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Homework Help: Cauchy's Integral Theorem Problem

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data

    In each of the following problems, evaluate the integral of the function over the given path. All paths are positively oriented (counterclockwise). In some cases Cauchy's theorem applies, and in some it does not.

    2. Relevant equations

    Cauchy's Theorem

    Let [tex]f[/tex] be differentiable on a simply connected domain G. Let [tex]\Gamma[/tex] be a closed path in G. Then

    [tex]\int_{\Gamma} f(z)dz = 0[/tex]

    3. The attempt at a solution

    Alright, let me see if I can explain this clearly. For the problem I am stuck on, [tex]f(z) = z^2 + Im(z)[/tex] and [tex]\Gamma[/tex] is the square with verticies [tex]0,-2i,2-2i,2[/tex]

    So this is basically a square in the lower right quadrant (positive real, negative imaginary) with the upper left point being the origin. Now, I had a different problem in which [tex]\Gamma[/tex] was given to be [tex]|z|= 2[/tex], and the function, [tex]f(z) = Re(z)[/tex], and so I was able to use Euler's formula to put the function in terms of [tex]cos(t) +isin(t)[/tex], and [tex]Re(z) = cos(t)[/tex]. Taking the derivative of z(t) is easy, so I was able to evaluate the integral and get the correct answer. Now, I didn't know why Cauchy's formula didn't apply in that problem, as isn't f(z) differentiable on and within the domain enclosed by [tex]\Gamma[/tex] ? This is where I was confused at first.

    But back to the current problem, [tex]\Gamma[/tex] is not the equation for a circle, so I do not know how I am supposed to interpret Im(z), or how to determine if it is differentiable on and within the domain enclosed by [tex]\Gamma[/tex] (the square).

    Can anyone clarify or help me understand how to parametrize the equation, and to help me determine if Cauchy's equation applies here (i.e. whether the answer is zero, or whether I have to perform the integration)? Thanks for your help.
  2. jcsd
  3. Nov 18, 2007 #2


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    f(z) is differentiable in the sense of having partial derivative wrt x and y. It is NOT differentiable in the complex sense. Im(z) is not differentiable. Consider it's derivative along the curves z(t)=t and z(t)=i*t (for real t). They are different. Complex differentiability means it's derivative doesn't depend on the direction.
  4. Nov 18, 2007 #3
    Alright, that makes sense...but then how does one integrate Im(z)? Should I resolve it into real and imaginary components, i.e. a+ib, and integrate z^2 + ib? I am still confused about that part.

    I guess what I mean is, if Cauchy's theorem does not apply, I have to integrate. I figure I integrate along the 4 different paths (each a line segment of the square), but what am I integrating? f(z) is z^2 + Im(z), but then what is f'(z), and how does this become a function z(t)?
    Last edited: Nov 18, 2007
  5. Nov 19, 2007 #4


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    Once you have parametrized z(t) for the contour then dz=z'(t)*dt. So your integral is just f(z(t))*z'(t)*dt. Which is just an ordinary integral with real and imaginary parts.
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