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Math Help - Complex Differentiation

  1. #1
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    Complex Differentiation

    Hello,

    I'm trying to find the set of all points z such that the function f is differentiable. I am having trouble comprehending the steps one needs to take. I have been playing around with the Cauchy-Riemann equations but to little success. The two different functions I am having trouble with are:

    1) f(x + iy) = 2x + ixy^2

    2) f(z) = cos(z*), where z* is z conjugate

    Thank you for your time.
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  2. #2
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    Well, what trouble did you run into after using C-R equations?

    Also, are you searching for domains where the function is differentiable or analytic?
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  3. #3
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    Thanks for the reply!

    To be honest, I have never fully understood the C-R equations so I keep ending up go in circles. I just attempted to use that to little success because I have read that that is the proper method.

    And the domains I am looking for are where the function is differentiable. Is analytic when the function is differentiable everywhere?

    Thanks again.
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  4. #4
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    1) C-R equations: let f(x,y) = u(x,y) + iv(x,y). Then:
    a) f is differentiable at (x_0, y_0) if u_x(x_0, y_0) = v_y(x_0, y_0) and u_y(x_0, y_0) = -v_x(x_0, y_0)
    b) f is analytic at (x_0, y_0) if it is differentiable there, and it is also differentiable at a small neighborhood around (x_0, y_0)

    I'll do the first one:

    f(x,y) = 2x + ixy^2, so u = 2x, \ v = xy^2
    And so u_x = 2, \ v_y = 2y
    u_y = 0, -v_x = -1

    So you can see that u_y is never equal to -v_x, and so the function is nowhere differentiable.

    For the second one, there is a theorem which can be used to solve but I'm not sure you've learned it - it says that f is analytic in z_0 if and only if \displaystyle \frac{ \partial f}{ \partial \bar{z}} =0.

    2) To further illustrate the point about the difference between a differentiable function and an analytic one --
    Let f(x,y) = (x-iy)(2 - x^2 - y^2).
    Applying the C-R condition gives you the result that f is differentiable only on the unit circle (verify it!).
    But since it is differentiable only on the unit circle, you don't have any points which have a neighborhood on which the function is differentiable, and so it is nowhere analytic.
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