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Is my methodology for checking differentiability and analyticity correct?

  1. Mar 31, 2012 #1
    1. The problem statement, all variables and given/known data
    State the Cauchy-Riemann equations and use them to show that the function defined by f(z) = |z|^2 is differentiable only at z = 0. Find f′(0). Where is f analytic?

    3. The attempt at a solution

    f(z) = |z|^2 = (x^2 + y^2)

    [itex]\frac{du}{dx}[/itex] = 2x, [itex]\frac{dv}{dy}[/itex] = 0

    [itex]\frac{du}{dy}[/itex] = 2y, [itex]\frac{dv}{dx}[/itex] = 0

    So x and y must equal 0 for the C.R equations to hold, thus z = 0 + 0 = 0

    So this proves they are only differentiable when z = 0.

    f'(0) = 2|0| = 0

    I'm not sure about the analytic part, f(z) = |z|^2 would be a parabola, so if it's only differentiable when z = 0, i.e. at the origin then it can't be analytic, because an e-neighbourhood at that point will yield points not on the parabolic line, therefore it's not analytic. Is this right or wrong?
     
  2. jcsd
  3. Mar 31, 2012 #2
    I just found out that f'(z) = [itex]\frac{du}{dx}[/itex] + i[itex]\frac{dv}{dx}[/itex]

    So according to this, f'(z) = 2x + 0 = 2x

    Which is 2(Re(z)), not the same thing as what I said earlier about f'(z) = 2|z| = 2[itex]\sqrt{x^2 + y^2}[/itex]

    In either case f'(0) = 0
     
  4. Mar 31, 2012 #3
    Hmm, just went over it again and realised, that if it's differentiable only when x = 0, and y = 0, then it's only differentiable when it's on the axes, and hence not analytic because an e-neighbourhood of any size on the axis is a point outside the axis, and hence not differentiable.
     
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