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Complex Contour Integral Problem, meaning

  1. Dec 23, 2014 #1
    1. The problem statement, all variables and given/known data

    First, lets take a look at the complex line integral.

    What is the geometry of the complex line integral?

    If we look at the real line integral GIF:

    [2]: http://en.wikipedia.org/wiki/File:Line_integral_of_scalar_field.gif

    The real line integral is a path, but then you make a 3d figure, and it is the area under the 3d shape.

    What about for complex integral?

    And

    In the problem from Schaum's Outline:

    [1]: http://i.stack.imgur.com/U65As.png

    This is an interesting complex analysis problem; **The figure on the bottom left is what is being referred to,Fig7-10.**

    How is in the solution:

    $$\int_{BDEFG} = \int_{0}^{2\pi} \frac{(Re^{i\theta})^{p-1}iRe^{i\theta} d\theta}{1 + Re^{i\theta}}$$

    How is the integral around the $$BDEFG$$, the same as the area from$$ 0 \to 2\pi$$

    Thanks.

    2. Relevant equations

    Above

    3. The attempt at a solution

    $$\int_{BDEFG} = \int_{0}^{2\pi} \frac{(Re^{i\theta})^{p-1}iRe^{i\theta} d\theta}{1 + Re^{i\theta}}$$

    How is the integral around the $BDEFG$, the same as the area from 0 to $2\pi$
     
  2. jcsd
  3. Dec 23, 2014 #2

    Zondrina

    User Avatar
    Homework Helper

    You can interpret the line integral of ##f(x,y) \geq 0## as an area. If ##f(x,y) \geq 0##, then ##\oint_C f(x,y) \space dr## is the area of one side of the "curtain" whose base is ##C## and whose height above any ##(x,y)## is ##f(x,y)##. Integrating along the whole path would give you the area of the curtain.

    In the case where ##C## is located between two points on an axis, such as the line segment that joins ##(a,0)## to ##(b,0)##, then the line integral is actually just a regular single integral:

    $$\int_C f(x,y) \space dr = \int_a^b f(x,0) \space dx$$

    Which is simply the area under the function. A similar case is observed for ##C := (0, c) \rightarrow (0, d)##:

    $$\int_C f(x,y) \space dr = \int_c^d f(0,y) \space dy$$

    Complex numbers are really 2-D vectors, and so they are analogous to 2-D vector fields. The line integral of a 2-D vector field corresponds to the real part of the line integral of the conjugate of the complex function.

    Usually the most intuitive way to view this is to think about the work done by a force field in moving a particle along a curve from one point to another.
     
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