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Integrate f(z) = Re(z) along contour of the square

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Integrate f(z) = Re(z) = [itex] \frac{z+\bar{z}}{2}[/itex] along the contour of the square {z = x + iy | |x| ≤ 1, |y| ≤ 1} with counterclockwise rotation.

    2. Relevant equations

    Cj: φj(t) = z0 + (z1 - z0)t

    C1: 1 -i + 2it
    C2: 1 +i - 2t
    C3: -1 +i - 2it
    C4: -1 -i + 2t

    3. The attempt at a solution

    If I'm not mistaken, because of linearity, I can simply integrate the real parts of parametrizations.
    Last edited: Mar 15, 2015
  2. jcsd
  3. Mar 15, 2015 #2

    Ray Vickson

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    Try it and see what you get.

    For TeX/LaTex commands: the command to get ##\bar{z}## is "\ b a r {z}" (no spaces). Also, the command "a^ bc " (no spaces) produces ##a^bc##. If you want more than one symbol in a superscript or a subscript, use { }, like this: 'a ^ {b c}" no spaces; that gives ##a^{bc}##.
  4. Mar 15, 2015 #3
    Thanks. I fixed the z-bar. I'm not quite sure that I'm handling everything correctly.

    ∫ f(z) dz = ∫ f ° φ(t)φ' (t) dt

    I get 0, 0, 0, and -1. I think this is somehow related to the fact that it's a closed curve with one loop around the origin and the area of the curve itself.
    Last edited: Mar 15, 2015
  5. Mar 15, 2015 #4

    Ray Vickson

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    If I were doing it I would stay away from parametrization via t, and just deal directly with ##\int f(z)\, dz## as a standard limit like
    [tex] \lim_{||P|| \to 0} \sum_{z_i \in P} f(z_i) (z_{i+1} - z_i). [/tex]
    Here, ##P = P_n = [z_1, z_2,... z_n]## is a partition of the integration path and ##||P||## denotes something like ##\max_i |z_{i+1} - z_i|.## Basically, ##dz = dx + i dy##. If you use a parametrization ##z = \varphi(t)##, you need to be careful about keeping the "velocity" constant; otherwise, a ##dt## in one part of the path could give a different ##|dz|## than the same ##dt## in another part of the path. That could lead to things like violations in Cauchy's Theorem, etc. Of course, parametrization would be OK if it were done very carefully, but then it might be more trouble that it is worth.
  6. Mar 15, 2015 #5
    Ah, that's right. It's easy when the velocity is one.

    Oh, I forgot to mention that I found the dt for each parametrization.

    So, something like this - (1)(2i) + (1)(-2) + (-1)(-2i) + (1)(2)?

    Using both approaches, I get zero. I guess that it means sense because it's a closed curve.
    Last edited: Mar 15, 2015
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