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Real number calculus vs complex calculus

  1. Jan 19, 2015 #1
    I have started studying complex integration recently and i just can't seem to get the things in my head .
    the biggest problem i am facing is that :
    when solving real number integrals the area under the curve of the function is what integration means ...
    but i can't seem to find an analogy between this and complex integration ......
     
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  3. Jan 19, 2015 #2

    jedishrfu

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  4. Jan 19, 2015 #3

    micromass

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    Consult the book "visual complex analysis" by Needham.

    Also, it is a variant of path integration, so you should first be comfortable with that. Path integrals can be easily interpreted as an area under a curve. http://en.wikipedia.org/wiki/Line_integral#mediaviewer/File:Line_integral_of_scalar_field.gif
     
  5. Jan 19, 2015 #4
    i thought complex plane is a 2 dimensional plane just like our cartesian plane ... so why can't i just take the area under the curve in complex plane here too ?
    and while approaching non complex integration why don't we integrate for any two points on the plane with an arbitrary path ? ( sorry if i am wrong about everything i said , i am still a noob at complex integration )
     
  6. Jan 19, 2015 #5

    Svein

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    Yes and no. Yes it is a 2-dimensional cartesian plane when viewed a certain way. No because it is a generalization of the real line when viewed another way.

    Example: In the 2-dimensional cartesian plane you can define addition and subtraction of points easily, but there is no straightforward way of defining a way to multiply two points and get a new point. In the complex plane, multiplication is defined from the beginning.

    As I said, the complex plane is a generalization of the real line. Complex functions, though, have stricter requirements than real functions. For a real function to be differentiable, the right- and left-hand derivative must both exist and be equal. For a complex function, the derivatives must be equal no matter how we approach the point in question. Such functions are called analytic, and they have several interesting properties, one of them being that the integral from one point to another is not dependent on the path used. This again means that the integral of an analytic function along a closed curve is zero.

    Integration is one of the most important tools in complex analysis and the basis for several important theorems.
     
  7. Jan 19, 2015 #6

    lavinia

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    If a function in the plane were real valued then one could draw its graph in three dimensions. It would look like a surface and it definitely makes sense to talk about the volume underneath it (not area). But the only real valued continuous complex differentiable functions are constants. So for them computing these volumes is uninteresting.

    You should try to convince yourself that complex differentiable functions that are real valued are constants. It is not hard.
     
  8. Jan 20, 2015 #7

    lavinia

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    There is nothing to stop you from taking the area under a curve if the function is real valued. If it is complex valued the idea makes no sense.
     
    Last edited: Jan 20, 2015
  9. Jan 20, 2015 #8
    but why doesn't it make sense when the function is complex valued ?
     
  10. Jan 20, 2015 #9

    lavinia

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    A complex number is not a height.

    A complex line integral can be thought of as two regular integrals added together (after multiplying one of them by i) to get a complex number
     
  11. Feb 5, 2015 #10

    mathwonk

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    i agree one should think in terms of path integrals, not area. consider the path integral of -ydx/(x^2+y^2) + xdy/(x^2+y^2) around a loop missing the origin. If I remembered the formula correctly, this is just dtheta, so the integral gives you the change in angle of a ray emanating from the origin, as you go around this curve, i.e. 2pi times the winding number. this can be computed in stages as actual angel changes for continuously chosen branches of the angle function theta.

    in general, every complex analytic function f is locally of form f = dg for some analytic function g, so integrating f along a path is computed from the changes in value of these locally given g's in stages along this path.

    the catchy-residue formula tells you that the integral of a local quotient of analytic functions is always equivalent to just computing winding numbers and residues at points where the denominator equals zero.
     
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