z} + \frac1{3!} + \ldots. Now replace z by z–2 to get \displaystyle e^{-2}\frac{e^z}{(z-2)^3} = \frac{e^{z-2}}{(z-2)^3} = \frac1{(z-2)^3} + \frac1{(z-2)^2} + \frac1{2!(z-2)} + \frac1{3!} + \ldots.
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    Thanks dude.
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