1. Joined
    08 Oct '06
    Moves
    24000
    21 Apr '10 10:452 edits
    Lets say f(x) = 1. If we integrate f over some finite domain, we get the length of that domain. Lets extend this to the complex plane and say f(z) = 1. If we integrate f around the unit circle centered at the intersection of the real and imaginary axes we get 0. I do not understand the meaning of this integral. I would have expected to get the circumference of the circle: 2*pi. What is going on here?

    Is it because when multiplying f(z)dz dz is negative in the real and imaginary parts at just the right places to give 0?
  2. Germany
    Joined
    27 Oct '08
    Moves
    3087
    21 Apr '10 11:381 edit
    You are calculating the residual of the function f(z) = 1. This function has no poles and the residual is zero.

    The real analog of this intergration would be to integrate f(x) = 1 from -1 to 1, and then from 1 back to -1 (obviously giving zero). Maybe this analog is a bit handwaving, but just to get you a feel of what you're doing.
  3. Amarillo , TX
    Joined
    10 May '07
    Moves
    15893
    30 Apr '10 05:37
    Since you are dealing with both real and imaginary, positive and negative the result will be a net zero.
  4. silicon valley
    Joined
    27 Oct '04
    Moves
    101289
    03 May '10 18:10
    dang. i've forgotten all that 🙁.
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