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Cauchy-Riemann equation

  1. Nov 2, 2012 #1
    Recently I read about Cauchy-Riemann equations and I got a doubt in that.

    I can understand the derivation of it but I couldn't understand the basic assumption with which you derive that.

    Why should the limit give the same value for both real and for imaginary axis?

    I hope you can understand the question. If you can't understand the question just reply.

    Thanks a lot
     
  2. jcsd
  3. Nov 2, 2012 #2

    Bacle2

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    It comes down to the fact that for the limit to exist, it should exist from every direction

    you approach a point. In particular, it must exist as you approach along the real

    axis as well as when you approach along the imaginary axis, and the two limits

    must be equal.
     
  4. Nov 2, 2012 #3
    That point is valid if you are finding just the limit as in

    [itex] \lim_{x \to a} f(x,b) [/itex] to be same as
    [itex] \lim_{y \to b} f(a,y) [/itex]

    But it is not necessary that the differentiation with respect to real and imaginary axis had to be same, right? I mean it's like partial derivative where derivative with respect to one axes (real in this case) need not be same as the derivative with respect to another axes (imaginary axes).

    thanks a lot for your help.
     
  5. Nov 2, 2012 #4

    Bacle2

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    But you are finding a limit; the derivative is a limit , the quotient limit

    [f(z+zo)-f(z)]/(z-zo) as z→zo . But z can approach zo along _every possible

    complex direction. Then f'(z) exists when this limit exist,so that the limit must

    exist along any direction along which you approximate zo, and for the limit to

    exist, it must be the same no-matter how you approximate zo. In particular,

    (re Cauchy-Riemann) , the approximation along the x-axis and the y-axis must

    exist, and must agree with each other.
     
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